If $Y \sim geometric(P)$ and $P \sim \mathcal B(2, 1)$ how to compute $E(Y)$ and marginal pmf of $Y$?

Question

$$Y \sim Geometric(P)\\ P \sim \mathcal B(2, 1)$$

I'm trying to compute $E[Y]$ without finding marginal distribution of $Y$. I need some hints here. I also need to find the pmf of $Y$. My approach is as follows:

Integrate from 0, 1 over $p: f(p)(1-f(p))^{(y-1)}$ where $f(p)$ is the pdf of $\mathcal B(a, b)$. This leads to some messy algebra. Am I on the right track?

For the expected value of $Y$, use the [Law of Total Expectation](http://en.wikipedia.org/wiki/Law_of_total_expectation). — assumednormal, Sep 26 '12 at 16:30
how does the law of total expectation apply in a context where there are no conditionals? — fl4nders, Sep 26 '12 at 17:03
@fl4nders, Zen has answered your question. It's important to be precise about which random variable follows a given distribution. $Y|P=p$ is random and follows a Geometric distribution with parameter $p$. $Y$ is also a random variable but the problem has not specified its distribution. — assumednormal, Sep 26 '12 at 18:23
@Max right--this is exactly how the problem was given, in its confusing state. thanks for suggestions. — fl4nders, Sep 26 '12 at 18:44

score 5 · Answer 1 · answered Sep 26 '12 at 17:47

About your comment:

"How does the law of total expectation apply in a context where there are no conditionals?".

You don't see the conditional distribution because a more precise statement of the problem is $$Y\mid P=p\sim \textrm{Geo}(p) \qquad \textrm{and} \qquad P\sim \textrm{Be}(2,1) \, .$$

Now that you see the conditional distribution, just do the integral: $$\textrm{E}[Y]=\int_0^1 \textrm{E}[Y\mid P=p]\,f(p)\,dp \, .$$

Similarly for the pmf: $$p(y)=\int_0^1 \textrm{Pr}(Y=y\mid P=p)\,f(p)\,dp \, .$$

This is all that you need.

If $Y \sim geometric(P)$ and $P \sim \mathcal B(2, 1)$ how to compute $E(Y)$ and marginal pmf of $Y$?

1 Answers1