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I'm having some trouble on how to tackle the following problem

$X_1$ is a random variable with probability density $f(x)$ in the range $[0,1]$. A value of $X_1$ is picked, call its value $p$. A coin is played $n$ times with a probability $p$ to come up heads in each time. Calculate the expected value of the number of $k$ heads in the $n$ plays in the following cases:

  1. Each coin toss is independent and the $p$ value is the same for all of them. Find an expression for $E [k]$ in the case of a general $f(x)$

  2. Find $E[k]$ if $f(x)$ is uniform over $[0,1]$

  3. Find $E[k]$ if $f(x)$ is uniform over $[0,1]$ and a new $p$ value is picked before each coin flip.

I'm not sure I'm interpreting this correctly and honestly I think it's a little bit confusing.

If $p$ is fixed, the PMF would be the binomial distribution. In the case of a general $f(x)$, I assume I've to first derive a posterior distribution for $X_1$, where $f(x)$ is the prior. I'd proceed by finding the likelilhood based on the information that the coin was flipped $n$ times with a probability $p$ to come up heads. Then I could find the distribution for the next $m$ plays and calculate the expected value for this case. Here starts the trouble for me - it's asking for the expected value of the same $n$ plays I'm using to construct the likelihood. Because of that I'm not sure my approach is correct.

Also, I would appreciate some insight in the case where $X_1$ is picked before each coin flip.

Thanks.

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  • $\begingroup$ So in (1), from the binomial distribution, the expectation is $E[k]=E[np]=nE[X_1]$. Similarly for (2), though you can actually find $E[X_1]$ $\endgroup$
    – Henry
    Jun 4, 2020 at 14:13
  • $\begingroup$ Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. $\endgroup$ Jun 4, 2020 at 14:16
  • $\begingroup$ There is nothing Bayesian about this. Hint: 2 & 3 are examples of compound-distributions, which are somewhat confusingly sometimes called "mixtures", so you may have seen them under that name. Does that help? $\endgroup$ Jun 4, 2020 at 14:18
  • $\begingroup$ @StephanKolassa This definitely helps. Just to make things clear, I can derive an expression for the distribution of k by marginalizing over the binomial distribution and f(p), is that correct? But what happens when I have to deal with multiple parameters as in 3 (which I presume is the case because there's no guarantee p is the same each time I flip a coin) ? Is the mathematical generalization just a multiple integral? $\endgroup$ Jun 4, 2020 at 22:44

1 Answer 1

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I think this is like a conditional probability and marginal probability problem. First, for RV $X_1$, we have pdf $f(x)$ and range [0,1]. Then given '$X_1$' we have $k$ obeying a binomial distribution with pmd $g(k/p)=\binom nk p^k(1-p)^{n-k}$.

Now $E(k)=E(E(k|X_1))=nE(X_1)$

$E(X_1)=\int_{0}^{1}xf(x)dx$

$E(k)=n\int_{0}^{1}xf(x)dx$

When $f(x)$ is uniform. $E(k)=n\int_{0}^{1}x\frac{1}{1}dx=n(\frac{1}{2}-0)=\frac{n}{2}$

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