Conditional expectation when more than certain number of heads are observed out of 'n' trials

Question

Suppose a coin is tossed $n$ times and it is then observed that the numbers of head (the success event) appeared is at least $k<n$. I'm interested in knowing the conditional expectation of no. of heads. If $h$ denotes the no. of heads then what is $E[h|h\geq k]$? For this, we need to know the conditional pmf of $h$. I have two thoughts here:

(i) Conjecturing $\gamma$ as the success probability, calculate the conditional pdf. Here I have the following expression: \begin{equation} E[h|h\geq k]= \bigg(\frac{1}{p(h\geq k)}\bigg) \big[k.p(h=k)+(k+1).p(h=k+1)+ ....+n.p(h=n)\big] \end{equation} The relevant probabilities involved are nothing but binomial probabilities and are calculated using the conjecture $\gamma$. Does this result above have a simpler expression?

(ii) Use MLE method to infer the probability of success first and then use this inferred value to perform (i). Now, I looked hard, I could not find an MLE for this. I know for an observed event $(h=k)$, the MLE is $\hat{\gamma}=k/n$ but couldn't find the same for the event $(h \geq k)$.

Please help.

Answer 1

You need to specify a prior for the "p". For binomial distribution, the conjugate and popular prior for "p" is beta distribution. You can then apply Bayes theorem to get the posterior distribution of p, given n and k, and also conditional expectation.