2022-03-22

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asked 2022-04-22

Given you have an independent random sample $X}_{1},{X}_{2},\dots ,{X}_{n$ of a Bernoulli random variable with parameter p, estimate the variance of the maximum likelihood estimator of p using the Cramer-Rao lower bound for the variance

So, with large enough sample size, I know the population mean of the estimator $\hat{P}$ will be p, and the variance will be:

$Var\left[\hat{P}\right]=\frac{1}{nE\left[\right((\partial /\partial p)\mathrm{ln}\hspace{0.17em}{f}_{x}\left(X\right){)}^{2}]}$

Now I'm having some trouble calculating the variance of $\hat{P}$, this is what I have so far:

since the probability function of $\stackrel{\u2015}{X}$ is binomial, we have:

${f}_{x}\left(\overline{X}\right)=\left(\genfrac{}{}{0ex}{}{n}{\sum _{i=1}^{n}{X}_{i}}\right)*{p}^{\sum _{i=1}^{n}{X}_{i}}*(1-p{)}^{n-\sum _{i=1}^{n}{X}_{i}}$

so: $\mathrm{ln}\hspace{0.17em}{f}_{X}\left(X\right)=\mathrm{ln}\left(\left(\genfrac{}{}{0ex}{}{n}{\sum _{i=1}^{n}{X}_{i}}\right)\right)+\sum _{i=1}^{n}{X}_{i}ln\left(p\right)+\hspace{0.17em}(n-\sum _{i=1}^{n}{X}_{i})\mathrm{ln}(1-p)$

and: $\frac{\partial \mathrm{ln}\hspace{0.17em}{f}_{X}\left(X\right)}{\partial p}=\frac{{\sum}_{i=1}^{n}{X}_{i}}{p}-\frac{(n-{\sum}_{i=1}^{n}{X}_{i})}{(1-p)}=\frac{n\overline{X}}{p}-\frac{(n-n\overline{X})}{(1-p)}$

and: $(\frac{\partial ln\hspace{0.17em}{f}_{X}\left(X\right)}{\partial p}{)}^{2}=(\frac{n\overline{X}}{p}-\frac{(n-n\overline{X})}{(1-p)}{)}^{2}=\frac{{n}^{2}{p}^{2}-2{n}^{2}p\overline{X}+{n}^{2}{\overline{X}}^{2}}{{p}^{2}(1-p{)}^{2}}$

since $E\left[{\stackrel{\u2015}{X}}^{2}\right]={\mu}^{2}+\frac{{\sigma}^{2}}{n}$, and for a Bernoulli random variable $E\left[X\right]=\mu =p=E\left[\stackrel{\u2015}{X}\right]$ and $Var\left[X\right]={\sigma}^{2}=p(1-p)$:

$E\left[{\left(\frac{\partial \mathrm{ln}:\left\{{f}_{X}\left(X\right)\right\}}{\partial p}\right)}^{2}\right]=\frac{{n}^{2}{p}^{2}-2{n}^{2}pE\left[\stackrel{\u2015}{X}\right]+{n}^{2}E\left[{\stackrel{\u2015}{X}}^{2}\right]}{{p}^{2}{(1-p)}^{2}}=\frac{{n}^{2}{p}^{2}-2{n}^{2}{p}^{2}+{n}^{2}({p}^{2}+\frac{p(1-p)}{n})}{{p}^{2}{(1-p)}^{2}}=\frac{np(1-p)}{{p}^{2}{(1-p)}^{2}}=\frac{n}{p(1-p)}$

Therefore, $Var\left[\hat{P}\right]=\frac{1}{nE\left[\right((\partial /\partial p)\mathrm{ln}\hspace{0.17em}{f}_{x}\left(X\right){)}^{2}]}=\frac{1}{n\frac{n}{p(1-p)}}=\frac{p(1-p)}{{n}^{2}}$

However, I believe the true value I should have come up with is $\frac{p(1-p)}{n}$.

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asked 2022-04-19

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