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. 🙂

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