Normal laws

[math] \newcommand{\mathds}{\mathbb}[/math]

6a. Gauss integral

In this chapter we discuss the basics of probability theory, as an application of the methods developed in chapter 5. You might have heard or noticed that most measurements in real life lead to a bell-shaped curve, depending on a parameter [math]h \gt 0[/math], measuring the height of the bell, that we will assume here to appear above the point [math]x=0[/math].

Some further experiments and study show that the bell-shaped curve must appear in fact as a variation of the function [math]e^{-x^2}[/math]. Thus, we can expect the formula of this curve to be as follows, with [math]\lambda \gt 0[/math] being related to the height [math]h \gt 0[/math]:

[[math]] f_\lambda(x)=C_\lambda\cdot e^{-\lambda x^2} [[/math]]

But, the problem is, what is the constant [math]C_\lambda[/math]? Since the total probability for our event to appear is 1, this constant must be the inverse of the integral of [math]e^{-\lambda x^2}[/math]:

[[math]] C_\lambda=\left(\int_\mathbb Re^{-\lambda x^2}dx\right)^{-1} [[/math]]

Thus, we are led into computing the above integral. By doing a change of variables we can assume [math]\lambda=1[/math], and so we are led into the following question:

[[math]] \int_\mathbb Re^{-x^2}dx=\,? [[/math]]

And suprise here, all the methods that we know from basic 1-variable calculus fail, in relation with this question. The function [math]e^{-x^2}[/math] has no computable primitive, and all tricks that we know, namely change of variables, partial integration, and more, all fail.

Fortunately multivariable calculus comes to the rescue, and we have:

Theorem

We have the following formula,

[[math]] \int_\mathbb Re^{-x^2}dx=\sqrt{\pi} [[/math]]

called Gauss integral formula.

Show Proof

Let [math]I[/math] be the above integral. By using polar coordinates, we obtain:

[[math]] \begin{eqnarray*} I^2 &=&\int_\mathbb R\int_\mathbb Re^{-x^2-y^2}dxdy\\ &=&\int_0^{2\pi}\int_0^\infty e^{-r^2}rdrdt\\ &=&2\pi\int_0^\infty\left(-\frac{e^{-r^2}}{2}\right)'dr\\ &=&2\pi\left[0-\left(-\frac{1}{2}\right)\right]\\ &=&\pi \end{eqnarray*} [[/math]]

Thus, we are led to the formula in the statement.

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As a main application of the Gauss formula, we can now formulate:

Definition

The normal law of parameter [math]1[/math] is the following measure:

[[math]] g_1=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx [[/math]]

More generally, the normal law of parameter [math]t \gt 0[/math] is the following measure:

[[math]] g_t=\frac{1}{\sqrt{2\pi t}}e^{-x^2/2t}dx [[/math]]

These are also called Gaussian distributions, with “g” standing for Gauss.

We should mention that these laws are traditionally denoted [math]\mathcal N(0,1)[/math] and [math]\mathcal N(0,t)[/math], but since we will be doing here in this book all kinds of probability, namely real and complex, discrete and continuous, and so on, we will have to deal with lots of interesting probability laws, and we will be using simplified notations for them. Let us mention as well that the normal laws traditionally have 2 parameters, the mean and the variance. Here we do not need the mean, all our theory in this book using centered laws.

Observe that the above laws have indeed mass 1, as they should. This follows indeed from the Gauss formula, which gives, with [math]x=\sqrt{2t}\,y[/math]:

[[math]] \begin{eqnarray*} \int_\mathbb R e^{-x^2/2t}dx &=&\int_\mathbb R e^{-y^2}\sqrt{2t}\,dy\\ &=&\sqrt{2t}\int_\mathbb R e^{-y^2}dy\\ &=&\sqrt{2t}\times\sqrt{\pi}\\ &=&\sqrt{2\pi t} \end{eqnarray*} [[/math]]

Generally speaking, the normal laws appear as bit everywhere, in real life. The reasons behind this phenomenon come from the Central Limit Theorem (CLT), that we will explain in a moment. At the level of basic facts now, we first have:

Proposition

We have the variance formula

[[math]] V(g_t)=t [[/math]]

valid for any [math]t \gt 0[/math].

Show Proof

The first moment is 0, because our normal law [math]g_t[/math] is centered. As for the second moment, this can be computed as follows:

[[math]] \begin{eqnarray*} M_2 &=&\frac{1}{\sqrt{2\pi t}}\int_\mathbb Rx^2e^{-x^2/2t}dx\\ &=&\frac{1}{\sqrt{2\pi t}}\int_\mathbb R(tx)\left(-e^{-x^2/2t}\right)'dx\\ &=&\frac{1}{\sqrt{2\pi t}}\int_\mathbb Rte^{-x^2/2t}dx\\ &=&t \end{eqnarray*} [[/math]]

We conclude from this that the variance is [math]V=M_2=t[/math].

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More generally now, we have the following result:

Theorem

The moments of the normal law are the numbers

[[math]] M_k(g_t)=t^{k/2}\times k!! [[/math]]

where the double factorials are by definition given by

[[math]] k!!=1\cdot 3\cdot 5\ldots(k-1) [[/math]]

with the convention [math]k!!=0[/math] when [math]k[/math] is odd.

Show Proof

We have the following computation:

[[math]] \begin{eqnarray*} M_k &=&\frac{1}{\sqrt{2\pi t}}\int_\mathbb Rx^ke^{-x^2/2t}dx\\ &=&\frac{1}{\sqrt{2\pi t}}\int_\mathbb R(tx^{k-1})\left(-e^{-x^2/2t}\right)'dx\\ &=&\frac{1}{\sqrt{2\pi t}}\int_\mathbb Rt(k-1)x^{k-2}e^{-x^2/2t}dx\\ &=&t(k-1)\times\frac{1}{\sqrt{2\pi t}}\int_\mathbb Rx^{k-2}e^{-x^2/2t}dx\\ &=&t(k-1)M_{k-2} \end{eqnarray*} [[/math]]

On the other hand, we have [math]M_0=1[/math], [math]M_1=0[/math]. Thus by recurrence, the even moments vanish, and the odd moments are given by the formula in the statement.

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We can improve the above result, as follows:

Theorem

The moments of the normal law are the numbers

[[math]] M_k(g_t)=t^{k/2}|P_2(k)| [[/math]]

where [math]P_2(k)[/math] is the set of pairings of [math]\{1,\ldots,k\}[/math].

Show Proof

Let us count the pairings of [math]\{1,\ldots,k\}[/math]. In order to have such a pairing, we must pair [math]1[/math] with one of [math]2,\ldots,k[/math], and then use a pairing of the remaining [math]k-2[/math] points. Thus, we have the following recurrence formula:

[[math]] |P_2(k)|=(k-1)|P_2(k-2)| [[/math]]

Thus [math]|P_2(k)|=k!![/math], and we are led to the conclusion in the statement.

■

We are done done yet, and here is one more improvement:

Theorem

The moments of the normal law are the numbers

[[math]] M_k(g_t)=\sum_{\pi\in P_2(k)}t^{|\pi|} [[/math]]

where [math]P_2(k)[/math] is the set of pairings of [math]\{1,\ldots,k\}[/math], and [math]|.|[/math] is the number of blocks.

Show Proof

This follows indeed from Theorem 6.5, because the number of blocks of a pairing of [math]\{1,\ldots,k\}[/math] is trivially [math]k/2[/math], independently of the pairing.

■

We will see later on that many other interesting probability distributions are subject to similar formulae regarding their moments, involving partitions, and a lot of interesting combinatorics. Discussing this will be in fact a main theme of the present book.

6b. Central limits

The fundamental result in probability is the Central Limit Theorem (CLT), and our next task will be that of explaining this. Let us start with:

Definition

Let [math]X[/math] be a probability space, that is to say, a space with a probability measure, and with the corresponding integration denoted [math]\mathbb E[/math], and called expectation.

• The random variables are the real functions as follows:
  [[math]] f\in L^\infty(X) [[/math]]
• The moments of such a variable are the following numbers:
  [[math]] M_k(f)=\mathbb E(f^k) [[/math]]
• The law of such a variable is the measure given by:
  [[math]] M_k(f)=\int_\mathbb Rx^kd\mu_f(x) [[/math]]

Here the fact that [math]\mu_f[/math] exists indeed is not trivial. By linearity, we would like to have a real probability measure making hold the following formula, for any [math]P\in\mathbb R[X][/math]:

[[math]] \mathbb E(P(f))=\int_\mathbb RP(x)d\mu_f(x) [[/math]]

By using a continuity argument, it is enough to have this formula for the characteristic functions [math]\chi_I[/math] of the arbitrary measurable sets of real numbers [math]I\subset\mathbb R[/math]:

[[math]] \mathbb E(\chi_I(f))=\int_\mathbb R\chi_I(x)d\mu_f(x) [[/math]]

Thus, we would like to have a measure [math]\mu_f[/math] such that:

[[math]] \mathbb P(f\in I)=\mu_f(I) [[/math]]

But this latter formula can serve as a definition for [math]\mu_f[/math], with the axioms of real probability measures being trivially satisfied, and so we are done.

Next in line, we need to talk about independence. Once again with the idea of doing things a bit abstractly, the definition here is as follows:

Definition

Two variables [math]f,g\in L^\infty(X)[/math] are called independent when

[[math]] \mathbb E(f^kg^l)=\mathbb E(f^k)\cdot\mathbb E(g^l) [[/math]]

happens, for any [math]k,l\in\mathbb N[/math].

Once again, this definition hides some non-trivial things. Indeed, by linearity, we would like to have a formula as follows, valid for any polynomials [math]P,Q\in\mathbb R[X][/math]:

[[math]] \mathbb E(P(f)Q(g))=\mathbb E(P(f))\cdot\mathbb E(Q(g)) [[/math]]

By continuity, it is enough to have this formula for characteristic functions [math]\chi_I,\chi_J[/math] of the arbitrary measurable sets of real numbers [math]I,J\subset\mathbb R[/math]:

[[math]] \mathbb E(\chi_I(f)\chi_J(g))=\mathbb E(\chi_I(f))\cdot\mathbb E(\chi_J(g)) [[/math]]

Thus, we are led to the usual definition of independence, namely:

[[math]] \mathbb P(f\in I,g\in J)=\mathbb P(f\in I)\cdot\mathbb P(g\in J) [[/math]]

All this might seem a bit abstract, but in practice, the idea is of course that [math]f,g[/math] must be independent, in an intuitive, real-life sense.

Here is now our first result, regarding this notion of independence:

Proposition

Assuming that [math]f,g\in L^\infty(X)[/math] are independent, we have

[[math]] \mu_{f+g}=\mu_f*\mu_g [[/math]]

where [math]*[/math] is the convolution of real probability measures.

Show Proof

We have the following computation, using the independence of [math]f,g[/math]:

[[math]] \begin{eqnarray*} M_k(f+g) &=&\mathbb E((f+g)^k)\\ &=&\sum_l\binom{k}{l}\mathbb E(f^lg^{k-l})\\ &=&\sum_l\binom{k}{l}M_l(f)M_{k-l}(g) \end{eqnarray*} [[/math]]

On the other hand, by using the Fubini theorem, we have as well:

[[math]] \begin{eqnarray*} \int_\mathbb Rx^kd(\mu_f*\mu_g)(x) &=&\int_{\mathbb R\times\mathbb R}(x+y)^kd\mu_f(x)d\mu_g(y)\\ &=&\sum_l\binom{k}{l}\int_\mathbb Rx^kd\mu_f(x)\int_\mathbb Ry^ld\mu_g(y)\\ &=&\sum_l\binom{k}{l}M_l(f)M_{k-l}(g) \end{eqnarray*} [[/math]]

Thus the measures [math]\mu_{f+g}[/math] and [math]\mu_f*\mu_g[/math] have the same moments:

[[math]] M_k(\mu_{f+g})=M_k(\mu_f*\mu_g) [[/math]]

We conclude that these two measures coincide, as stated.

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Here is now our second result, which is something more advanced, providing us with some efficient tools for the study of the independence:

Theorem

Assuming that [math]f,g\in L^\infty(X)[/math] are independent, we have

[[math]] F_{f+g}=F_fF_g [[/math]]

where [math]F_f(x)=\mathbb E(e^{ixf})[/math] is the Fourier transform.

Show Proof

We have the following computation, using Proposition 6.9 and Fubini:

[[math]] \begin{eqnarray*} F_{f+g}(x) &=&\int_\mathbb Re^{ixy}d\mu_{f+g}(y)\\ &=&\int_\mathbb Re^{ixy}d(\mu_f*\mu_g)(y)\\ &=&\int_{\mathbb R\times\mathbb R}e^{ix(y+z)}d\mu_f(y)d\mu_g(z)\\ &=&\int_\mathbb Re^{ixy}d\mu_f(y)\int_\mathbb Re^{ixz}d\mu_g(z)\\ &=&F_f(x)F_g(x) \end{eqnarray*} [[/math]]

Thus, we are led to the conclusion in the statement.

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Getting now to the normal laws, we have the following key result:

Theorem

We have the following formula, valid for any [math]t \gt 0[/math]:

[[math]] F_{g_t}(x)=e^{-tx^2/2} [[/math]]

In particular, the normal laws satisfy [math]g_s*g_t=g_{s+t}[/math], for any [math]s,t \gt 0[/math].

Show Proof

The Fourier transform formula can be established as follows:

[[math]] \begin{eqnarray*} F_{g_t}(x) &=&\frac{1}{\sqrt{2\pi t}}\int_\mathbb Re^{-y^2/2t+ixy}dy\\ &=&\frac{1}{\sqrt{2\pi t}}\int_\mathbb Re^{-(y/\sqrt{2t}-\sqrt{t/2}ix)^2-tx^2/2}dy\\ &=&\frac{1}{\sqrt{2\pi t}}\int_\mathbb Re^{-z^2-tx^2/2}\sqrt{2t}dz\\ &=&\frac{1}{\sqrt{\pi}}e^{-tx^2/2}\int_\mathbb Re^{-z^2}dz\\ &=&\frac{1}{\sqrt{\pi}}e^{-tx^2/2}\cdot\sqrt{\pi}\\ &=&e^{-tx^2/2} \end{eqnarray*} [[/math]]

As for the last assertion, this follows from the fact that [math]\log F_{g_t}[/math] is linear in [math]t[/math].

■

We are now ready to state and prove the CLT, as follows:

Theorem (CLT)

Given random variables [math]f_1,f_2,f_3,\ldots\in L^\infty(X)[/math] which are i.i.d., centered, and with variance [math]t \gt 0[/math], we have, with [math]n\to\infty[/math], in moments,

[[math]] \frac{1}{\sqrt{n}}\sum_{i=1}^nf_i\sim g_t [[/math]]

where [math]g_t[/math] is the Gaussian law of parameter [math]t[/math], having as density [math]\frac{1}{\sqrt{2\pi t}}e^{-y^2/2t}dy[/math].

Show Proof

We use the Fourier transform, which is by definition given by:

[[math]] F_f(x)=\mathbb E(e^{ixf}) [[/math]]

In terms of moments, we have the following formula:

[[math]] \begin{eqnarray*} F_f(x) &=&\mathbb E\left(\sum_{k=0}^\infty\frac{(ixf)^k}{k!}\right)\\ &=&\sum_{k=0}^\infty\frac{(ix)^k\mathbb E(f^k)}{k!}\\ &=&\sum_{k=0}^\infty\frac{i^kM_k(f)}{k!}\,x^k \end{eqnarray*} [[/math]]

Thus, the Fourier transform of the variable in the statement is:

[[math]] \begin{eqnarray*} F(x) &=&\left[F_f\left(\frac{x}{\sqrt{n}}\right)\right]^n\\ &=&\left[1-\frac{tx^2}{2n}+O(n^{-2})\right]^n\\ &\simeq&\left[1-\frac{tx^2}{2n}\right]^n\\ &\simeq&e^{-tx^2/2} \end{eqnarray*} [[/math]]

But this latter function being the Fourier transform of [math]g_t[/math], we obtain the result.

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This was for our basic presentation of probability theory, which was of course rather quick. For more on all this, we refer to any standard probability book, such as Feller , or Durrett . But we will be back to this, on several occasions.

Needless to say, you can also learn some good probability from physicists, or other scientists. In fact, probability theory was accepted only recently, in the late 20th century, as a respectable branch of mathematics, and if there are some scientists who have taken probability seriously, and this since ever, these are the physicists.

So, in what regards now physics books, there is a bit of probability a bit everywhere, but especially in thermodynamics and statistical mechanics. And here, for getting introduced to the subject, you can go with Feynman , or with the classical book of Fermi , or with the more recent book of Schroeder . And for more, a standard reference for learning statistical mechanics is the book of Huang .

6c. Spherical integrals

Let us discuss now the computation of the arbitrary integrals over the sphere. We will need a technical result extending the formulae from chapter 5, as follows:

Theorem

We have the following formula,

[[math]] \int_0^{\pi/2}\cos^pt\sin^qt\,dt=\left(\frac{\pi}{2}\right)^{\varepsilon(p)\varepsilon(q)}\frac{p!!q!!}{(p+q+1)!!} [[/math]]

where [math]\varepsilon(p)=1[/math] if [math]p[/math] is even, and [math]\varepsilon(p)=0[/math] if [math]p[/math] is odd, and where

[[math]] m!!=(m-1)(m-3)(m-5)\ldots [[/math]]

with the product ending at [math]2[/math] if [math]m[/math] is odd, and ending at [math]1[/math] if [math]m[/math] is even.

Show Proof

Let [math]I_{pq}[/math] be the integral in the statement. In order to do the partial integration, observe that we have:

[[math]] \begin{eqnarray*} (\cos^pt\sin^qt)' &=&p\cos^{p-1}t(-\sin t)\sin^qt\\ &+&\cos^pt\cdot q\sin^{q-1}t\cos t\\ &=&-p\cos^{p-1}t\sin^{q+1}t+q\cos^{p+1}t\sin^{q-1}t \end{eqnarray*} [[/math]]

By integrating between [math]0[/math] and [math]\pi/2[/math], we obtain, for [math]p,q \gt 0[/math]:

[[math]] pI_{p-1,q+1}=qI_{p+1,q-1} [[/math]]

Thus, we can compute [math]I_{pq}[/math] by recurrence. When [math]q[/math] is even we have:

[[math]] \begin{eqnarray*} I_{pq} &=&\frac{q-1}{p+1}\,I_{p+2,q-2}\\ &=&\frac{q-1}{p+1}\cdot\frac{q-3}{p+3}\,I_{p+4,q-4}\\ &=&\frac{q-1}{p+1}\cdot\frac{q-3}{p+3}\cdot\frac{q-5}{p+5}\,I_{p+6,q-6}\\ &=&\vdots\\ &=&\frac{p!!q!!}{(p+q)!!}\,I_{p+q} \end{eqnarray*} [[/math]]

But the last term comes from the formulae in chapter 5, and we obtain the result:

[[math]] \begin{eqnarray*} I_{pq} &=&\frac{p!!q!!}{(p+q)!!}\,I_{p+q}\\ &=&\frac{p!!q!!}{(p+q)!!}\left(\frac{\pi}{2}\right)^{\varepsilon(p+q)}\frac{(p+q)!!}{(p+q+1)!!}\\ &=&\left(\frac{\pi}{2}\right)^{\varepsilon(p)\varepsilon(q)}\frac{p!!q!!}{(p+q+1)!!} \end{eqnarray*} [[/math]]

Observe that this gives the result for [math]p[/math] even as well, by symmetry. Indeed, we have [math]I_{pq}=I_{qp}[/math], by using the following change of variables:

[[math]] t=\frac{\pi}{2}-s [[/math]]

In the remaining case now, where both [math]p,q[/math] are odd, we can use once again the formula [math]pI_{p-1,q+1}=qI_{p+1,q-1}[/math] established above, and the recurrence goes as follows:

[[math]] \begin{eqnarray*} I_{pq} &=&\frac{q-1}{p+1}\,I_{p+2,q-2}\\ &=&\frac{q-1}{p+1}\cdot\frac{q-3}{p+3}\,I_{p+4,q-4}\\ &=&\frac{q-1}{p+1}\cdot\frac{q-3}{p+3}\cdot\frac{q-5}{p+5}\,I_{p+6,q-6}\\ &=&\vdots\\ &=&\frac{p!!q!!}{(p+q-1)!!}\,I_{p+q-1,1} \end{eqnarray*} [[/math]]

In order to compute the last term, observe that we have:

[[math]] \begin{eqnarray*} I_{p1} &=&\int_0^{\pi/2}\cos^pt\sin t\,dt\\ &=&-\frac{1}{p+1}\int_0^{\pi/2}(\cos^{p+1}t)'\,dt\\ &=&\frac{1}{p+1} \end{eqnarray*} [[/math]]

Thus, we can finish our computation in the case [math]p,q[/math] odd, as follows:

[[math]] \begin{eqnarray*} I_{pq} &=&\frac{p!!q!!}{(p+q-1)!!}\,I_{p+q-1,1}\\ &=&\frac{p!!q!!}{(p+q-1)!!}\cdot\frac{1}{p+q}\\ &=&\frac{p!!q!!}{(p+q+1)!!} \end{eqnarray*} [[/math]]

Thus, we obtain the formula in the statement, the exponent of [math]\pi/2[/math] appearing there being [math]\varepsilon(p)\varepsilon(q)=0\cdot 0=0[/math] in the present case, and this finishes the proof.

■

We can now integrate over the spheres, as follows:

Theorem

The polynomial integrals over the unit sphere [math]S^{N-1}_\mathbb R\subset\mathbb R^N[/math], with respect to the normalized, mass [math]1[/math] measure, are given by the following formula,

[[math]] \int_{S^{N-1}_\mathbb R}x_1^{k_1}\ldots x_N^{k_N}\,dx=\frac{(N-1)!!k_1!!\ldots k_N!!}{(N+\Sigma k_i-1)!!} [[/math]]

valid when all exponents [math]k_i[/math] are even. If an exponent is odd, the integral vanishes.

Show Proof

Assume first that one of the exponents [math]k_i[/math] is odd. We can make then the following change of variables, which shows that the integral in the statement vanishes:

[[math]] x_i\to-x_i [[/math]]

Assume now that all the exponents [math]k_i[/math] are even. As a first observation, the result holds indeed at [math]N=2[/math], due to the formula from Theorem 6.13, which reads:

[[math]] \begin{eqnarray*} \int_0^{\pi/2}\cos^pt\sin^qt\,dt &=&\left(\frac{\pi}{2}\right)^{\varepsilon(p)\varepsilon(q)}\frac{p!!q!!}{(p+q+1)!!}\\\ &=&\frac{p!!q!!}{(p+q+1)!!} \end{eqnarray*} [[/math]]

Indeed, this formula computes the integral in the statement over the first quadrant. But since the exponents [math]p,q\in\mathbb N[/math] are assumed to be even, the integrals over the other quadrants are given by the same formula, so when averaging we obtain the result.

In the general case now, where the dimension [math]N\in\mathbb N[/math] is arbitrary, the integral in the statement can be written in spherical coordinates, as follows:

[[math]] I=\frac{2^N}{A}\int_0^{\pi/2}\ldots\int_0^{\pi/2}x_1^{k_1}\ldots x_N^{k_N}J\,dt_1\ldots dt_{N-1} [[/math]]

Here [math]A[/math] is the area of the sphere, [math]J[/math] is the Jacobian, and the [math]2^N[/math] factor comes from the restriction to the [math]1/2^N[/math] part of the sphere where all the coordinates are positive. According to our computations in chapter 5, the normalization constant in front of the integral is:

[[math]] \frac{2^N}{A}=\left(\frac{2}{\pi}\right)^{[N/2]}(N-1)!! [[/math]]

As for the unnormalized integral, this is given by:

[[math]] \begin{eqnarray*} I'=\int_0^{\pi/2}\ldots\int_0^{\pi/2} &&(\cos t_1)^{k_1} (\sin t_1\cos t_2)^{k_2}\\ &&\vdots\\ &&(\sin t_1\sin t_2\ldots\sin t_{N-2}\cos t_{N-1})^{k_{N-1}}\\ &&(\sin t_1\sin t_2\ldots\sin t_{N-2}\sin t_{N-1})^{k_N}\\ &&\sin^{N-2}t_1\sin^{N-3}t_2\ldots\sin^2t_{N-3}\sin t_{N-2}\\ &&dt_1\ldots dt_{N-1} \end{eqnarray*} [[/math]]

By rearranging the terms, we obtain:

[[math]] \begin{eqnarray*} I' &=&\int_0^{\pi/2}\cos^{k_1}t_1\sin^{k_2+\ldots+k_N+N-2}t_1\,dt_1\\ &&\int_0^{\pi/2}\cos^{k_2}t_2\sin^{k_3+\ldots+k_N+N-3}t_2\,dt_2\\ &&\vdots\\ &&\int_0^{\pi/2}\cos^{k_{N-2}}t_{N-2}\sin^{k_{N-1}+k_N+1}t_{N-2}\,dt_{N-2}\\ &&\int_0^{\pi/2}\cos^{k_{N-1}}t_{N-1}\sin^{k_N}t_{N-1}\,dt_{N-1} \end{eqnarray*} [[/math]]

Now by using the above-mentioned formula at [math]N=2[/math], this gives:

[[math]] \begin{eqnarray*} I' &=&\frac{k_1!!(k_2+\ldots+k_N+N-2)!!}{(k_1+\ldots+k_N+N-1)!!}\left(\frac{\pi}{2}\right)^{\varepsilon(N-2)}\\ &&\frac{k_2!!(k_3+\ldots+k_N+N-3)!!}{(k_2+\ldots+k_N+N-2)!!}\left(\frac{\pi}{2}\right)^{\varepsilon(N-3)}\\ &&\vdots\\ &&\frac{k_{N-2}!!(k_{N-1}+k_N+1)!!}{(k_{N-2}+k_{N-1}+l_N+2)!!}\left(\frac{\pi}{2}\right)^{\varepsilon(1)}\\ &&\frac{k_{N-1}!!k_N!!}{(k_{N-1}+k_N+1)!!}\left(\frac{\pi}{2}\right)^{\varepsilon(0)} \end{eqnarray*} [[/math]]

Now let [math]F[/math] be the part involving the double factorials, and [math]P[/math] be the part involving the powers of [math]\pi/2[/math], so that [math]I'=F\cdot P[/math]. Regarding [math]F[/math], by cancelling terms we have:

[[math]] F=\frac{k_1!!\ldots k_N!!}{(\Sigma k_i+N-1)!!} [[/math]]

As in what regards [math]P[/math], by summing the exponents, we obtain [math]P=\left(\frac{\pi}{2}\right)^{[N/2]}[/math]. We can now put everything together, and we obtain:

[[math]] \begin{eqnarray*} I &=&\frac{2^N}{A}\times F\times P\\ &=&\left(\frac{2}{\pi}\right)^{[N/2]}(N-1)!!\times\frac{k_1!!\ldots k_N!!}{(\Sigma k_i+N-1)!!}\times\left(\frac{\pi}{2}\right)^{[N/2]}\\ &=&\frac{(N-1)!!k_1!!\ldots k_N!!}{(\Sigma k_i+N-1)!!} \end{eqnarray*} [[/math]]

Thus, we are led to the conclusion in the statement.

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We have the following useful generalization of the above formula:

Theorem

We have the following integration formula over the sphere [math]S^{N-1}_\mathbb R\subset\mathbb R^N[/math], with respect to the normalized measure, valid for any exponents [math]k_i\in\mathbb N[/math],

[[math]] \int_{S^{N-1}_\mathbb R}|x_1^{k_1}\ldots x_N^{k_N}|\,dx=\left(\frac{2}{\pi}\right)^{\Sigma(k_1,\ldots,k_N)}\frac{(N-1)!!k_1!!\ldots k_N!!}{(N+\Sigma k_i-1)!!} [[/math]]

with [math]\Sigma=[odds/2][/math] if [math]N[/math] is odd and [math]\Sigma=[(odds+1)/2][/math] if [math]N[/math] is even, where “odds” denotes the number of odd numbers in the sequence [math]k_1,\ldots,k_N[/math].

Show Proof

As before, the formula holds at [math]N=2[/math], due to Theorem 6.13. In general, the integral in the statement can be written in spherical coordinates, as follows:

[[math]] I=\frac{2^N}{A}\int_0^{\pi/2}\ldots\int_0^{\pi/2}x_1^{k_1}\ldots x_N^{k_N}J\,dt_1\ldots dt_{N-1} [[/math]]

Here [math]A[/math] is the area of the sphere, [math]J[/math] is the Jacobian, and the [math]2^N[/math] factor comes from the restriction to the [math]1/2^N[/math] part of the sphere where all the coordinates are positive. The normalization constant in front of the integral is, as before:

[[math]] \frac{2^N}{A}=\left(\frac{2}{\pi}\right)^{[N/2]}(N-1)!! [[/math]]

As for the unnormalized integral, this can be written as before, as follows:

[[math]] \begin{eqnarray*} I' &=&\int_0^{\pi/2}\cos^{k_1}t_1\sin^{k_2+\ldots+k_N+N-2}t_1\,dt_1\\ &&\int_0^{\pi/2}\cos^{k_2}t_2\sin^{k_3+\ldots+k_N+N-3}t_2\,dt_2\\ &&\vdots\\ &&\int_0^{\pi/2}\cos^{k_{N-2}}t_{N-2}\sin^{k_{N-1}+k_N+1}t_{N-2}\,dt_{N-2}\\ &&\int_0^{\pi/2}\cos^{k_{N-1}}t_{N-1}\sin^{k_N}t_{N-1}\,dt_{N-1} \end{eqnarray*} [[/math]]

Now by using the formula at [math]N=2[/math], we get:

[[math]] \begin{eqnarray*} I' &=&\frac{\pi}{2}\cdot\frac{k_1!!(k_2+\ldots+k_N+N-2)!!}{(k_1+\ldots+k_N+N-1)!!}\left(\frac{2}{\pi}\right)^{\delta(k_1,k_2+\ldots+k_N+N-2)}\\ &&\frac{\pi}{2}\cdot\frac{k_2!!(k_3+\ldots+k_N+N-3)!!}{(k_2+\ldots+k_N+N-2)!!}\left(\frac{2}{\pi}\right)^{\delta(k_2,k_3+\ldots+k_N+N-3)}\\ &&\vdots\\ &&\frac{\pi}{2}\cdot\frac{k_{N-2}!!(k_{N-1}+k_N+1)!!}{(k_{N-2}+k_{N-1}+k_N+2)!!}\left(\frac{2}{\pi}\right)^{\delta(k_{N-2},k_{N-1}+k_N+1)}\\ &&\frac{\pi}{2}\cdot\frac{k_{N-1}!!k_N!!}{(k_{N-1}+k_N+1)!!}\left(\frac{2}{\pi}\right)^{\delta(k_{N-1},k_N)} \end{eqnarray*} [[/math]]

In order to compute this quantity, let us denote by [math]F[/math] the part involving the double factorials, and by [math]P[/math] the part involving the powers of [math]\pi/2[/math], so that we have:

[[math]] I'=F\cdot P [[/math]]

Regarding [math]F[/math], there are many cancellations there, and we end up with:

[[math]] F=\frac{k_1!!\ldots k_N!!}{(\Sigma k_i+N-1)!!} [[/math]]

As in what regards [math]P[/math], the [math]\delta[/math] exponents on the right sum up to the following number:

[[math]] \Delta(k_1,\ldots,k_N)=\sum_{i=1}^{N-1}\delta(k_i,k_{i+1}+\ldots+k_N+N-i-1) [[/math]]

In other words, with this notation, the above formula reads:

[[math]] \begin{eqnarray*} I' &=&\left(\frac{\pi}{2}\right)^{N-1}\frac{k_1!!k_2!!\ldots k_N!!}{(k_1+\ldots+k_N+N-1)!!}\left(\frac{2}{\pi}\right)^{\Delta(k_1,\ldots,k_N)}\\ &=&\left(\frac{2}{\pi}\right)^{\Delta(k_1,\ldots,k_N)-N+1}\frac{k_1!!k_2!!\ldots k_N!!}{(k_1+\ldots+k_N+N-1)!!}\\ &=&\left(\frac{2}{\pi}\right)^{\Sigma(k_1,\ldots,k_N)-[N/2]}\frac{k_1!!k_2!!\ldots k_N!!}{(k_1+\ldots+k_N+N-1)!!} \end{eqnarray*} [[/math]]

Here the formula relating [math]\Delta[/math] to [math]\Sigma[/math] follows from a number of simple observations, the first of which being the act that, due to obvious parity reasons, the sequence of [math]\delta[/math] numbers appearing in the definition of [math]\Delta[/math] cannot contain two consecutive zeroes. Together with [math]I=(2^N/V)I'[/math], this gives the formula in the statement.

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Summarizing, we have complete results for the integration over the spheres, with the answers involving various multinomial type coefficients, defined in terms of factorials, or of double factorials. All these formulae are of course very useful, in practice.

As a basic application of all this, we have the following result:

Theorem

The moments of the hyperspherical variables are

[[math]] \int_{S^{N-1}_\mathbb R}x_i^kdx=\frac{(N-1)!!k!!}{(N+k-1)!!} [[/math]]

and the normalized hyperspherical variables

[[math]] y_i=\frac{x_i}{\sqrt{N}} [[/math]]

become normal and independent with [math]N\to\infty[/math].

Show Proof

We have two things to be proved, the idea being as follows:

(1) The formula in the statement follows from the general integration formula over the sphere, established above. Indeed, this formula gives:

[[math]] \int_{S^{N-1}_\mathbb R}x_i^kdx=\frac{(N-1)!!k!!}{(N+k-1)!!} [[/math]]

Now observe that with [math]N\to\infty[/math] we have the following estimate:

[[math]] \begin{eqnarray*} \int_{S^{N-1}_\mathbb R}x_i^kdx &=&\frac{(N-1)!!}{(N+k-1)!!}\times k!!\\ &\simeq&N^{k/2}k!!\\ &=&N^{k/2}M_k(g_1) \end{eqnarray*} [[/math]]

Thus, the variables [math]y_i=\frac{x_i}{\sqrt{N}}[/math] become normal with [math]N\to\infty[/math].

(2) As for the asymptotic independence result, this is standard as well, once again by using Theorem 6.14, for computing mixed moments, and taking the [math]N\to\infty[/math] limit.

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As a comment here, all this might seem quite specialized. However, we will see later on that all this is related to linear algebra, and more specifically to the fine study of the group [math]O_N[/math] formed by the orthogonal matrices. But more on this later.

6d. Complex spheres

Let us discuss now the complex analogues of all the above. We must first introduce the complex analogues of the normal laws, and this can be done as follows:

Definition

The complex Gaussian law of parameter [math]t \gt 0[/math] is

[[math]] G_t=law\left(\frac{1}{\sqrt{2}}(a+ib)\right) [[/math]]

where [math]a,b[/math] are independent, each following the law [math]g_t[/math].

The combinatorics of these laws is a bit more complicated than in the real case, and we will be back to this in a moment. But to start with, we have:

Theorem

The complex Gaussian laws have the property

[[math]] G_s*G_t=G_{s+t} [[/math]]

for any [math]s,t \gt 0[/math], and so they form a convolution semigroup.

Show Proof

This follows indeed from the real result, for the usual Gaussian laws, established in above, by taking real and imaginary parts.

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We have as well the following complex analogue of the CLT:

Theorem (CCLT)

Given complex random variables [math]f_1,f_2,f_3,\ldots\in L^\infty(X)[/math], which are i.i.d., centered, and with variance [math]t \gt 0[/math], we have, with [math]n\to\infty[/math], in moments,

[[math]] \frac{1}{\sqrt{n}}\sum_{i=1}^nf_i\sim G_t [[/math]]

where [math]G_t[/math] is the complex Gaussian law of parameter [math]t[/math].

Show Proof

This follows indeed from the real CLT, established above, simply by taking the real and imaginary parts of all the variables involved.

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Regarding now the moments, things are a bit more complicated than before, because our variables are now complex instead of real. In order to deal with this issue, we will use “colored moments”, which are the expectations of the “colored powers”, with these latter powers being defined by the following formulae, and multiplicativity:

[[math]] f^\emptyset=1\quad,\quad f^\circ=f\quad,\quad f^\bullet=\bar{f} [[/math]]

With these conventions made, the result is as follows, with a pairing of a colored integer [math]k=\circ\bullet\bullet\circ\ldots[/math] being called matching when it pairs [math]\circ[/math] symbols with [math]\bullet[/math] symbols:

Theorem

The moments of the complex normal law are the numbers

[[math]] M_k(G_t)=\sum_{\pi\in\mathcal P_2(k)}t^{|\pi|} [[/math]]

where [math]\mathcal P_2(k)[/math] are the matching pairings of [math]\{1,\ldots,k\}[/math], and [math]|.|[/math] is the number of blocks.

Show Proof

This can be done in several steps, as follows:

(1) We recall from the above that the moments of the real Gaussian law [math]g_1[/math], with respect to integer exponents [math]k\in\mathbb N[/math], are the following numbers:

[[math]] m_k=|P_2(k)| [[/math]]

(2) We will show here that in what concerns the complex Gaussian law [math]G_1[/math], a similar result holds. Numerically, we will prove that we have the following formula, where a colored integer [math]k=\circ\bullet\bullet\circ\ldots[/math] is called uniform when it contains the same number of [math]\circ[/math] and [math]\bullet[/math]\,, and where [math]|k|\in\mathbb N[/math] is the length of such a colored integer:

[[math]] M_k=\begin{cases} (|k|/2)!&(k\ {\rm uniform})\\ 0&(k\ {\rm not\ uniform}) \end{cases} [[/math]]

Now since the matching partitions [math]\pi\in\mathcal P_2(k)[/math] are counted by exactly the same numbers, and this for trivial reasons, we will obtain the formula in the statement, namely:

[[math]] M_k=|\mathcal P_2(k)| [[/math]]

(3) This was for the plan. In practice now, we must compute the moments, with respect to colored integer exponents [math]k=\circ\bullet\bullet\circ\ldots[/math]\,, of the variable in the statement:

[[math]] c=\frac{1}{\sqrt{2}}(a+ib) [[/math]]

As a first observation, in the case where such an exponent [math]k=\circ\bullet\bullet\circ\ldots[/math] is not uniform in [math]\circ,\bullet[/math]\,, a rotation argument shows that the corresponding moment of [math]c[/math] vanishes. To be more precise, the variable [math]c'=wc[/math] can be shown to be complex Gaussian too, for any [math]w\in\mathbb C[/math], and from [math]M_k(c)=M_k(c')[/math] we obtain [math]M_k(c)=0[/math], in this case.

(4) In the uniform case now, where [math]k=\circ\bullet\bullet\circ\ldots[/math] consists of [math]p[/math] copies of [math]\circ[/math] and [math]p[/math] copies of [math]\bullet[/math]\,, the corresponding moment can be computed as follows:

[[math]] \begin{eqnarray*} M_k &=&\int(c\bar{c})^p\\ &=&\frac{1}{2^p}\int(a^2+b^2)^p\\ &=&\frac{1}{2^p}\sum_s\binom{p}{s}\int a^{2s}\int b^{2p-2s}\\ &=&\frac{1}{2^p}\sum_s\binom{p}{s}(2s)!!(2p-2s)!!\\ &=&\frac{1}{2^p}\sum_s\frac{p!}{s!(p-s)!}\cdot\frac{(2s)!}{2^ss!}\cdot\frac{(2p-2s)!}{2^{p-s}(p-s)!}\\ &=&\frac{p!}{4^p}\sum_s\binom{2s}{s}\binom{2p-2s}{p-s} \end{eqnarray*} [[/math]]

(5) In order to finish now the computation, let us recall that we have the following formula, coming from the generalized binomial formula, or from the Taylor formula:

[[math]] \frac{1}{\sqrt{1+t}}=\sum_{k=0}^\infty\binom{2k}{k}\left(\frac{-t}{4}\right)^k [[/math]]

By taking the square of this series, we obtain the following formula:

[[math]] \begin{eqnarray*} \frac{1}{1+t} &=&\sum_{ks}\binom{2k}{k}\binom{2s}{s}\left(\frac{-t}{4}\right)^{k+s}\\ &=&\sum_p\left(\frac{-t}{4}\right)^p\sum_s\binom{2s}{s}\binom{2p-2s}{p-s} \end{eqnarray*} [[/math]]

Now by looking at the coefficient of [math]t^p[/math] on both sides, we conclude that the sum on the right equals [math]4^p[/math]. Thus, we can finish the moment computation in (4), as follows:

[[math]] M_p=\frac{p!}{4^p}\times 4^p=p! [[/math]]

(6) As a conclusion, if we denote by [math]|k|[/math] the length of a colored integer [math]k=\circ\bullet\bullet\circ\ldots[/math]\,, the moments of the variable [math]c[/math] in the statement are given by:

[[math]] M_k=\begin{cases} (|k|/2)!&(k\ {\rm uniform})\\ 0&(k\ {\rm not\ uniform}) \end{cases} [[/math]]

On the other hand, the numbers [math]|\mathcal P_2(k)|[/math] in the statement are given by exactly the same formula. Indeed, in order to have matching pairings of [math]k[/math], our exponent [math]k=\circ\bullet\bullet\circ\ldots[/math] must be uniform, consisting of [math]p[/math] copies of [math]\circ[/math] and [math]p[/math] copies of [math]\bullet[/math], with:

[[math]] p=\frac{|k|}{2} [[/math]]

But then the matching pairings of [math]k[/math] correspond to the permutations of the [math]\bullet[/math] symbols, as to be matched with [math]\circ[/math] symbols, and so we have [math]p![/math] such matching pairings. Thus, we have exactly the same formula as for the moments of [math]c[/math], and this finishes the proof.

■

There are of course many other possible proofs for the above result, which are all instructive, and some further theory as well, that can be developed for the complex normal variables, which is very interesting too. We refer here to Feller , or Durrett . We will be back to this, on several occasions, in what follows.

In practice, we also need to know how to compute joint moments of independent normal variables. We have here the following result, to be heavily used later on:

Theorem (Wick formula)

Given independent variables [math]f_i[/math], each following the complex normal law [math]G_t[/math], with [math]t \gt 0[/math] being a fixed parameter, we have the formula

[[math]] \mathbb E\left(f_{i_1}^{k_1}\ldots f_{i_s}^{k_s}\right)=t^{s/2}\#\left\{\pi\in\mathcal P_2(k)\Big|\pi\leq\ker(i)\right\} [[/math]]

where [math]k=k_1\ldots k_s[/math] and [math]i=i_1\ldots i_s[/math], for the joint moments of these variables.

Show Proof

This is something well-known, and the basis for all possible computations with complex normal variables, which can be proved in two steps, as follows:

(1) Let us first discuss the case where we have a single variable [math]f[/math], which amounts in taking [math]f_i=f[/math] for any [math]i[/math] in the formula in the statement. What we have to compute here are the moments of [math]f[/math], with respect to colored integer exponents [math]k=\circ\bullet\bullet\circ\ldots\,[/math], and the formula in the statement tells us that these moments must be:

[[math]] \mathbb E(f^k)=t^{|k|/2}|\mathcal P_2(k)| [[/math]]

But this is the formula in Theorem 6.20, so we are done with this case.

(2) In general now, when expanding the product [math]f_{i_1}^{k_1}\ldots f_{i_s}^{k_s}[/math] and rearranging the terms, we are left with doing a number of computations as in (1), and then making the product of the expectations that we found. But this amounts in counting the partitions in the statement, with the condition [math]\pi\leq\ker(i)[/math] there standing for the fact that we are doing the various type (1) computations independently, and then making the product.

■

The above statement is one of the possible formulations of the Wick formula, and there are in fact many more formulations, which are all useful. Here is an alternative such formulation, which is quite popular, and that we will often use in what follows:

Theorem (Wick formula 2)

Given independent variables [math]f_i[/math], each following the complex normal law [math]G_t[/math], with [math]t \gt 0[/math] being a fixed parameter, we have the formula

[[math]] \mathbb E\left(f_{i_1}\ldots f_{i_k}f_{j_1}^*\ldots f_{j_k}^*\right)=t^k\#\left\{\pi\in S_k\Big|i_{\pi(r)}=j_r,\forall r\right\} [[/math]]

for the non-vanishing joint moments of these variables.

Show Proof

This follows from the usual Wick formula, from Theorem 6.21. With some changes in the indices and notations, the formula there reads:

[[math]] \mathbb E\left(f_{I_1}^{K_1}\ldots f_{I_s}^{K_s}\right)=t^{s/2}\#\left\{\sigma\in\mathcal P_2(K)\Big|\sigma\leq\ker(I)\right\} [[/math]]

Now observe that we have [math]\mathcal P_2(K)=\emptyset[/math], unless the colored integer [math]K=K_1\ldots K_s[/math] is uniform, in the sense that it contains the same number of [math]\circ[/math] and [math]\bullet[/math] symbols. Up to permutations, the non-trivial case, where the moment is non-vanishing, is the case where the colored integer [math]K=K_1\ldots K_s[/math] is of the following special form:

[[math]] K=\underbrace{\circ\circ\ldots\circ}_k\ \underbrace{\bullet\bullet\ldots\bullet}_k [[/math]]

So, let us focus on this case, which is the non-trivial one. Here we have [math]s=2k[/math], and we can write the multi-index [math]I=I_1\ldots I_s[/math] in the following way:

[[math]] I=i_1\ldots i_k\ j_1\ldots j_k [[/math]]

With these changes made, the above usual Wick formula reads:

[[math]] \mathbb E\left(f_{i_1}\ldots f_{i_k}f_{j_1}^*\ldots f_{j_k}^*\right)=t^k\#\left\{\sigma\in\mathcal P_2(K)\Big|\sigma\leq\ker(ij)\right\} [[/math]]

The point now is that the matching pairings [math]\sigma\in\mathcal P_2(K)[/math], with [math]K=\circ\ldots\circ\bullet\ldots\bullet\,[/math], of length [math]2k[/math], as above, correspond to the permutations [math]\pi\in S_k[/math], in the obvious way. With this identification made, the above modified usual Wick formula becomes:

[[math]] \mathbb E\left(f_{i_1}\ldots f_{i_k}f_{j_1}^*\ldots f_{j_k}^*\right)=t^k\#\left\{\pi\in S_k\Big|i_{\pi(r)}=j_r,\forall r\right\} [[/math]]

Thus, we have reached to the formula in the statement, and we are done.

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Finally, here is one more formulation of the Wick formula, which is useful as well:

Theorem (Wick formula 3)

Given independent variables [math]f_i[/math], each following the complex normal law [math]G_t[/math], with [math]t \gt 0[/math] being a fixed parameter, we have the formula

[[math]] \mathbb E\left(f_{i_1}f_{j_1}^*\ldots f_{i_k}f_{j_k}^*\right)=t^k\#\left\{\pi\in S_k\Big|i_{\pi(r)}=j_r,\forall r\right\} [[/math]]

for the non-vanishing joint moments of these variables.

Show Proof

This follows from our second Wick formula, from Theorem 6.22, simply by permuting the terms, as to have an alternating sequence of plain and conjugate variables. Alternatively, we can start with Theorem 6.21, and then perform the same manipulations as in the proof of Theorem 6.22, but with the exponent being this time as follows:

[[math]] K=\underbrace{\circ\bullet\circ\bullet\ldots\ldots\circ\bullet}_{2k} [[/math]]

Thus, we are led to the conclusion in the statement.

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In relation now with the spheres, we first have the following variation of the integration formula in Theorem 6.15, dealing this time with integrals over the complex sphere:

Theorem

We have the following integration formula over the complex sphere [math]S^{N-1}_\mathbb C\subset\mathbb R^N[/math], with respect to the normalized measure,

[[math]] \int_{S^{N-1}_\mathbb C}|z_1|^{2l_1}\ldots|z_N|^{2l_N}\,dz=4^{\sum l_i}\frac{(2N-1)!l_1!\ldots l_n!}{(2N+\sum l_i-1)!} [[/math]]

valid for any exponents [math]l_i\in\mathbb N[/math]. As for the other polynomial integrals in [math]z_1,\ldots,z_N[/math] and their conjugates [math]\bar{z}_1,\ldots,\bar{z}_N[/math], these all vanish.

Show Proof

Consider an arbitrary polynomial integral over [math]S^{N-1}_\mathbb C[/math], written as follows:

[[math]] I=\int_{S^{N-1}_\mathbb C}z_{i_1}\bar{z}_{i_2}\ldots z_{i_{2l-1}}\bar{z}_{i_{2l}}\,dz [[/math]]

(1) By using transformations of type [math]p\to\lambda p[/math] with [math]|\lambda|=1[/math], we see that [math]I[/math] vanishes, unless each [math]z_a[/math] appears as many times as [math]\bar{z}_a[/math] does, and this gives the last assertion.

(2) Assume now that we are in the non-vanishing case. Then the [math]l_a[/math] copies of [math]z_a[/math] and the [math]l_a[/math] copies of [math]\bar{z}_a[/math] produce by multiplication a factor [math]|z_a|^{2l_a}[/math], so we have:

[[math]] I=\int_{S^{N-1}_\mathbb C}|z_1|^{2l_1}\ldots|z_N|^{2l_N}\,dz [[/math]]

Now by using the standard identification [math]S^{N-1}_\mathbb C\simeq S^{2N-1}_\mathbb R[/math], we obtain:

[[math]] \begin{eqnarray*} I &=&\int_{S^{2N-1}_\mathbb R}(x_1^2+y_1^2)^{l_1}\ldots(x_N^2+y_N^2)^{l_N}\,d(x,y)\\ &=&\sum_{r_1\ldots r_N}\binom{l_1}{r_1}\ldots\binom{l_N}{r_N}\int_{S^{2N-1}_\mathbb R}x_1^{2l_1-2r_1}y_1^{2r_1}\ldots x_N^{2l_N-2r_N}y_N^{2r_N}\,d(x,y) \end{eqnarray*} [[/math]]

(3) By using the formula in Theorem 6.14, we obtain:

[[math]] \begin{eqnarray*} &&I\\ &=&\sum_{r_1\ldots r_N}\binom{l_1}{r_1}\ldots\binom{l_N}{r_N}\frac{(2N-1)!!(2r_1)!!\ldots(2r_N)!!(2l_1-2r_1)!!\ldots (2l_N-2r_N)!!}{(2N+2\sum l_i-1)!!}\\ &=&\sum_{r_1\ldots r_N}\binom{l_1}{r_1}\ldots\binom{l_N}{r_N} \frac{(2N-1)!(2r_1)!\ldots (2r_N)!(2l_1-2r_1)!\ldots (2l_N-2r_N)!}{(2N+\sum l_i-1)!r_1!\ldots r_N!(l_1-r_1)!\ldots (l_N-r_N)!} \end{eqnarray*} [[/math]]

(4) We can rewrite the sum on the right in the following way:

[[math]] \begin{eqnarray*} &&I\\ &=&\sum_{r_1\ldots r_N}\frac{l_1!\ldots l_N!(2N-1)!(2r_1)!\ldots (2r_N)!(2l_1-2r_1)!\ldots (2l_N-2r_N)!}{(2N+\sum l_i-1)!(r_1!\ldots r_N!(l_1-r_1)!\ldots (l_N-r_N)!)^2}\\ &=&\sum_{r_1}\binom{2r_1}{r_1}\binom{2l_1-2r_1}{l_1-r_1}\ldots\sum_{r_N}\binom{2r_N}{r_N}\binom{2l_N-2r_N}{l_N-r_N}\frac{(2N-1)!l_1!\ldots l_N!}{(2N+\sum l_i-1)!}\\ &=&4^{l_1}\times\ldots\times 4^{l_N}\times\frac{(2N-1)!l_1!\ldots l_N!}{(2N+\sum l_i-1)!} \end{eqnarray*} [[/math]]

Thus, we obtain the formula in the statement.

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Regarding now the hyperspherical variables, investigated in the above in the real case, we have similar results for the complex spheres, as follows:

Theorem

The rescaled coordinates on the complex sphere [math]S^{N-1}_\mathbb C[/math],

[[math]] w_i=\frac{z_i}{\sqrt{N}} [[/math]]

become complex Gaussian and independent with [math]N\to\infty[/math].

Show Proof

We have two assertions to be proved, the idea being as follows:

(1) The assertion about the laws follows as in the real case, by using this time Theorem 6.24 as a main technical ingredient.

(2) As for the independence result, this follows as well as in the real case, by using this time the Wick formula as a main technical ingredient.

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As a conclusion to all this, we have now a good level in linear algebra, and probability. And this can only open up a whole new set of perspectives, on what further books can be read. Here is a guide, to what can be learned, as a continuation:

(1) Before anything, talking analysis, you should imperatively learn some Fourier analysis and differential equations, topics not covered in this book. With good places here being the books of Arnold , , , , Evans and Rudin , .

(2) You can also embark on the reading of some tough, ultimate physics, say from Landau-Lifshitz , , , . Normally these are books which are quite strong on mathematics, but with your knowledge from here, you will certainly survive.

(3) You can also start reading whatever fancy physics that you might like, such as astrophysics and cosmology from Ryden , or Weinberg , , quantum information from Bengtsson-\.Zyczkowski or Nielsen-Chuang , and so on.

(4) And for algebra and probability, stay with us. The story is far from being over with what we learned, and dozens of further interesting things to follow. We still have 250 more pages, and there will be algebra and probability in them, that is promised.

General references

Banica, Teo (2024). "Linear algebra and group theory". arXiv:2206.09283 [math.CO].