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Let $X_1,X_2,X_3 ...X_n$ be a random sample from Bernoulli distribution with parameter $\theta$.Find UMVUE of $\theta(1-\theta)$.

I know that $T=\sum_{i=1}^{n}X_i$ is complete sufficient statistic for our paramenter $\theta$. I am trying to find out function of T which is a unique unbiased estimator of $\theta(1-\theta)$. Now $T\sim Bin(n,\theta)$

$E(T^2)-E(T)^2=V(T)$

$n\theta(1-\theta)+n^2\theta^2-n^2\theta^2=n\theta(1-\theta)$

$\implies\dfrac{1}{n}(T^2-\bar{T}^2)$ is umvue of $\theta(1-\theta)$

If I have sample size $n=10$ with observations $1,1,1,1,1,0,0,0,0,0$ obtain the value of this estimator.

Now I am stuck at this point that is $T^2$ is $T^2=\sum_{i=1}^{n}X_i^2$ or $T^2=(\sum_{i=1}^{n}X_i)^2$. Can someone help me and tell at what point I am doing things wrongly?

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An unbiased estimator of $\theta$ is $\frac{T}{n} $ where $T=\sum\limits_{k=1}^n X_k$.

From your approach that $\operatorname{Var}_{\theta}(T)=\mathrm E_{\theta}(T^2)-(\mathrm E_{\theta}(T))^2=n\theta(1-\theta)$, it follows that an unbiased estimator of $\theta^2$ is $\frac{T(T-1)}{n(n-1)}$. You do not get unbiased estimator of $\theta(1-\theta)$ directly from this step.

An unbiased estimator of $\theta(1-\theta)$ is therefore $\frac{T}{n}-\frac{T(T-1)}{n(n-1)}$, which is also the UMVUE.

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  • $\begingroup$ If $T \sim Bin(n, \theta)$ then T is random variable right ? $\endgroup$ – Daman deep Jan 25 at 10:24
  • $\begingroup$ Of course it is. $\endgroup$ – StubbornAtom Jan 25 at 10:26
  • $\begingroup$ estimator I used $\dfrac{1}{n}(T^2-\bar{T}^2)$ if I take expectation I get same result how come mine is not umvue ? $\dfrac{E(T^2)-E(E(T)^2)}{n}=\dfrac{n\theta(1-\theta)+n^2\theta^2-n^2\theta^2}{n}=\theta(1-\theta)$ but umvue is unique. Can you tell me whats wrong ? I didn't get it . $\endgroup$ – Daman deep Jan 25 at 10:32
  • $\begingroup$ $\bar T=\frac{\sum_{i=1}^{n}X_i}{n}$ sorry I didn't specify that . $\endgroup$ – Daman deep Jan 25 at 10:36
  • $\begingroup$ No need to define things like $\bar T$; your notation is confusing. You appear to be doing something like $E(T^2)-(E(T))^2=n\theta(1-\theta)\implies\left[E(\frac{1}{n}(T^2-(E(T))^2))\right]=\theta(1-\theta)$, and hence concluding that $\frac{1}{n}(T^2-(E(T))^2)$ is UMVUE. This is not correct because $E(T)=n\theta$, so that your proposed estimator is not even a statistic (it depends on the parameter). $\endgroup$ – StubbornAtom Jan 25 at 10:47

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