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The moment generating functions of two independent variables $X$ and $Y$ are $M_X(t)=\exp(2e^t-2)$ and $M_Y(t)=\left(\frac34e^t+\frac14\right)^{10}$. What are (a) $P(X+Y=2)$; (b) $P(XY=0)$; (c) $E[XY]$?

For (a), I did it in two ways that yield different answers.

First method: $M_{X+Y}(t)=\exp(2e^t-2)\left(\frac34e^t+\frac14\right)^{10}$, so $$P(X+Y=2)=\frac{d^2}{d(e^t)^2}M_{X+Y}(t)\big|_{e^t=0}=\frac{467}{524288e^2}.$$

Second method: $X$ is Poisson with parameter $2$ and $Y$ is Binomial with parameters $(10,\frac34)$. So $$P(X+Y=2)=\sum_{i=0}^2e^{-2}\frac{2^i}{i!}{10\choose 2-i}\left(\frac34\right)^{2-i}\left(\frac14\right)^{8+i}=\frac{467}{1048576e^2}.$$

Where did I go wrong?

For (b) and (c), can you check my solutions?

(b) $P(XY=0)=P(X=0)+P(Y=0)-P(X=0,Y=0)=e^{-2}+\frac1{4^{10}}-\frac{e^{-2}}{4^{10}}$

(c) $E[XY]=E[X]E[Y]=2\left(10\cdot\frac34\right)=15$

Thanks in advance.

  • 0
    You can calculate the moments of a random variable by evaluating the appropriate order derivative of the generating function at 0, but I do not think you can obtain the probability directly this way. Also, why are you evaluating at $e^{t}=0$ and not $t=0$? Think about the statement $e^{t}=0$. Maybe that is a typo. For part (a), consider directly calculating the inverse Fourier transform of the product.2012-04-09
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    I think I read somewhere something similar to the first method, but never mind if it's wrong. I don't know what is a Fourier transform, but is my second method correct?2012-04-09
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    The first method to solve (a) is a mangled version of the fact that, for every nonnegative integer valued random variable $Z$, $$P(Z=2)=\frac12g''_Z(0)$$ where $g_Z$ denotes the **simple** generating function of $Z$, defined as $$g_Z(s)=E(s^Z)=\sum_{n=0}^\infty P(Z=n)s^n$$ This remark explains where the strange formula involving $e^t=0$ in your post comes from, since $$g_Z(e^t)=M_Z(t)$$ It also explains why your numerical value is missing a factor $2$.2017-09-18

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