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#1
01-28-2007, 12:53 PM
 daaaave David Revelle Join Date: Feb 2006 Posts: 2,985
Dave Revelle's video problem solving session

I will be teaching an online seminar for Exam 1/P with the Infinite Actuary beginning with the August sitting of the exam. In addition, I will post problems here most weeks along with links to video solutions. The topics of these problems will correspond to the recommended study schedule for the seminar once it gets underway, but before that happens the topics will be fairly arbitrarily.

To start, here are some problems on moment generating functions. I've chosen this topic partly because it is relatively late on most people's study schedules and the February sitting is getting close. These questions don't cover everything on the subject, so I will revisit mgf's some time in the future. If anyone has requests for other topics, let me know.

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1. Suppose the moment generating function for X is given by
$M_X(t)=\frac{4}{(2-t)^2}$ for t<2. Find the variance of X.

Spoiler:
Var X=1/2. Intermediate steps may include $E X=1$ and $E\left(X^2\right)=3/2$.

**************************************************

2. Let $Y=2X-3$, where $X$ is a random variable whose moment generating function is
$M_X(t)=\exp\left[5e^t-5\right].$ Find $M_Y(t).$

Spoiler:
$M_Y(t)=\exp\left[5e^{2t}-5-3t\right]$

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3. Suppose X and Y are random variables whose joint density is given by
$f(x,y)=2e^{-x-y}$ for $0 and $f(x,y)=0$ otherwise. Find $M_{X,Y}(t,s)=E e^{tX+sY}$ for $s<1$ and $s+t<2$.

Spoiler:
$M_{X,Y}(t,s)=\frac{2}{1-s}\cdot \frac{1}{2-s-t}$

************************************************** **

Video solutions available at the Infinite Actuary
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Last edited by daaaave; 02-12-2007 at 07:43 AM..
#2
02-04-2007, 02:56 PM
 daaaave David Revelle Join Date: Feb 2006 Posts: 2,985

Here are this week's problems. Video solutions are available at The Infinite Actuary
.
1. Suppose that $X,Y,$ and $Z$ are random variables whose moment generating functions are given by
$\begin{eqnarray*} M_X(t)&=&\frac{1}{(1-3t)^2}\\ M_Y(t)&=&\frac{1}{(1-3t)^3}\\ M_Z(t)&=&\frac{1}{(1-3t)^4} \end{eqnarray*}$

Find the coefficient of variation of W=X+Y+Z.

Spoiler:
1/3

2. A random variable $X$ has a discrete distribution with $P[X=0]=1/2, P[X=1]=P[X=2]=1/4.$ Find the moment generating function of $X^2.$

Spoiler:
$\frac{1}{2} + \frac{e}{4} + \frac{e^4}{4}$

3. A bivariate random variable $(X_1,X_2)$ has joint moment generating function
$M_{X_1,X_2}(t_1,t_2)=\exp\left[t_1^2+t_2^2-t_1t_2\right].$ Find Cov(X_1,X_2).

Spoiler:
-1

Detailed video solutions are now available at The Infinite Actuary.
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Last edited by daaaave; 02-12-2007 at 07:43 AM..
#3
02-11-2007, 04:48 PM
 daaaave David Revelle Join Date: Feb 2006 Posts: 2,985

This week's problems are based on old practice material from before the 2005 syllabus change that I think is still relevant. As usual, numerical answers are here and video solutions will be available at The Infinite Actuary.

1. [S '01 #36] A town in the shape of a square with each side measuring 4 has an industrial plant at its center. The industrial plant is polluting the air such that the concentration of pollutants at each location $(x,y)$ in the town can be modeled by the function
$C(x,y)=22,500(8-x^2-y^2)$ for $-2 \leq x \leq 2$ and $-2 \leq y \leq 2$ .

Calculate the average pollution concentration over the entire town.

Note: the average concentration is the same thing as asking about the expected value of C(X,Y) if (X,Y) are uniformly chosen over the town.

Spoiler:
120,000

2. (#7 from 1999 practice exam for the old syllabus)

As part of the underwriting process for insurance, each prospective policyholder is tested for high blood pressure.

Let X represent the number of tests completed when the first person with high blood pressure is found. The expected value of X is 12.5.

Calculate the probability that the sixth person tested is the first one with high blood pressure.

Spoiler:

(11.5/12.5)^5 (1/12.5)=.053

3. (Based on #10 from 1999 practice exam for the old syllabus, modified to avoid an ambiguity in the problem.) An insurance policy covers the two employees of ABC Company. The policy will reimburse ABC
for no more than one loss per employee in a year. It reimburses the full amount of the loss up to an annual company-wide maximum of 8000.

The probability of an employee incurring a loss in a year is 40%. The probability that an employee incurs a loss is independent of the other employee’s losses.

The amount of each loss is uniformly distributed on [1000, 5000].

Given that at least one of the employees has incurred a loss in excess of 2000, determine the probability that losses will exceed reimbursements.

Spoiler:
2/51

Video Solutions are available at The Infinite Actuary We've changed the URL from previous weeks and also added QuickTime versions of the videos.
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Last edited by daaaave; 02-12-2007 at 07:42 AM..
#4
02-17-2007, 03:20 PM
 daaaave David Revelle Join Date: Feb 2006 Posts: 2,985

Here are this week's problems. The second problem is about covariance matrices because of the recent threads on the topic (see here and here), but while it is a legitimate topic, the fact is that some of the textbooks on the syllabus don't cover it and I think it is unlikely to come up. A variation of question #2 that is more likely to appear on the exam would be something in which you are told Var(X_i)=3 for all i, and Cov(X_i,X_j)=-2 for all $i \neq j$.

1. Let $N$ be a randomly chosen integer with $1\leq N \leq 1,000$. What is the probability that $N$ is not divisible
by 7, 11, or 13?

Spoiler:
0.720

2. The covariance matrix of $(X_1,X_2,X_3)$ is given by
$\left( \begin{array}{c c c} 2 & -1 & -2 \\ -1 & 3 & 2 \\ -2 & 2 & 4 \end{array} \right)$
If $E X_i=i$ for all $i$, find $E[(X+Y+Z)^2]$

Spoiler:
43

3. Let $U$ and $V$ be independent, continuous uniform random variables on the interval [1,5]. Find
$\Pr[\min\{U,V\} <2 \mid \max\{U,V\}>2 ]$.

Spoiler:
6/15

Video solutions are available at The Infinite Actuary
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Last edited by daaaave; 02-18-2007 at 02:22 PM..
#5
02-25-2007, 01:22 PM
 daaaave David Revelle Join Date: Feb 2006 Posts: 2,985

Since the February sitting of the exam was this past week, I am going to take a couple weeks off from posting practice problems so I can finish the practice exams that I am working on.
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#6
02-05-2008, 04:57 PM
 H.R. Paperstacks Member Join Date: Oct 2006 Location: Real America Studying for FM College: Morehouse Alumni Favorite beer: Samuel Jackson Posts: 14,308

Hey Daave,

On question 9c in Lesson 3 of Continuous distributions, couldn't that be taken as waiting exactly 45 minutes rather than at least 45 minutes. I nearly tried to do this as a Poisson until I thought about the section this was from.
#7
02-08-2009, 12:11 PM
 efh0888 Member Join Date: Jan 2009 Posts: 49

In general, how do you know when using the shortcut that the second derivative of the ln(Mx(t)) at t=0 equals V(x) is advantageous to finding the first two moments and plugging into V(x) = E(x^2) - [E(x)]^2?
#8
02-08-2009, 08:36 PM
 daaaave David Revelle Join Date: Feb 2006 Posts: 2,985

If M(t) is of the from e^(something), then it will usually be easier to work with ln(M(t)). For example, if $M(t)=e^{5t^2-3t}$ or $M(t)=e^{e^{5t}-5}$ then $\ln M(t)$ will be easier to work with.

It is also probably easier to use ln(M(t)) when you have M(t)=something^power, such as $M(t)=\frac{1}{(1-3t)^5}$ because ln(M(t)) will have a nice derivative to work with.

The main time when you might not want to work with ln(M(t)) is when M(t) is a sum of pieces. Logs work better with products than sums, and if we have something like $M(t)=\frac{1}{4}[1+e^t + e^{3t}+e^{8t}]$ then M'(t) and M''(t) will be nice, but ln(M(t)) and its derivatives will be messy.
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#9
02-09-2009, 10:18 AM
 efh0888 Member Join Date: Jan 2009 Posts: 49

Thanks Dave. I guess I was more worried about when not to use it as opposed to when to use it. So thank you for including that in your response.
#10
03-17-2009, 03:08 PM
 campbell Mary Pat Campbell SOA AAA Join Date: Nov 2003 Location: NY Studying for duolingo and coursera Favorite beer: Murphy's Irish Stout Posts: 84,209 Blog Entries: 6

Info on watching the videos on an iPhone or iPod Touch:
http://www.actuarialoutpost.com/actu...d.php?t=161231
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