Question

Suppose that Y1, Y2,..., Yn denote a random sample of size n from a Poisson distribution with mean λ. Consider λˆ 1 = (Y1 + Y2)/2 and λˆ 2 = Y . Derive the efficiency of λˆ 1 relative to λˆ 2.

Answer 1

Answer:

The answer is "$\bold{\frac{2}{n}}$".

Step-by-step explanation:

considering $Y_1, Y_2,........, Y_n$ signify a random Poisson distribution of the sample size of n which means is λ.

$E(Y_i)= \lambda \ \ \ \ \ and \ \ \ \ \ Var(Y_i)= \lambda$

Let assume that,  

$\hat \lambda_i = \frac{Y_1+Y_2}{2}$

multiply the above value by Var on both sides:

$Var (\hat \lambda_1 )= Var(\frac{Y_1+Y_2}{2} )$

            $=\frac{1}{4}(Var (Y_1)+Var (Y_2))\\\\=\frac{1}{4}(\lambda+\lambda)\\\\=\frac{1}{4}( 2\lambda)\\\\=\frac{\lambda}{2}$

now consider $\hat \lambda_2$ = $\bar Y$

$Var (\hat \lambda_2 )= Var(\bar Y )$

             $=Var \{ \frac{\sum Y_i}{n}\}$

             $=\frac{1}{n^2}\{\sum_{i}^{}Var(Y_i)\}\\\\=\frac{1}{n^2}\{ n \lambda \}\\\\=\frac{\lambda }{n}$

For calculating the efficiency divides the $\hat \lambda_1 \ \ \ and \ \ \ \hat \lambda_2$ value:

Formula:

$\bold{Efficiency = \frac{Var(\lambda_2)}{Var(\lambda_1)}}$

                  $=\frac{\frac{\lambda}{n}}{\frac{\lambda}{2}}\\\\= \frac{\lambda}{n} \times \frac {2} {\lambda}\\\\ \boxed{= \frac{2}{n}}$