Variance of a sequence of bernouli(p) trials where p is drawn from uniform distribution [0,1]

Question

A number p is drawn from the interval [0,1] according to the uniform distribution, and then a sequence of independent Bernoulli trials is performed, each with success probability p. What is the variance of the number of successes in k trials? Note k is a deterministic number.

In this question, I tried solving by calculating the expected value of p which is $\frac{1}{2}$ as p ~ U[0,1]. Then took a random variable M = No. of successes in k trials Since M ~ Binomial(k,p). Therefore, calculated the variance to be V(M) = $k\times{p}\times{(1-p)} = \frac{k}{4}$. The Answer is $ \frac{k\times{(k+2)}}{12}$. Not sure where I am going wrong.

Answer 1

This is an application of the Law of total variance: https://en.wikipedia.org/wiki/Law_of_total_variance

Let $P = Unif(0,1)$ and $Y = Binom(k,p)$

First let's note that:

• $E[P] = \frac{1}{2}$
• $Var[P] = \frac{1}{12}$
• $E[P^2] = Var[P] + E[P]^2 = \frac{1}{3}$

Then the law states that $Var(Y) = E[Var(Y|P)] + Var[E(Y|P)]$

• $Var(Y|X=p) = kp(1-p)$ Because Y is Binomial

  $E[Var(Y|X)] = k*E[p(1-p)]= k*(E[p]-E[p^2]) = k(\frac{1}{2}-\frac{1}{3}) = k\frac{1}{6}$

• $E(Y|X) = kp$ Because Y is Binomial

  $Var[E(Y|X)] = k^2*Var(p) = k^2\frac{1}{12}$

Finally:

$Var(Y) = k\frac{1}{6} + k^2\frac{1}{12}$

$Var(Y) = k(\frac{2}{12} + k\frac{1}{12})$

$Var(Y) = k(\frac{k+2}{12})$

Answer 2

This is a special case of the beta-binomial distribution, with $k$ trials and $\alpha=\beta=1$.