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Jun 02'22

A loss variable [math]L[/math] has a density function that is proportional to

[[math]] x^{\alpha -1}e^{-x/\theta}. [[/math]]

The parameters [math]\theta [/math] and [math]\alpha [/math] are random with the following joint distribution

[math]\alpha = 1 [/math] [math]\alpha = 2[/math]
[math]\theta = 500 [/math] 0.25 0.35
[math]\theta = 1000 [/math] 0.15 0.25

Determine the standard deviation of [math]L[/math] to the nearest integer.

  • 298
  • 1,088
  • 1,220
  • 1,279
  • 1,565
Jun 02'22

The joint density function for the random variables [math]X,Y [/math] equals

[[math]] f_{X,Y}(x,y) = \begin{cases} cxy^3, y^2 \lt x \lt y, 0 \lt y \lt 1 \\ 0, \, \textrm{Otherwise} \end{cases} [[/math]]

for a constant [math]c[/math]. Determine the marginal density of [math]2Y^{1/2}[/math] given [math]X=1/2[/math].

  • [[math]] g(z)= \begin{cases} \frac{z^7}{6}, \sqrt{2} \lt z \lt 2^{3/4} \\ 0, \, \textrm{Otherwise} \end{cases} [[/math]]
  • [[math]] g(z)= \begin{cases} \frac{64z^3}{3}, \frac{1}{2} \lt z \lt \frac{1}{\sqrt{2}} \\ 0, \, \textrm{Otherwise} \end{cases} [[/math]]
  • [[math]] g(z)= \begin{cases} z^3, \sqrt{2} \lt z \lt 2^{3/4} \\ 0, \, \textrm{Otherwise} \end{cases} [[/math]]
  • [[math]] g(z)= \begin{cases} \frac{255z^7}{1688}, \frac{1}{2} \lt z \lt \frac{1}{\sqrt{2}} \\ 0, \, \textrm{Otherwise} \end{cases} [[/math]]
  • [[math]] g(z)= \begin{cases} \frac{2^{7/2}z^{5/2}}{5}, 0 \lt z \lt 2 \\ 0, \, \textrm{Otherwise} \end{cases} [[/math]]
Jun 02'22

You are given the following about a portfolio of risks:

  • Risks are classified into three classes: 10% belong to class A, 30% belong to class B and 60% belong to class C.
  • Losses for each risk are uniformly distributed on an interval [a,b] with a and b dependent on class:
Class a b
A 0 1,500
B 500 2,300
C 100 1,000

Determine the expected loss for a randomly selected risk given that the loss is greater than 1,000.

  • 1,065
  • 1,238
  • 1,313
  • 1,587
  • 1,597
Jun 02'22

You are given the following:

  • Claim frequency and claim size are independent
  • Monthly claim frequency is Poisson distributed with mean 3
  • The claim size distribution is uniform on [0,1000]

If [math]S[/math] is the annual loss, determine the variance of [math]S[/math]

  • 8,000,000
  • 10,500,000
  • 11,000,000
  • 12,000,000
  • 13,000,000
May 05'23

Let [math]X[/math] and [math]Y[/math] be continuous random variables with joint density function

[[math]] f(x,y) = \begin{cases} 24xy, \,\, 0 \lt x \lt 1 \,\, \textrm{and} \,\, 0 \lt y \lt 1-x \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Calculate [math] \operatorname{P}[Y \lt X | X = 1/3][/math]

  • 1/27
  • 2/27
  • 1/4
  • 1/3
  • 4/9

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

May 05'23

Once a fire is reported to a fire insurance company, the company makes an initial estimate, [math]X[/math], of the amount it will pay to the claimant for the fire loss. When the claim is finally settled, the company pays an amount, [math]Y[/math], to the claimant. The company has determined that [math]X[/math] and [math]Y[/math] have the joint density function

[[math]] f(x,y) = \begin{cases} \frac{2}{x^2(x-1)}y^{-(2x-1)/(x-1)}, \,\, x \gt1, y \gt 1 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Given that the initial claim estimated by the company is 2, calculate the probability that the final settlement amount is between 1 and 3.

  • 1/9
  • 2/9
  • 1/3
  • 2/3
  • 8/9

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

May 05'23

The stock prices of two companies at the end of any given year are modeled with random variables [math]X[/math] and [math]Y[/math] that follow a distribution with joint density function

[[math]] f(x,y) = \begin{cases} 2x, \,\, 0 \lt x \lt 1, x \lt y \lt x +1 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Determine the conditional variance of [math]Y[/math] given that [math]X = x[/math].

  • 1/12
  • 7/6
  • [math]x + 1/2 [/math]
  • [math]x^2 - 1/6[/math]
  • [math]x^2 +x + 1/3[/math]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

May 05'23

An actuary determines that the annual number of tornadoes in counties P and Q are jointly distributed as follows:

Q= 0 Q=1 Q=2 Q=3
P=0 0.12 0.06 0.05 0.02
P=1 0.13 0.15 0.12 0.03
P=2 0.05 0.15 0.10 0.02

Calculate the conditional variance of the annual number of tornadoes in county Q, given that there are no tornadoes in county P.

  • 0.51
  • 0.84
  • 0.88
  • 0.99
  • 1.76

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

May 05'23

You are given the following information about [math]N[/math], the annual number of claims for a randomly selected insured:

[[math]] \operatorname{P}[N = 0] = \frac{1}{2}, \, \operatorname{P}[N = 1] = \frac{1}{3}, \, \operatorname{P}[N \gt1] = \frac{1}{6} [[/math]]

Let [math]S[/math] denote the total annual claim amount for an insured. When [math]N = 1 [/math], [math]S[/math] is exponentially distributed with mean 5. When [math]N \gt 1 [/math], [math]S[/math] is exponentially distributed with mean 8.

Calculate [math]\operatorname{P}(4 \lt S \lt 8) [/math]

  • 0.04
  • 0.08
  • 0.12
  • 0.24
  • 0.25

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

May 06'23

The joint probability density for [math]X[/math] and [math]Y[/math] is

[[math]] f(x,y) = \begin{cases} 2e^{-(x + 2y)}, \,\, x \gt 0, y \gt 0\\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Calculate the variance of [math]Y[/math] given that [math]X \gt 3 [/math] and [math]Y \gt 3 [/math].

  • 0.25
  • 0.50
  • 1.00
  • 3.25
  • 3.50

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.