Answer to Question #280366 in Statistics and Probability for Sana

Question

 Let 𝑋1, … , 𝑋𝑛 be a random sample from the Bernoulli distribution, 

say P[X=1] = θ=1-P[X=0]. (a) Find the CRLB for the variance of unbiased estimators of 

θ(1-θ). (b) Find the UMVUE of θ(1-θ) if such exists.

Answer 1

$a)$

For a Bernoulli random variable,

$f(x,\theta)=\theta^x(1-\theta)^{1-x},\space x=0,1$

To find the CRLB, we proceed as follows.

$lnf(x,\theta)=ln(\theta^x(1-\theta)^{1-x})=xln\theta+(1-x)ln(1-\theta)$

${\delta\over \delta\theta}(lnf(x,\theta))={x\over\theta}-{(1-x)\over (1-\theta)}$

$E({\delta\over \delta\theta}(lnf(x,\theta)))^2=E({x\over\theta}-{(1-x)\over (1-\theta)})^2={1\over(\theta(1-\theta))^2}E(x-\theta)^2={1\over\theta(1-\theta)}$

Now,

$\Tau(\theta)=\theta(1-\theta)$and $\Tau'(\theta)=1-2\theta$

Therefore,

$var(\Tau(\theta))=var(\theta(1-\theta))={(\Tau'(\theta))^2\over n\times E({\delta\over \delta\theta}(lnf(x,\theta)))^2 }={(1-2\theta)^2\over n\times{1\over(\theta(1-\theta))}}={(1-2\theta)^2\times \theta(1-\theta)\over n}={\theta\over n}(1-5\theta+8\theta^2-4\theta^3)$

Thus, the CRLB is given by,

$var(\theta(1-\theta))\ge{\theta\over n}(1-5\theta+8\theta^2-4\theta^3)$

$b)$

Let $T=X_1+X_2+...+X_n$. Therefore, $T=X_1+X_2+...+X_n$ is a complete sufficient statistic . By the Lehmann-Scheffe theorem, if we can find a function of $T$ whose expectation is $\theta(1-\theta)$, it is an UMVUE.

In any set up, the sample variance ${1\over(n-1)}\sum(X_i-\bar{X})^2$, is an unbiased estimate of the variance. Since $X^2=X$ for Bernoulli random variables,

${1\over(n-1)}\sum(X_i-\bar{X})^2={1\over(n-1)}(\sum X_i^2-n\bar{X}^2)$

$={1\over(n-1)}(\sum X_i-n\bar{X}^2)$

$={T(n-T)\over n(n-1)}$

Hence, ${T(n-T)\over n(n-1)}$ is an UMVUE for the variance, $\theta(1-\theta)$