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Let $X$ be a uniform random number in $0..n$. Its expected value is $n/2$.

Next, let $Y$ obey a binomial distribution with $X$ trials and success probability $p$. So now we have a distribution where one of the parameters is itself a random variable. I'm not sure how this is called, but it reminds me of composite functions.

Can I compute the expected value of $Y$ simply as $X\cdot p$, and subsitute $n/2$ for $X$, yielding an expected value of $np/2$ for $Y$? If not, how do we compute the expected value of a distribution that has a parameter that is a random variable itself?

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  • $\begingroup$ I suppose you want $X$ to be uniform on $1,\ldots,n$ because a binomial distribution with $0$ trials would just be the degenerate random variable $0$. Or are you okay with that? $\endgroup$ – Math1000 Nov 16 '19 at 2:28
  • $\begingroup$ By the way, what this is "called" is that $Y$, conditional on $X=n$ has $\mathrm{Bin}(n,p)$ distribution. With a slight abuse of notation, we can write $Y\mid X=n\sim\mathrm{Bin}(n,p)$. $\endgroup$ – Math1000 Nov 16 '19 at 2:31
  • $\begingroup$ @Math1000 I think 0 trials should be a valid input to the binomial distribution. Of course, in that case the random variable will always be zero, but that's okay. Anyway, it's just an example, and I hope an answer will be generically enough so that I can apply it to arbitrary interval bounds. :) $\endgroup$ – user1494080 Nov 16 '19 at 2:36
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    $\begingroup$ In that case, $$ \mathbb E[Y] = \sum_{k=0}^n \mathbb E[Y\mid X=k]\mathbb P(X=k) = \sum_{k=0}^n kp\cdot\frac1{n+1} = \frac{np}2, $$ as expected. I will write this as an answer as well. $\endgroup$ – Math1000 Nov 16 '19 at 2:40
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By the law of total expectation we compute $$ \mathbb E[Y] = \sum_{k=0}^n \mathbb E[Y\mid X=k]\mathbb P(X=k) = \sum_{k=0}^n kp\cdot\frac1{n+1} = \frac{np}2, $$ as expected.

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  • $\begingroup$ Thank you, just to be sure I understood the prinicple: Let $W$ be an additional uniform random number in $0..m$ and let $X$ now be uniform in $0..W$. Then the expected value of this chain of three distributions would be $\Sigma_{j=0}^m\Sigma_{k=0}^j E[Y | X = k]P(X=k | W = j)P(W=j) = pm/4$. Is that correct? $\endgroup$ – user1494080 Nov 16 '19 at 3:35
  • $\begingroup$ That should really be asked as a separate question, in my opinion. But I am not sure that your equation holds true. I have never seen such a question so I would have to defer to someone with more expertise, sorry. $\endgroup$ – Math1000 Nov 16 '19 at 4:01

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