Let $X_1,X_2,\ldots,X_n\,, n > 2$ be a random sample from the binomial distribution $b(1, \theta)$.

I have shown $$Y_1 = X_1 + X_2 + \cdots + X_n$$ is a complete sufficient statistic for $\theta$ and $$Y_2 = \frac{X_1 + X_2}{2}$$ is unbiased estimator of $\theta$.

The question ask also to find $$E(Y_2\mid Y_1 = y_1)$$

So I am thinking to find the conditional distribution then find the expectation.

\begin{align} P(Y_2=y_2\mid Y_1=y_1)&=\frac{P(Y_2=y_2,Y_1=y_1)}{P(Y_1=y_1)} \\\\&=\frac{P(Y_2=y_2)P(Y_1-Y_2=y_1-y_2)}{P(Y_1=y_1)} \end{align}

But I do not know the distribution of $Y_2 = (X_1 + X_2)/2$.

The question from Introduction to Mathematical Statistics Hogg 7ed page 401 problem 11.

  • $\begingroup$ Please learn some basic MathJax here for typsetting math on this site. $\endgroup$ – StubbornAtom Sep 29 '18 at 7:14

$P(Y_2 = y_2) = P(X_1 + X_2 = 2 y_2) = \binom{2}{2y_2} \theta^{2 y_2} (1-\theta)^{2 - 2y_2}$ for $y_2 \in \{0, 1/2, 1\}$.


You do not need the distribution of $Y_2$ to find the conditional expectation.

Since $\sum_{i=1}^n X_i\sim b(n,\theta)$ , and $X_1$ and $\sum_{i=2}^n X_i$ are independently distributed, we directly get

\begin{align} E(X_1\mid Y_1=y_1)&=P(X_1=1\mid Y_1=y_1) \\\\&=\frac{P(X_1=1)P\left(\sum_{i=2}^n X_i=y_1-1\right)}{P\left(\sum_{i=1}^n X_i=y_1\right)} \\\\&=\frac{\theta\binom{n-1}{y_1-1}\theta^{y_1-1}(1-\theta)^{n-y_1}}{\binom{n}{y_1}\theta^{y_1}(1-\theta)^{n-y_1}} \\\\&=\frac{\binom{n-1}{y_1-1}}{\binom{n}{y_1}}\\\\&=\frac{y_1}{n}=\bar x \end{align}

So, once again,

\begin{align} E\left(\frac{X_1+X_2}{2}\mid Y_1=y_1\right)&=\frac{1}{2}\left(E(X_1\mid Y_1=y_1)+E(X_2\mid Y_1=y_1)\right) \\&=\frac{1}{2}(\bar x+\bar x)=\bar x \end{align}

  • 1
    $\begingroup$ Of course, symmetry arguments (more precisely, exchangeability) give instantly the result (as explained several times on the site). $\endgroup$ – Did Sep 29 '18 at 18:08

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