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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.

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    $\begingroup$ Add the 'self-study' tag. $\endgroup$ Feb 27 '19 at 15:01
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    $\begingroup$ 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 $(1-p)^2$. $\endgroup$ Feb 27 '19 at 15:04
  • $\begingroup$ @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? $\endgroup$
    – user11128
    Feb 28 '19 at 11:52

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