# Sufficient Statistic and MLE

Suppose $$X_1, \dots, X_n \sim B(1,p)$$. Show that a sufficient statistic for $$\theta = (1-p)^2$$ is $$T(x) = \sum X_i$$ and that the MLE for $$\theta$$ is $$(1-\frac{1}{n}T)^2$$.

I am having a lot of difficulty reproducing this seemingly simple result.

It is not clear to me whether the likelihood $$f(x_1,\dots,x_n|p)$$ is a product of $$n$$ Bernoulli trials (i.e. order matters) or whether it should be a binomial distribution and we just care about the results. Given all of the info in the question I opted for the former and said

$$f(x_1, \dots, x_n) = \Pi_i p^{x_i} (1-p)^{1-x_i} = p^{\sum x_i} (1-p)^{n-\sum x_i}$$

For part 1:

I tried factorising this to pull out a factor of $$(1-p)^2$$ but wasn't sure how to identify $$T(x)$$ from the remaining stuff. Instead, since I am not being asked to find $$T(x)$$ but rather just confirm what it is, I decided to try to should $$P(X=x|T=t)$$ was independent of $$\theta$$. We have

$$P(X_1=x_1, \dots, X_n=x_n|\sum X_i=t) = \frac{P(X_1=x_1, \dots, X_n=x_n \text{ and } \sum X_i=t) }{P(\sum X_i=t) } = \frac{1}{\binom{n}{t}}$$ which is independent of $$\theta$$.

Can somebody please confirm this is ok and also let me know if it is possible via factorisation?

For part 2:

I tried differentiation of the log likelihood using the chain rule:

$$\log{f} = \sum x_i \log{p} + (n-\sum x_i) \log{(1-p)}$$

$$\frac{d \log{f}}{d \theta} = \frac{d \log{f}}{d p} \frac{dp}{d \theta} = \left( \frac{\sum x_i}{p} - \frac{n-\sum x_i}{1-p}\right) \left(1 - \frac{1}{2 \sqrt{\theta}} \right)$$

However the solution to this is $$\theta = \frac{1}{4}$$ which is clearly incorrect and is making me question all of the above work? Would appreciate some help on this.

Thanks.

• Add the 'self-study' tag. Feb 27 '19 at 15:01
• A sufficient statistic for $p$ is also a sufficient statistic for any function of $p$. For part 2, find MLE of $p$ and hence conclude about MLE of $(1-p)^2$. Feb 27 '19 at 15:04
• @StubbornAtom Thank you - I got it now. To follow up on part 1, I completed it by using the factorisation theorem to demonstrate $\sum x_i$ is sufficient for $\theta$. However, would my method of showing the conditional probability is independent of $\theta$ also work? Or would you advise that I always just use factorisation to find a sufficient statistic due to the conditional probability being typically difficult to evaluate? Feb 28 '19 at 11:52