91Ó°ÊÓ

It is given that the random variable X follows a Poisson distribution with parameter \(\lambda .\)

Therefore, its pdf is:

\(P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}},x = 0,1 \ldots \)

An estimator \(\delta \left( X \right)\,\,of\,\,g\left( \theta \right)\) is unbiased if \(E\left( {\delta \left( X \right)} \right) = g\left( \theta \right)\) for all possible values of \(\theta \) .

Now, we have to show \(E\left( {\delta \left( X \right)} \right) = {e^\lambda }\)

This implies that we have to find \(\delta \left( x \right)\)such that it satisfies the following equation,

\(\sum\limits_{x = 0}^\infty {\frac{{\delta \left( x \right){e^{ - \lambda }}{\lambda ^x}}}{{x!}}} = {e^\lambda }\)

Multiply the equation on both sides by \({e^\lambda }\)

\(\sum\limits_{x = 0}^\infty {\frac{{\delta \left( x \right){\lambda ^x}}}{{x!}}} = {e^{2\lambda }}\)

\(\begin{aligned}{} \Rightarrow \frac{{\delta \left( 0 \right){\lambda ^0}}}{{0!}} + \ldots + \frac{{\delta \left( n \right){\lambda ^n}}}{{n!}} &= {e^{2\lambda }}\\ \Rightarrow \frac{{\delta \left( 0 \right){\lambda ^0}}}{{0!}} + \ldots + \frac{{\delta \left( n \right){\lambda ^n}}}{{n!}} &= \frac{{{{\left( {2\lambda } \right)}^0}}}{{0!}} + \ldots + \frac{{{{\left( {2\lambda } \right)}^n}}}{{n!}}\\ \Rightarrow \delta \left( x \right) &= {2^x}\end{aligned}\)

Therefore, the value of the unbiased estimator \(\delta \left( x \right) = {2^x}\)