Normal laws

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

This article was automatically generated from a tex file and may contain conversion errors. If permitted, you may login and edit this article to improve the conversion.

1a. Probability theory

Generally speaking, probability theory is best learned by flipping coins and throwing dice. At a more advanced level, which is playing cards, we have:

Theorem

The probabilities at poker are as follows:

• One pair: [math]0.533[/math].
• Two pairs: [math]0.120[/math].
• Three of a kind: [math]0.053[/math].
• Full house: [math]0.006[/math].
• Straight: [math]0.005[/math].
• Four of a kind: [math]0.001[/math].
• Flush: [math]0.000[/math].
• Straight flush: [math]0.000[/math].

Show Proof

Let us consider indeed our deck of 32 cards, [math]7,8,9,10,J,Q,K,A[/math]. The total number of possibilities for a poker hand is:

[[math]] \binom{32}{5} =\frac{32\cdot 31\cdot 30\cdot 29\cdot 28}{2\cdot 3\cdot 4\cdot 5} =32\cdot 31\cdot 29\cdot 7 [[/math]]

(1) For having a pair, the number of possibilities is:

[[math]] N =\binom{8}{1}\binom{4}{2}\times\binom{7}{3}\binom{4}{1}^3 =8\cdot 6\cdot 35\cdot 64 [[/math]]

Thus, the probability of having a pair is:

[[math]] P =\frac{8\cdot 6\cdot 35\cdot 64}{32\cdot 31\cdot 29\cdot 7} =\frac{6\cdot 5\cdot 16}{31\cdot 29} =\frac{480}{899} =0.533 [[/math]]

(2) For having two pairs, the number of possibilities is:

[[math]] N =\binom{8}{2}\binom{4}{2}^2\times\binom{24}{1} =28\cdot 36\cdot 24 [[/math]]

Thus, the probability of having two pairs is:

[[math]] P =\frac{28\cdot 36\cdot 24}{32\cdot 31\cdot 29\cdot 7} =\frac{36\cdot 3}{31\cdot 29} =\frac{108}{899} =0.120 [[/math]]

(3) For having three of a kind, the number of possibilities is:

[[math]] N =\binom{8}{1}\binom{4}{3}\times\binom{7}{2}\binom{4}{1}^2 =8\cdot 4\cdot 21\cdot 16 [[/math]]

Thus, the probability of having three of a kind is:

[[math]] P =\frac{8\cdot 4\cdot 21\cdot 16}{32\cdot 31\cdot 29\cdot 7} =\frac{3\cdot 16}{31\cdot 29} =\frac{48}{899} =0.053 [[/math]]

(4) For having full house, the number of possibilities is:

[[math]] N =\binom{8}{1}\binom{4}{3}\times\binom{7}{1}\binom{4}{2} =8\cdot 4\cdot 7\cdot 6 [[/math]]

Thus, the probability of having full house is:

[[math]] P =\frac{8\cdot 4\cdot 7\cdot 6}{32\cdot 31\cdot 29\cdot 7} =\frac{6}{31\cdot 29} =\frac{6}{899} =0.006 [[/math]]

(5) For having a straight, the number of possibilities is:

[[math]] N =4\left[\binom{4}{1}^4-4\right] =16\cdot 63 [[/math]]

Thus, the probability of having a straight is:

[[math]] P =\frac{16\cdot 63}{32\cdot 31\cdot 29\cdot 7} =\frac{9}{2\cdot 31\cdot 29} =\frac{9}{1798} =0.005 [[/math]]

(6) For having four of a kind, the number of possibilities is:

[[math]] N =\binom{8}{1}\binom{4}{4}\times\binom{7}{1}\binom{4}{1} =8\cdot 7\cdot 4 [[/math]]

Thus, the probability of having four of a kind is:

[[math]] P =\frac{8\cdot 7\cdot 4}{32\cdot 31\cdot 29\cdot 7} =\frac{1}{31\cdot 29} =\frac{1}{899} =0.001 [[/math]]

(7) For having a flush, the number of possibilities is:

[[math]] N =4\left[\binom{8}{4}-4\right] =4\cdot 66 [[/math]]

Thus, the probability of having a flush is:

[[math]] P =\frac{4\cdot 66}{32\cdot 31\cdot 29\cdot 7} =\frac{33}{4\cdot 31\cdot 29\cdot 7} =\frac{9}{25172} =0.000 [[/math]]

(8) For having a straight flush, the number of possibilities is:

[[math]] N=4\cdot 4 [[/math]]

Thus, the probability of having a straight flush is:

[[math]] P =\frac{4\cdot 4}{32\cdot 31\cdot 29\cdot 7} =\frac{1}{2\cdot 31\cdot 29\cdot 7} =\frac{1}{12586} =0.000 [[/math]]

Thus, we have obtained the numbers in the statement.

■

Summarizing, probability is basically about binomials and factorials, and ultimately about numbers. We will see later that, in connection with more advanced questions, of continuous nature, some standard calculus comes into play as well.

Let us discuss now the general theory. The fundamental result in probability is the Central Limit Theorem (CLT), and our first task will be that of explaining this. With the idea in mind of doing things a bit abstractly, our starting point will be:

Definition

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

• The random variables are the real functions [math]f\in L^\infty(X)[/math].
• The moments of such a variable are the numbers [math]M_k(f)=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 well-known. 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]] E(P(f))=\int_\mathbb RP(x)d\mu_f(x) [[/math]]

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

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

But this latter formula, which reads [math]P(f\in I)=\mu_f(I)[/math], can serve as a definition for [math]\mu_f[/math], and we are done. Alternatively, assuming some familiarity with measure theory, [math]\mu_f[/math] is the push-forward of the probability measure on [math]X[/math], via the function [math]f:X\to\mathbb R[/math].

Next in line, we need to talk about independence. We can do this as follows:

Definition

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

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

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

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

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

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

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

[[math]] P(f\in I,g\in J)=P(f\in I)\,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. As a first result now, we have:

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) &=&E((f+g)^k)\\ &=&\sum_r\binom{k}{r}E(f^rg^{k-r})\\ &=&\sum_r\binom{k}{r}M_r(f)M_{k-r}(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_r\binom{k}{r}\int_\mathbb Rx^rd\mu_f(x)\int_\mathbb Ry^{k-r}d\mu_g(y)\\ &=&\sum_r\binom{k}{r}M_r(f)M_{k-r}(g) \end{eqnarray*} [[/math]]

Thus [math]\mu_{f+g}[/math] and [math]\mu_f*\mu_g[/math] have the same moments, so they coincide, as desired.

■

Here is now a second result on independence, which is something more advanced:

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)=E(e^{ixf})[/math] is the Fourier transform.

Show Proof

We have the following computation, using Proposition 1.4 and Fubini:

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

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

■

This was for the foundations of probability theory, quickly explained. For further reading, a classical book is Feller . A nice, more modern book is Durrett .

1b. Central limits

The main result in classical probability is the Central Limit Theorem (CLT), that we will explain now. Let us first discuss the normal distributions, that we will see later to appear as limiting laws in the CLT. We will need the following standard result:

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 integral in the statement. By using polar coordinates, namely [math]x=r\cos t[/math], [math]y=r\sin t[/math], with the corresponding Jacobian being [math]r[/math], we have:

[[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}r\,drdt\\ &=&2\pi\int_0^\infty\left(-\frac{e^{-r^2}}{2}\right)'dr\\ &=&\pi \end{eqnarray*} [[/math]]

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

■

We can now introduce the normal distributions, as follows:

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.

The above laws are usually denoted [math]\mathcal N(0,1)[/math] and [math]\mathcal N(0,t)[/math], but since we will be doing in this book all kinds of probability, we will use simplified notations for all our measures. Let us mention as well that the normal laws traditionally have 2 parameters, the mean and the variance, but here we will not need the mean, all our theory using centered laws. Finally, observe that the above laws have indeed mass 1, as they should, due to:

[[math]] \int_\mathbb R e^{-x^2/2t}dx =\int_\mathbb R e^{-y^2}\sqrt{2t}\,dy =\sqrt{2\pi t} [[/math]]

Generally speaking, the normal laws appear as bit everywhere, in real life. The reasons for this come from the Central Limit Theorem (CLT), that we will explain in a moment, after developing some more general theory. As a first result, we 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].

■

Here is another result, which is widely useful in practice:

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^{-z^2/2t+ixz}dz\\ &=&\frac{1}{\sqrt{2\pi t}}\int_\mathbb Re^{-(z/\sqrt{2t}-\sqrt{t/2}\,iz)^2-tx^2/2}dz\\ &=&\frac{1}{\sqrt{2\pi t}}\int_\mathbb Re^{-y^2-tx^2/2}\sqrt{2t}\,dy\\ &=&\frac{1}{\sqrt{\pi}}e^{-tx^2/2}\int_\mathbb Re^{-y^2}dy\\ &=&e^{-tx^2/2} \end{eqnarray*} [[/math]]

As for [math]g_s*g_t=g_{s+t}[/math], this follows via Theorem 1.5, [math]\log F_{g_t}[/math] being linear in [math]t[/math].

■

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

Theorem (CLT)

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

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

with [math]n\to\infty[/math], in moments.

Show Proof

In terms of moments, the Fourier transform is given by:

[[math]] \begin{eqnarray*} F_f(x) &=&E\left(\sum_{r=0}^\infty\frac{(ixf)^r}{r!}\right)\\ &=&\sum_{r=0}^\infty\frac{(ix)^rE(f^r)}{r!}\\ &=&\sum_{r=0}^\infty\frac{i^rM_r(f)}{r!}\,x^r \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&e^{-tx^2/2} \end{eqnarray*} [[/math]]

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

■

Let us discuss now some further properties of the normal law. We first have:

Proposition

The even moments of the normal law are the numbers

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

where [math]k!!=(k-1)(k-3)(k-5)\ldots\,[/math], and the odd moments vanish.

Show Proof

We have the following computation, valid for any integer [math]k\in\mathbb N[/math]:

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

Now recall from the proof of Proposition 1.8 that we have [math]M_0=1[/math], [math]M_1=0[/math]. Thus by recurrence, we are led to the formula in the statement.

■

We have the following alternative formulation of the above result:

Proposition

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 the numbers [math]2,\ldots,k[/math], and then use a pairing of the remaining [math]k-2[/math] numbers. Thus, we have the following recurrence formula:

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

As for the initial data, this is [math]P_1=0[/math], [math]P_2=1[/math]. Thus, we are led to the result.

■

We are not done yet, and here is one more improvement of the above:

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 Proposition 1.12, 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 in this book that many other interesting probability distributions are subject to similar formulae regarding their moments, involving partitions.

1c. Spherical integrals

In a purely mathematical context, the simplest way of recovering the normal laws is by looking at the coordinates over the real spheres [math]S^{N-1}_\mathbb R[/math], in the [math]N\to\infty[/math] limit. To start with, at [math]N=2[/math] the sphere is the unit circle [math]\mathbb T[/math], and with [math]z=e^{it}[/math] the coordinates are [math]\cos t,\sin t[/math]. Let us first integrate powers of these coordinates. We have here:

Proposition

We have the following formulae,

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

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

Show Proof

Let us call [math]I_k[/math] the integral on the left in the statement. In order to compute it, we use partial integration. We have the following formula:

[[math]] \begin{eqnarray*} (\cos^kt\sin t)' &=&k\cos^{k-1}t(-\sin t)\sin t+\cos^kt\cos t\\ &=&(k+1)\cos^{k+1}t-k\cos^{k-1}t \end{eqnarray*} [[/math]]

By integrating between [math]0[/math] and [math]\pi/2[/math], we obtain the following formula:

[[math]] (k+1)I_{k+1}=kI_{k-1} [[/math]]

Thus we can compute [math]I_k[/math] by recurrence, and we obtain in this way:

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

The initial data being [math]I_0=\pi/2[/math] and [math]I_1=1[/math], we obtain the result. As for the second formula, this follows from the first one, with the change of variables [math]t=\pi/2-s[/math].

■

More generally now, we have the following result:

Theorem

We have the following formula,

[[math]] \int_0^{\pi/2}\cos^rt\sin^st\,dt=\left(\frac{\pi}{2}\right)^{\varepsilon(r)\varepsilon(s)}\frac{r!!s!!}{(r+s+1)!!} [[/math]]

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

Show Proof

Let us call [math]I_{rs}[/math] the integral in the statement. In order to do the partial integration, observe that we have the following formula:

[[math]] \begin{eqnarray*} (\cos^rt\sin^st)' &=&r\cos^{r-1}t(-\sin t)\sin^st+\cos^rt\cdot s\sin^{s-1}t\cos t\\ &=&-r\cos^{r-1}t\sin^{s+1}t+s\cos^{r+1}t\sin^{s-1}t \end{eqnarray*} [[/math]]

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

[[math]] rI_{r-1,s+1}=sI_{r+1,s-1} [[/math]]

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

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

But the last term comes from Proposition 1.14, and we obtain the result:

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

Observe that this gives the result for [math]r[/math] even as well, by symmetry. In the remaining case now, where both the exponents [math]r,s[/math] are odd, we can use once again the formula [math]rI_{r-1,s+1}=sI_{r+1,s-1}[/math] found above, and the recurrence goes as follows:

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

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

[[math]] \begin{eqnarray*} I_{r1} &=&\int_0^{\pi/2}\cos^rt\sin t\,dt\\ &=&-\frac{1}{r+1}\int_0^{\pi/2}(\cos^{r+1}t)'\,dt\\ &=&\frac{1}{r+1} \end{eqnarray*} [[/math]]

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

■

In order to deal now with the higher spheres, we will use spherical coordinates:

Theorem

We have spherical coordinates in [math]N[/math] dimensions,

[[math]] \begin{cases} x_1\!\!\!&=\ r\cos t_1\\ x_2\!\!\!&=\ r\sin t_1\cos t_2\\ \vdots\\ x_{N-1}\!\!\!&=\ r\sin t_1\sin t_2\ldots\sin t_{N-2}\cos t_{N-1}\\ x_N\!\!\!&=\ r\sin t_1\sin t_2\ldots\sin t_{N-2}\sin t_{N-1} \end{cases} [[/math]]

the corresponding Jacobian being given by the following formula:

[[math]] J(r,t)=r^{N-1}\sin^{N-2}t_1\sin^{N-3}t_2\,\ldots\,\sin^2t_{N-3}\sin t_{N-2} [[/math]]

Show Proof

The fact that we have indeed spherical coordinates is clear. Regarding now the Jacobian, by developing over the last column, we have:

[[math]] \begin{eqnarray*} J_N &=&r\sin t_1\ldots\sin t_{N-2}\sin t_{N-1}\times \sin t_{N-1}J_{N-1}\\ &+&r\sin t_1\ldots \sin t_{N-2}\cos t_{N-1}\times\cos t_{N-1}J_{N-1}\\ &=&r\sin t_1\ldots\sin t_{N-2}(\sin^2 t_{N-1}+\cos^2 t_{N-1})J_{N-1}\\ &=&r\sin t_1\ldots\sin t_{N-2}J_{N-1} \end{eqnarray*} [[/math]]

Thus, we obtain the formula in the statement, by recurrence.

■

As a first application, we can compute the volume of the sphere:

Theorem

The volume of the unit sphere in [math]\mathbb R^N[/math] is given by

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

with our usual convention [math]m!!=(m-1)(m-3)(m-5)\ldots[/math] for double factorials.

Show Proof

If we denote by [math]Q[/math] the positive part of the sphere, obtained by cutting the sphere in [math]2^N[/math] parts, we have, by using Theorems 1.15 and 1.16 and Fubini:

[[math]] \begin{eqnarray*} \frac{V}{2^N} &=&\int_0^1\int_0^{\pi/2}\ldots\int_0^{\pi/2}r^{N-1}\sin^{N-2}t_1\ldots\sin t_{N-2}\,drdt_1\ldots dt_{N-1}\\ &=&\int_0^1r^{N-1}\,dr\int_0^{\pi/2}\sin^{N-2}t_1\,dt_1\ldots\int_0^{\pi/2}\sin t_{N-2}dt_{N-2}\int_0^{\pi/2}1dt_{N-1}\\ &=&\frac{1}{N}\times\left(\frac{\pi}{2}\right)^{[N/2]}\times\frac{(N-2)!!}{(N-1)!!}\cdot\frac{(N-3)!!}{(N-2)!!}\ldots\frac{2!!}{3!!}\cdot\frac{1!!}{2!!}\cdot1\\ &=&\left(\frac{\pi}{2}\right)^{[N/2]}\frac{1}{(N+1)!!} \end{eqnarray*} [[/math]]

Here we have used the following formula for computing the exponent of [math]\pi/2[/math], where [math]\varepsilon(r)=1[/math] if [math]r[/math] is even and [math]\varepsilon(r)=0[/math] if [math]r[/math] is odd, as in Theorem 1.15:

[[math]] \begin{eqnarray*} \varepsilon(0)+\varepsilon(1)+\varepsilon(2)+\ldots+\varepsilon(N-2) &=&1+0+1+0+\ldots+\varepsilon(N-2)\\ &=&\left[\frac{N-2}{2}\right]+1\\ &=&\left[\frac{N}{2}\right] \end{eqnarray*} [[/math]]

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

■

Let us discuss now the computation of the arbitrary polynomial integrals, over the spheres of arbitrary dimension. The result here is as follows:

Theorem

The spherical integral of [math]x_{i_1}\ldots x_{i_r}[/math] vanishes, unless each index [math]a\in\{1,\ldots,N\}[/math] appears an even number of times in the sequence [math]i_1,\ldots,i_r[/math]. We have

[[math]] \int_{S^{N-1}_\mathbb R}x_{i_1}\ldots x_{i_r}\,dx=\frac{(N-1)!!k_1!!\ldots k_N!!}{(N+\Sigma k_i-1)!!} [[/math]]

with [math]k_a[/math] being this number of occurrences.

Show Proof

In what concerns the first assertion, regarding vanishing when some multiplicity [math]k_a[/math] is odd, this follows via the change of variables [math]x_a\to-x_a[/math]. Regarding now the formula in the statement, assume that we are in the case [math]k_a\in 2\mathbb N[/math], for any [math]a\in\{1,\ldots,N\}[/math]. The integral in the statement can be written in spherical coordinates, as follows:

[[math]] I=\frac{2^N}{V}\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]]

In this formula [math]V[/math] is the volume 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 the formula in Theorem 1.17, the normalization constant is:

[[math]] \frac{2^N}{V} =\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 formula in Theorem 1.15, 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}+k_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 observe that the various double factorials multiply up to quantity in the statement, modulo a [math](N-1)!![/math] factor, and that the [math]\pi/2[/math] factors multiply up to:

[[math]] F=\left(\frac{\pi}{2}\right)^{[N/2]} [[/math]]

Thus by multiplying by the normalization constant, we obtain the result.

■

We can now recover the normal laws, geometrically, as follows:

Theorem

The moments of the hyperspherical variables are

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

and the rescaled variables [math]y_i=\sqrt{N}x_i[/math] become normal and independent with [math]N\to\infty[/math].

Show Proof

The moment formula in the statement follows from Theorem 1.18. As a consequence, with [math]N\to\infty[/math] we have the following estimate:

[[math]] \int_{S^{N-1}_\mathbb R}x_i^pdx \simeq N^{-p/2}\times p!! =N^{-p/2}M_p(g_1) [[/math]]

Thus, the rescaled variables [math]\sqrt{N}x_i[/math] become normal with [math]N\to\infty[/math], as claimed. As for the proof of the asymptotic independence, this is standard too, once again by using the formula in Theorem 1.18. Indeed, the joint moments of [math]x_1,\ldots,x_N[/math] are given by:

[[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)!!} \simeq N^{-\Sigma k_i}\times k_1!!\ldots k_N!! [[/math]]

By rescaling, the joint moments of the variables [math]y_i=\sqrt{N}x_i[/math] are given by:

[[math]] \int_{S^{N-1}_\mathbb R}y_1^{k_1}\ldots y_N^{k_N}\,dx\simeq k_1!!\ldots k_N!! [[/math]]

Thus, we have multiplicativity, and so independence with [math]N\to\infty[/math], as claimed.

■

As a last result about the normal laws, we can recover these as well in connection with rotation groups. Indeed, we have the following reformulation of Theorem 1.19:

Theorem

We have the integration formula

[[math]] \int_{O_N}U_{ij}^p\,dU=\frac{(N-1)!!p!!}{(N+p-1)!!} [[/math]]

and the rescaled variables [math]V_{ij}=\sqrt{N}U_{ij}[/math] become normal and independent with [math]N\to\infty[/math].

Show Proof

We use the basic fact that the rotations [math]U\in O_N[/math] act on the points of the real sphere [math]z\in S^{N-1}_\mathbb R[/math], with the stabilizer of [math]z=(1,0,\ldots,0)[/math] being the subgroup [math]O_{N-1}\subset O_N[/math]. In algebraic terms, this gives an identification as follows:

[[math]] S^{N-1}_\mathbb R=O_N/O_{N-1} [[/math]]

In functional analytic terms, this result provides us with an embedding as follows, for any [math]i[/math], which makes correspond the respective integration functionals:

[[math]] C(S^{N-1}_\mathbb R)\subset C(O_N)\quad,\quad x_i\to U_{1i} [[/math]]

With this identification made, the result follows from Theorem 1.19.

■

We will see later, following , , that the relation between the orthogonal group [math]O_N[/math] and the normal laws goes well beyond Theorem 1.20. And we will see as well, following , and related papers, that there are also “free versions” of all this.

1d. Complex variables

We have seen so far a number of interesting results regarding the normal laws, and their geometric interpretation. As a last topic for this chapter, let us discuss now the complex analogues of all this. To start with, we have the following definition:

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].

As in the real case, these measures form convolution semigroups:

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, namely [math]g_s*g_t=g_{s+t}[/math], established in Theorem 1.9, simply by taking real and imaginary parts.

■

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

Theorem (CCLT)

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

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

with [math]n\to\infty[/math], in moments.

Show Proof

This follows indeed from the real CLT, established in Theorem 1.10, simply by taking the real and imaginary parts of all variables involved.

■

Regarding now the moments, the situation is more complicated than in the real case, because in order to have good results, we have to deal with both the complex variables, and their conjugates. Let us formulate the following definition:

Definition

The moments a complex variable [math]f\in L^\infty(X)[/math] are the numbers

[[math]] M_k=E(f^k) [[/math]]

depending on colored integers [math]k=\circ\bullet\bullet\circ\ldots\,[/math], with the conventions

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

and multiplicativity, in order to define the colored powers [math]f^k[/math].

Observe that, since [math]f,\bar{f}[/math] commute, we can permute terms, and restrict the attention to exponents of type [math]k=\ldots\circ\circ\circ\bullet\bullet\bullet\bullet\ldots\,[/math], if we want to. However, our result about the complex Gaussian laws, and other complex laws, later on, will actually look better without doing is, so we will use Definition 1.24 as stated. We first have:

Theorem

The moments of the complex normal law are given by

[[math]] M_k(G_t)=\begin{cases} t^pp!&(k\ {\rm uniform, of\ length}\ 2p)\\ 0&(k\ {\rm not\ uniform}) \end{cases} [[/math]]

where [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].

Show Proof

We must compute the moments, with respect to colored integer exponents [math]k=\circ\bullet\bullet\circ\ldots[/math]\,, of the variable from Definition 1.21, namely:

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

We can assume that we are in the case [math]t=1[/math], and the proof here goes as follows:

(1) As a first observation, in the case where our exponent [math]k=\circ\bullet\bullet\circ\ldots[/math] is not uniform, a standard rotation argument shows that the corresponding moment of [math]f[/math] vanishes. To be more precise, the variable [math]f'=wf[/math] is complex Gaussian too, for any complex number [math]w\in\mathbb T[/math], and from [math]M_k(f)=M_k(f')[/math] we obtain [math]M_k(f)=0[/math], in this case.

(2) In the uniform case now, where the exponent [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(f\bar{f})^p\\ &=&\frac{1}{2^p}\int(a^2+b^2)^p\\ &=&\frac{1}{2^p}\sum_r\binom{p}{r}\int a^{2r}\int b^{2p-2r}\\ &=&\frac{1}{2^p}\sum_r\binom{p}{r}(2r)!!(2p-2r)!!\\ &=&\frac{1}{2^p}\sum_r\frac{p!}{r!(p-r)!}\cdot\frac{(2r)!}{2^rr!}\cdot\frac{(2p-2r)!}{2^{p-r}(p-r)!}\\ &=&\frac{p!}{4^p}\sum_r\binom{2r}{r}\binom{2p-2r}{p-r} \end{eqnarray*} [[/math]]

(3) 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_{q=0}^\infty\binom{2q}{q}\left(\frac{-t}{4}\right)^q [[/math]]

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

[[math]] \begin{eqnarray*} \frac{1}{1+t} &=&\sum_{qr}\binom{2q}{q}\binom{2r}{r}\left(\frac{-t}{4}\right)^{q+r}\\ &=&\sum_p\left(\frac{-t}{4}\right)^p\sum_r\binom{2r}{r}\binom{2p-2r}{p-r} \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 (2), as follows:

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

We are therefore led to the conclusion in the statement.

■

As before with the real Gaussian laws, a better-looking statement is in terms of partitions. Given a colored integer [math]k=\circ\bullet\bullet\circ\ldots\,[/math], we say that a pairing [math]\pi\in P_2(k)[/math] is matching when it pairs [math]\circ-\bullet[/math] symbols. With this convention, we have the following result:

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 is a reformulation of Theorem 1.25. Indeed, we can assume that we are in the case [math]t=1[/math], and here we know from Theorem 1.25 that the moments are:

[[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] are given by exactly the same formula. Indeed, in order to have a matching pairing 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=|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 pairings. Thus, we have the same formula as for the moments of [math]f[/math], and we are led to the conclusion in the statement.

■

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]] 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, where [math]\pi\leq\ker i[/math] means that the indices of [math]i[/math] must fit into the blocks of [math]\pi[/math], in the obvious way.

Show Proof

This is something well-known, which can be proved 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]] E(f^k)=t^{|k|/2}|\mathcal P_2(k)| [[/math]]

But this is the formula in Theorem 1.26, 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. We will be back to this in chapter 6 below, when discussing applications of the Wick formula. Getting back now to geometric aspects, in the spirit for what we did in the real case, we have:

Theorem

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

[[math]] \int_{S^{N-1}_\mathbb C}|z_1|^{2k_1}\ldots|z_N|^{2k_N}\,dz=\frac{(N-1)!k_1!\ldots k_n!}{(N+\sum k_i-1)!} [[/math]]

valid for any exponents [math]k_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_{2k-1}}\bar{z}_{i_{2k}}\,dz [[/math]]

By using transformations of type [math]p\to\lambda p[/math] with [math]|\lambda|=1[/math], we see that this integral [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. So, assume now that we are in the non-vanishing case. Then the [math]k_a[/math] copies of [math]z_a[/math] and the [math]k_a[/math] copies of [math]\bar{z}_a[/math] produce by multiplication a factor [math]|z_a|^{2k_a}[/math], so we have:

[[math]] I=\int_{S^{N-1}_\mathbb C}|z_1|^{2k_1}\ldots|z_N|^{2k_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)^{k_1}\ldots(x_N^2+y_N^2)^{k_N}\,d(x,y)\\ &=&\sum_{r_1\ldots r_N}\binom{k_1}{r_1}\ldots\binom{k_N}{r_N}\int_{S^{2N-1}_\mathbb R}x_1^{2k_1-2r_1}y_1^{2r_1}\ldots x_N^{2k_N-2r_N}y_N^{2r_N}\,d(x,y) \end{eqnarray*} [[/math]]

By using the formula in Theorem 1.18, we obtain:

[[math]] \begin{eqnarray*} I &=&\sum_{r_1\ldots r_N}\binom{k_1}{r_1}\ldots\binom{k_N}{r_N}\frac{(2N-1)!!(2r_1)!!\ldots(2r_N)!!(2k_1-2r_1)!!\ldots (2k_N-2r_N)!!}{(2N+2\sum k_i-1)!!}\\ &=&\sum_{r_1\ldots r_N}\binom{k_1}{r_1}\ldots\binom{k_N}{r_N}\frac{2^{N-1}(N-1)!\prod(2r_i)!/(2^{r_i}r_i!)\prod(2k_i-2r_i)!/(2^{k_i-r_i}(k_i-r_i)!)}{2^{N+\sum k_i-1}(N+\sum k_i-1)!}\\ &=&\sum_{r_1\ldots r_N}\binom{k_1}{r_1}\ldots\binom{k_N}{r_N} \frac{(N-1)!(2r_1)!\ldots (2r_N)!(2k_1-2r_1)!\ldots (2k_N-2r_N)!}{4^{\sum k_i}(N+\sum k_i-1)!r_1!\ldots r_N!(k_1-r_1)!\ldots (k_N-r_N)!} \end{eqnarray*} [[/math]]

Now observe that can rewrite this quantity in the following way:

[[math]] \begin{eqnarray*} I &=&\sum_{r_1\ldots r_N}\frac{k_1!\ldots k_N!(N-1)!(2r_1)!\ldots (2r_N)!(2k_1-2r_1)!\ldots (2k_N-2r_N)!}{4^{\sum k_i}(N+\sum k_i-1)!(r_1!\ldots r_N!(k_1-r_1)!\ldots (k_N-r_N)!)^2}\\ &=&\sum_{r_1}\binom{2r_1}{r_1}\binom{2k_1-2r_1}{k_1-r_1}\ldots\sum_{r_N}\binom{2r_N}{r_N}\binom{2k_N-2r_N}{k_N-r_N}\frac{(N-1)!k_1!\ldots k_N!}{4^{\sum k_i}(N+\sum k_i-1)!}\\ &=&4^{k_1}\times\ldots\times 4^{k_N}\times\frac{(N-1)!k_1!\ldots k_N!}{4^{\sum k_i}(N+\sum k_i-1)!}\\ &=&\frac{(N-1)!k_1!\ldots k_N!}{(N+\sum k_i-1)!} \end{eqnarray*} [[/math]]

Here we have used the following well-known identity, whose proof is standard:

[[math]] \sum_r\binom{2r}{r}\binom{2k-2r}{k-r}=4^k [[/math]]

Thus, we obtain the formula in the statement.

■

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

Theorem

The rescalings [math]\sqrt{N}z_i[/math] of the unit complex sphere coordinates

[[math]] z_i:S^{N-1}_\mathbb C\to\mathbb C [[/math]]

as well as the rescalings [math]\sqrt{N}U_{ij}[/math] of the unitary group coordinates

[[math]] U_{ij}:U_N\to\mathbb C [[/math]]

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

Show Proof

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

(1) According to the formula in Theorem 1.28, the polynomials integrals in [math]z_i,\bar{z}_i[/math] vanish, unless the number of [math]z_i,\bar{z}_i[/math] is the same. In this latter case these terms can be grouped together, by using [math]z_i\bar{z}_i=|z_i|^2[/math], and the relevant integration formula is:

[[math]] \int_{S^{N-1}_\mathbb C}|z_i|^{2k}\,dz=\frac{(N-1)!k!}{(N+k-1)!} [[/math]]

Now with [math]N\to\infty[/math], we obtain from this the following estimate:

[[math]] \int_{S^{N-1}_\mathbb C}|z_i|^{2k}dx \simeq N^{-k}\times k! [[/math]]

Thus, the rescaled variables [math]\sqrt{N}z_i[/math] become normal with [math]N\to\infty[/math], as claimed.

(2) As for the proof of the asymptotic independence, this is standard too, again by using Theorem 1.28. Indeed, the joint moments of [math]z_1,\ldots,z_N[/math] are given by:

[[math]] \begin{eqnarray*} \int_{S^{N-1}_\mathbb R}|z_1|^{2k_1}\ldots|z_N|^{2k_N}\,dx &=&\frac{(N-1)!k_1!\ldots k_n!}{(N+\sum k_i-1)!}\\ &\simeq&N^{-\Sigma k_i}\times k_1!\ldots k_N! \end{eqnarray*} [[/math]]

By rescaling, the joint moments of the variables [math]y_i=\sqrt{N}z_i[/math] are given by:

[[math]] \int_{S^{N-1}_\mathbb R}|y_1|^{2k_1}\ldots|y_N|^{2k_N}\,dx\simeq k_1!\ldots k_N! [[/math]]

Thus, we have multiplicativity, and so independence with [math]N\to\infty[/math], as claimed.

(3) Regarding the last assertion, we can use the basic fact that the rotations [math]U\in U_N[/math] act on the points of the sphere [math]z\in S^{N-1}_\mathbb C[/math], with the stabilizer of [math]z=(1,0,\ldots,0)[/math] being the subgroup [math]U_{N-1}\subset U_N[/math]. In algebraic terms, this gives an identification as follows:

[[math]] S^{N-1}_\mathbb C=U_N/U_{N-1} [[/math]]

In functional analytic terms, this result provides us with an embedding as follows, for any [math]i[/math], which makes correspond the respective integration functionals:

[[math]] C(S^{N-1}_\mathbb C)\subset C(U_N)\quad,\quad x_i\to U_{1i} [[/math]]

With this identification made, the result follows from (1,2).

■

As already mentioned in the real context, it is possible to get beyond such results, by using advanced group theory. We will be back to this, in chapter 4 below. It is also possible to formulate “free versions” of all the above, and we will do this later.

So long for the basics of probability theory, quickly explained. For further theory, the best is to go to a dedicated probability book, such as the one of Feller , or Durrett . Alternatively, you can learn good probability theory from the preliminary chapters of more specialized probability books, and with the comment here that, among probabilists, the random matrix people know well their job, and are very close to what we will be doing in this book. Well-known introductions to random matrices include the classical and delightful book by Mehta , the more modern book by Anderson, Guionnet and Zeitouni , the books by Bose , Mingo and Speicher , and many more.

Needless to say, you can also learn reliable probability theory from physicists, or other scientists. In fact, probability theory was fully 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.

General references

Banica, Teo (2024). "Calculus and applications". arXiv:2401.00911 [math.CO].