Let $X_{1},X_{2},\ldots,X_{n}$ be random sample from $X\sim\text{Binomial}(2,\theta)$.

(a) Find the least variance from the set of all unbiased estimators of $\theta$.

(b) Find a sufficient statistics of $\theta$.

(c) Obtain from this statistics an unbiased estimator to $\theta$.

(d) Verify if this estimator is efficient.


(a) To begin with, I started with \begin{align*} f(x|\theta) & = {{2}\choose{x}}\theta^{x}(1-\theta)^{2-x} = \frac{2}{x!(2-x)!}\theta^{x}(1-\theta)^{2-x} \Rightarrow\\\\ \ln f(x|\theta) & = \ln(2) - \ln(x!) - \ln[(2-x)!] + x\ln(\theta) + (2-x)\ln(1-\theta)\Rightarrow\\\\ \frac{\partial\ln f(x|\theta)}{\partial\theta} & = \frac{x}{\theta} + \frac{x-2}{1-\theta} \Rightarrow \frac{\partial^{2}\ln f(x|\theta)}{\partial\theta^{2}} = -\frac{x}{\theta^{2}} + \frac{x-2}{(1-\theta)^{2}} \Rightarrow\\\\ \textbf{E}\left(\frac{\partial^{2}\ln f(x|\theta)}{\partial\theta^{2}}\right) & = -\frac{2}{\theta} - \frac{2}{1-\theta} = -\frac{2}{\theta(1-\theta)} \Rightarrow I_{F}(\theta) = \frac{2}{\theta(1-\theta)} \end{align*}

(b) Observe that we can factor the likelihood function as follows:

\begin{align*} L(\textbf{x}|\theta) & = \prod_{k=1}^{n}f(x_{k}|\theta) = \prod_{k=1}^{n}{{2}\choose{x_{k}}}\theta^{x_{k}}(1-\theta)^{1-x_{k}}\\\\ & = h(x_{1},x_{2},\ldots,x_{k})g_{\theta}(T(x_{1},x_{2},\ldots,x_{n})) \end{align*}


\begin{align*} g_{\theta}(T(x_{1},x_{2},\ldots,x_{n})) = \theta^{\sum_{k=1}^{n}x_{k}}(1-\theta)^{n-\sum_{k=1}^{n}x_{k}} \end{align*}

Thefore $T(\textbf{x}) = \sum_{k=1}^{n}x_{k}$ is a sufficient statistics, as requested.

(c) To begin with, note that $\textbf{E}(X) = 2\theta$. Since $\textbf{E}(T(\textbf{x})) = 2n\theta$, we can set $\hat{\theta} = T(\textbf{x})/2n$.

(d) Let us start by determining $\textbf{Var}(\hat{\theta})$:

\begin{align*} \textbf{Var}(\hat{\theta}) = \textbf{Var}\left(\frac{1}{2n}\sum_{k=1}^{n}X_{k}\right) = \frac{1}{4n^{2}}\sum_{k=1}^{n}\textbf{Var}(X_{k}) = \frac{2n\theta(1-\theta)}{4n^{2}} = \frac{\theta(1-\theta)}{2n} \end{align*}

Therefore the efficiency of $\hat{\theta}$ is given by: \begin{align*} e(\hat{\theta}) = \frac{1}{\textbf{Var}(\hat{\theta})nI_{F}(\theta)} = 1 \end{align*}

Thus the estimator $\hat{\theta}$ is efficient.

  • $\begingroup$ You have edited your expressions to end up with $\frac{2}{\theta (1-\theta)} $ but already earlier did you flip a minus sign. The point where it starts to become incorrect is $\frac {x}{\theta} + \frac {x-2}{1-\theta} $ $\endgroup$ – Sextus Empiricus Apr 11 '19 at 18:11

You expected the efficiency to be $e(\hat\theta) = 1$ right?

Check why you do not end up with:

$$n\mathcal{I}_F(\theta) = - n \textbf{E}\left(\frac{\partial^{2}\ln f(x|\theta)}{\partial\theta^{2}}\right) = \frac{2n}{\theta(1-\theta)}$$

(especially see why you wrote $(x-2)$ in the third line)

For the rest it looks fine. Except you forgot a factor 2 in $\textbf{Var}(\hat{\theta}) = \textbf{Var}\left(\frac{1}{2n}\sum_{k=1}^{n}X_{k}\right)$

  • $\begingroup$ Thanks for the comment. But I still did not realize where is the mistake (third line). Could you please highlight it for me? $\endgroup$ – EagerToLearn Apr 11 '19 at 13:35
  • $\begingroup$ @user1337 $$\frac{\partial\ln f(x|\theta)}{\partial\theta} = \frac{\partial}{\partial\theta} \left( x \ln(\theta) + (2-x) \ln (1-\theta) \right)= \ldots $$ $\endgroup$ – Sextus Empiricus Apr 11 '19 at 13:59

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