# Exam P Practice Problem 5

Problem 5a
Suppose that $X$ and $Y$ are independent random variables with the following moment generating functions:

$\displaystyle M_X(t)=\frac{4}{9}+\frac{4}{9}e^t+\frac{1}{9}e^{2t} \ \ \ \ \ \ \ \ \ \ \ \ \ \ M_Y(t)=e^{\displaystyle -3+3e^t}$

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Find the probability $\displaystyle P(\frac{Y}{2}=X)$.

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Problem 5b
Suppose that $X$ and $Y$ are independent random variables with the following moment generating functions:

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$\displaystyle M_X(t)=\frac{1}{3-2e^t} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M_Y(t)=\biggl(\frac{1+3e^t}{4}\biggr)^2$

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Find the probability $P(Y=X-2)$.

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Solution is found below.

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Solution to Problem 5a
The mgf $M_X(t)$ is that of a binomial distribution with $n=2$ and $p=\frac{1}{3}$. The mgf $M_Y(t)$ is that of a Poisson distribution with parameter $\lambda=3$.

\displaystyle \begin{aligned}P(Y=2X)&=P(X=0,Y=0)+P(X=1,Y=2)+P(X=2,Y=4) \\&=\biggl(\frac{2}{3}\biggr)^2 \ e^{-3}+2 \biggl(\frac{1}{3}\biggr) \biggl(\frac{2}{3}\biggr) \ e^{-3} \ \frac{3^2}{2}+\biggl(\frac{1}{3}\biggr)^2 \ e^{-3} \ \frac{3^4}{4!} \\&=\frac{203}{72} \ e^{-3} \\&=0.1403718733 \end{aligned}

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Answer to Problem 5b

$\displaystyle \frac{1}{12}$
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### 5 responses

1. Part B — Help! I cannot figure out how to get the probability distribution for X. A hint would be very appreciated!

1. The skill to practice here is to recognize moment generating functions of some familiar distributions. For 5-B, one mgf is that of a geometric distribution. Hope this helps.

1. I still can’t see it. I see that Y is a discrete distribution once one expands the square. But X…? I assume this is the one you are saying is a geometric. My form for geometric is p/(1-qe^t) (This is using the E[x] = p/q form of the geometric.) The 3 in the denominator is what is throwing me off… Am I missing some algebra step that gets it to the desired form?

2. The mgf $\displaystyle M_X(t)=\frac{1}{3-2e^t}$ can be transformed as:

$\displaystyle M_X(t)=\frac{\frac{1}{3}}{1-\frac{2}{3}e^t}$.

3. Ok, I got it now… I didn’t see the algebra on the geometric distribution (as I had guessed was my problem). Thanks for the hint… and the great blog. 🙂