Could anyone help me with this problem, please? I have to calculate the expected value of the geometric distribution using the law of total expectation:
$$E(X) = E[E(X | Y )]$$
I think I have to consider $X$ being a random variable with geometric distribution and I know how to operate with this law, but the problem doesn't specify what I should consider $Y$ to be.
Thanks in advance and sorry for any mistakes in grammar, English is not my first language.
Edit: This problem comes from a lecture of my university in Probability II. My professor proposed this exercise saying that we should discover $Y$ by ourselves and then do the math to reach $1/p$ (the expected value of geometric distribution). Unfortunately, he didn't provide anymore context or tips :(
Edit2: What I did so far:
$ E[X/Y] =\\ \sum_{y=0}^1 \sum_{x=1}^\infty \ x\ P[X=x/Y=y]\ P[Y=y] = \\ \sum_{x=2}^\infty \ x\ P[X=x]\ +\ P[X=1] $
Hint: if $Y$ is the random variable, defined on the same probability space, as $1$ if $X$ succeeds on the first 'go' and as zero otherwise, you can determine $\Bbb E[X|Y]$ in terms of $\Bbb E[X]$. Assume $0<p<1$ (the $p=0$ case is forbidden and $p=1$ is trivial).
Motivation: this was just the simplest relevant thing I could think of. I think choosing $Y$ to be $1$ if $X$ succeeds within $n$ turns and $0$ otherwise would also work, but $n=1$ is by far the easiest to handle.
From your explicit calculation of $\Bbb E[X|Y]$, conclude algebraically that $\Bbb E[X]=1/p$.
Note: I think phrasing this in terms of the law of total expectation is silly. We could turn this argument into a direct and elegant proof that doesn't require the intermediary $Y$.