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So my question told me to find the $E[Y]$ and $V[Y]$ first and then find the PGF of $Y$ and state its distribution. Am I supposed to use the information from $E[Y]$ and $E[X]$ to derive the PGF or are they unrelated, because I am having trouble finding the PGF of $Y$.

Let $ X \sim exp(1) $ and $Y |X = x \sim Pn(x)$

From: $$E[E[Y|X=x]] = E[Y]$$ $$ E[X] = E[Y] $$ Since: $$E[X] = 1 $$ So I found that: $$E[Y] = E[X] = 1$$ and: $$V[Y] = E[V[Y|X]] +V[E[Y|X]]$$ $$V[Y] = E[X] + E[Y] = 2$$

And this is where I am stuck I don't know how to find the PGF of $Y$

YellowPillow
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1 Answers1

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MGF of $Y$ is ${\mathbb E} e^{tY}$. To find it, you may use the formula you cited,

$${\mathbb E} e^{tY} = \sum_x {\mathbb E} (e^{tY} | X=x) P(X=x) $$

EDIT: I notice that you actually wanted PGF, not MGF. Also, $X$ has a continuous distribution. So, the above formula modifies to

$${\mathbb E} z^Y = \int {\mathbb E} (z^Y | X=x) f(x) dx $$

PA6OTA
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