# $X_i$ follows Bernoulli distribution find UMVUE of $\theta(1-\theta)$

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?

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.

• If $T \sim Bin(n, \theta)$ then T is random variable right ? – Daman deep Jan 25 at 10:24
• Of course it is. – StubbornAtom Jan 25 at 10:26
• 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 . – Daman deep Jan 25 at 10:32
• $\bar T=\frac{\sum_{i=1}^{n}X_i}{n}$ sorry I didn't specify that . – Daman deep Jan 25 at 10:36
• 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). – StubbornAtom Jan 25 at 10:47