# Expectation of Sufficient Statistic

Consider $$X \sim B(n,p)$$ with pmf $$P(X=x) = {{n}\choose{x}} p^x (1-p)^{n-x}$$.

The general exponential form of an exponential family distribution is $$p(x|\theta) = f(x) g(\theta) e^{\phi(\theta)^T T(x)}$$. Writing the binomial distribution in this form, I get $$f(x) = {n \choose x}, g(p)=(1-p)^n, \phi(\theta)=\log{\frac{p}{1-p}}, T(x) = x$$.

In other words, $$x$$, the number of successes is a sufficient statistic for the probability of success on each Bernoulli trial, as is to be expected.

(1) the expected value of the sufficient statistic, $$_{p(x|\theta)}$$ (often called the "mean" or "moment" parameter of the distribution).

(2) the maximum likelihood value of the mean parameter in the above question

I have to be honest, I don't really understand the difference between the two questions. So far I have attempted this:

(1) $$\theta=p$$ so the likelihood $$P(x|p)$$ is the probability of getting p successes in n trials given a Bernoulli prob of p. i.e. $$P(x|p) = {n \choose x} p^x(1-p)^x$$.

Then $$ = \sum_x T(x)P(x|p) = \sum_x x {n \choose x} p^x(1-p)^x$$.

At this point, I'm not sure what to do.

(2) I know that for MLE I should find the maximum of $$log{L(p)}= log{P(x|p)} = \log{n \choose x} n+ n \log{1-p} + x \log{\frac{p}{1-p}}$$. If I differentiate and set to zero, I get that $$x = T(x) = np$$ which seems right as this is the expected value for $$X$$ following a binomial distribution.

Am I at all on the right lines here?

Hint #0: computing the expectation of a sufficient statistic (of which there is an infinity if there is one) is not equivalent or even related to finding the maximum likelihood of $$\theta$$ (which for one thing depends on the choice of parameterisation $$\theta$$)
Hint #1: check Wikipedia for the expectation of a Binomial $$\mathcal B(n,p)$$ random variable
Hint #2: solve the likelihood equation in $$p$$, not in $$x$$
• (1) compute the expectation of $X$ by identifying the distribution of $X$ (2) yes indeed – Xi'an Feb 15 '19 at 13:17
• So basically $E(X) = \sum_x x P(X=x) = \sum_x x {n \choose x} p^x (1-p)^{n-x} =... =np$? – user11128 Feb 15 '19 at 13:38
• I do not understand your question: two different functions $T$ have different expectations in most cases, but this is of no consequence as a statistic is not an estimator by default. – Xi'an Feb 15 '19 at 14:17