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#1
11-14-2017, 06:00 AM
 Nagsinde2002 SOA Join Date: May 2017 College: University of Wisconsin - Madison (1st year) Posts: 22
Joint PMFs

Hello. I have been struggling on this question.

Let X and Y be independent Geo(1/3) random variables. Let V =min(X,Y) and let W be defined as follows:
W = 0 if X< Y,
W=1 if X=Y
W=2 if X > Y.

Let p{V,W}(v,w) be the joint probability mass function of V and W. Compute p{V,W}(2,0).

I know that I have to make a pmf of V and W in terms of X and Y, but I am stuck on how to do so. Also, how would I use the geometric random variable pmf into this problem? If someone can get me started on this problem, that would be appreciated.
#2
11-14-2017, 10:09 AM
 ARodOmaha Member SOA Join Date: May 2016 Location: Omaha, NE College: University of Nebraska (alma mater) Favorite beer: Captain Morgan Posts: 200

Quote:
 Originally Posted by Nagsinde2002 Hello. I have been struggling on this question. Let X and Y be independent Geo(1/3) random variables. Let V =min(X,Y) and let W be defined as follows: W = 0 if X< Y, W=1 if X=Y W=2 if X > Y. Let p{V,W}(v,w) be the joint probability mass function of V and W. Compute p{V,W}(2,0). I know that I have to make a pmf of V and W in terms of X and Y, but I am stuck on how to do so. Also, how would I use the geometric random variable pmf into this problem? If someone can get me started on this problem, that would be appreciated.
I'll get you started and you can finish. Deciphering the last statement and working backwards, we need the probability that V=2 and W=0. If W=0 then we know X<Y. Also, since V=Min[X, Y], then we know that X=2 and Y>2.

To find the probability that X=2 and Y>2, that is the Pr(X=2)*Pr(Y>2). For Pr(X=2), plug into the Geometric distribution: (1-1/3)^(2-1) * (1/3) = 2/9. Now calculate Pr(Y>2) as the complement 1-P(Y<=2)....there are three options to consider. Multiply that result with 2/9.
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#3
11-14-2017, 01:43 PM
 Nagsinde2002 SOA Join Date: May 2017 College: University of Wisconsin - Madison (1st year) Posts: 22

Thanks for your help. I got the answer correct, which should be 8/81.
#4
11-20-2017, 10:12 AM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 361

The problem alone doesn't tell you whether geometric is counting failures or trials here. Your answer of $\frac{8}{81}$ is correct if you're counting trials. If you're counting failures it would be $\frac{32}{729}$.

Also just for fun and in case you find this useful (all true whether counting failures or trials):
The PMF of $W$ is
$
P(W=w)=\begin{cases}
\frac{2}{5}, & w\in\{0,2\}
\frac{1}{5}, & w=1
0, & \text{otherwise}
\end{cases}
$

$V$ is Geometric with parameter $\frac{5}{9}$

In general if $X$ and $Y$ are independent and geometric with parameters $p$ and $r$ respectively, then $\text{min}\{X,Y\}$ is geometric with parameter $1-(1-p)(1-r)$.

Last edited by Z3ta; 11-20-2017 at 10:28 AM..