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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
    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

1 Answers 1

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Let $Z$ a discrete random variable, then

$M_Z(t) = \sum_k e^{k \cdot t} \cdot \mathbb{P}[Z=k] \\ \Rightarrow \frac{d}{dt}M_Z(t) = \sum_k k \cdot e^{k \cdot t} \cdot \mathbb{P}[Z=k] \\ \Rightarrow \frac{d}{dt} \left( \frac{d}{dt} M_Z(t)-M_Z(t) \right) = \sum_k k \cdot (k-1) \cdot e^{t \cdot k} \cdot \mathbb{P}[Z=k] \\ \Rightarrow \frac{d}{dt} \left( \frac{d}{dt} M_Z(t)-M_Z(t) \right) \bigg|_{t=0} = 2 \cdot 1 \cdot \mathbb{P}[Z=2]$

If you apply this to $Z:=X+Y$ you get the correct result. So it's possibly to use the moment generating function to calculate the probability, but it's not that easy as you thought.

(Your first method is wrong, since the expression

$\frac{d^2}{d(e^t)^2} M_{X+Y}(t) \bigg|_{e^t=0}$

doesn't make sense at all: $e^t>0$ for all $t \in \mathbb{R}$.)

And yes, your second method is correct.

  • 0
    The first method to solve (a) is referring to the simple generating function of $X+Y$, not its exponential generating function, hence it is a mangled version of an actual result, see my comment on main.2017-09-18