# Conditional Expectations of Binomial Probability (Log-concave?)

I am stuck on the following problem from my research. A version of the problem is as follows.

We observe a binomial random variable X with n=2 and an unknown probability $\theta$ of success on each trial (so X is the number of successes in two trials). Our prior beliefs about p are given by some continuous distribution F with density f.

I am trying to prove or find a counter-example to the following conjecture:

$\mathbb{E}[\theta | X=1]^2 \geq \mathbb{E}[\theta | X=0] \mathbb{E}[\theta | X=2].$

I cannot seem to prove this but I also can't find a counterexample. It works for:

• a uniform prior: $\frac{1}{4} > \frac{1}{4}\frac{3}{4}$
• A Beta prior: $\frac{(\alpha +1)^2}{(\alpha +\beta +2)^2}-\frac{\alpha (\alpha +2)}{(\alpha +\beta +2)^2} = \frac{1}{(\alpha +\beta +2)^2} > 0$
• Every density with support on [0, 1] that I could make up

Some approaches that have not worked so far for proving the statement to be true:

• I can express all three conditional probabilities as functions of unconditional expectations and the inequality becomes $$\frac{(\mathbb{E}[\theta^2] - \mathbb{E}[\theta^3])^2}{(\mathbb{E}[\theta] - \mathbb{E}[\theta^2])^2} \geq \frac{\mathbb{E}[\theta^3] - 2\mathbb{E}[\theta^2] + \mathbb{E}[\theta]}{\mathbb{E}[\theta^2] - 2\mathbb{E}[\theta] + 1}\frac{\mathbb{E}[\theta^3]}{\mathbb{E}[\theta^2]}.$$ My idea was that some applications of Jensen's inequality would give me more purchase on this statement, which might be true but I certainly haven't gotten there.
• I have also tried working from the fact that $\mathbb{E}[\theta]$ is a probability-weighted average of the three terms in my conjecture.

Edit: It turns out another way of stating what I want to prove is that the conditional expectations of p given X are log-concave w.r.t. X. The binomial distribution is log-concave but I am not sure that implies the conditional expectations are log concave given any prior. I added that to the title to help people out.