Question

Suppose that\({{\bf{X}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{X}}_{\bf{n}}}\)form a random sample from the Poisson distribution with unknown mean θ, and let

\({\bf{Y = }}\sum\nolimits_{{\bf{i = 1}}}^{\bf{n}} {{{\bf{X}}_{\bf{i}}}} \).

a. Determine the value of a constant c such that the estimator\({{\bf{e}}^{{\bf{ - cY}}}}\)is an unbiased estimator of\({{\bf{e}}^{{\bf{ - \theta }}}}\).

b. Use the information inequality to obtain a lower bound for the variance of the unbiased estimator found in part (a).

Short answer

1. The value of constant c is \(\log \left( {\frac{n}{{n - 1}}} \right)\).
2. \(Var\left( {\exp \left( { - cY} \right)} \right) \ge \frac{{\theta \exp \left( { - 2\theta } \right)}}{n}\).

Step-by-step solution

Given information

Suppose \({X_1},...,{X_n}\)

\(Y = \sum\nolimits_{i = 1}^n {{X_i}} \)

Finding the value of constant c

a.

Since Y has a Poisson distribution with mean \(n\theta \)

Then it follows that,

\(\begin{align}E\left( {\exp \left( { - cY} \right)} \right) &= \sum\limits_{y = 0}^\infty {\frac{{\exp \left( { - cy} \right)\exp \left( { - n\theta } \right){{\left( {n\theta } \right)}^y}}}{{y!}}} \\ &= \exp \left( { - n\theta } \right)\sum\limits_{y = 0}^\infty {\frac{{{{\left( {n\theta \exp \left( { - c} \right)} \right)}^y}}}{{y!}}} \\ &= \exp \left( { - n\theta } \right)\exp \left( {n\theta \exp \left( { - c} \right)} \right)\\ &= \exp \left( {n\theta \left( {\exp \left( { - c} \right) - 1} \right)} \right)\end{align}\)

Since this expectation must be \(\exp \left( { - \theta } \right)\), it follows that

\(\begin{align}n\left( {\exp \left( { - c} \right) - 1} \right) &= - 1\\\left( {\exp \left( { - c} \right) - 1} \right) &= - \frac{1}{n}\\\exp \left( { - c} \right) &= - \frac{1}{n} + 1\\\exp \left( { - c} \right) &= \frac{{ - 1 + n}}{n}\\ - c &= \log \left( {\frac{{n - 1}}{n}} \right)\\c &= \log \left( {\frac{n}{{n - 1}}} \right)\end{align}\)

Therefore, the value of constant c is \(\log \left( {\frac{n}{{n - 1}}} \right)\).

Finding the lower bound of the variance

b.

Here,

\(f\left( {x\left| \theta \right.} \right) = \frac{{\exp \left( { - \theta } \right){\theta ^x}}}{{x!}}\)

So, taking log in both sides,

\(\begin{align}\log f\left( {x\left| \theta \right.} \right) &= \lambda \left( {x\left| \theta \right.} \right)\\ &= - \theta + x\log \theta - \log \left( {x!} \right)\end{align}\)

Then,

\(\begin{align}\lambda '\left( {x\left| \theta \right.} \right) &= - 1 + \frac{x}{\theta }\\\lambda ''\left( {x\left| \theta \right.} \right) &= - \frac{x}{{{\theta ^2}}}\end{align}\)

Then,

\(\begin{align}I\left( \theta \right) &= - {E_\theta }\left( {\lambda ''\left( {X\left| \theta \right.} \right)} \right)\\ &= \frac{{E\left( X \right)}}{{{\theta ^2}}}\\ &= \frac{1}{\theta }\end{align}\)

Since \(m\left( \theta \right) = \exp \left( { - \theta } \right)\)

So,\(Var\left( {\exp \left( { - cY} \right)} \right) \ge \frac{{\theta \exp \left( { - 2\theta } \right)}}{n}\)