Respuesta :
Answer:
The answer is "[tex]\bold{\frac{2}{n}}[/tex]".
Step-by-step explanation:
considering [tex]Y_1, Y_2,........, Y_n[/tex] signify a random Poisson distribution of the sample size of n which means is λ.
[tex]E(Y_i)= \lambda \ \ \ \ \ and \ \ \ \ \ Var(Y_i)= \lambda[/tex]
Let assume that,
[tex]\hat \lambda_i = \frac{Y_1+Y_2}{2}[/tex]
multiply the above value by Var on both sides:
[tex]Var (\hat \lambda_1 )= Var(\frac{Y_1+Y_2}{2} )[/tex]
[tex]=\frac{1}{4}(Var (Y_1)+Var (Y_2))\\\\=\frac{1}{4}(\lambda+\lambda)\\\\=\frac{1}{4}( 2\lambda)\\\\=\frac{\lambda}{2}\\[/tex]
now consider [tex]\hat \lambda_2[/tex] = [tex]\bar Y[/tex]
[tex]Var (\hat \lambda_2 )= Var(\bar Y )[/tex]
[tex]=Var \{ \frac{\sum Y_i}{n}\}[/tex]
[tex]=\frac{1}{n^2}\{\sum_{i}^{}Var(Y_i)\}\\\\=\frac{1}{n^2}\{ n \lambda \}\\\\=\frac{\lambda }{n}\\[/tex]
For calculating the efficiency divides the [tex]\hat \lambda_1 \ \ \ and \ \ \ \hat \lambda_2[/tex] value:
Formula:
[tex]\bold{Efficiency = \frac{Var(\lambda_2)}{Var(\lambda_1)}}[/tex]
[tex]=\frac{\frac{\lambda}{n}}{\frac{\lambda}{2}}\\\\= \frac{\lambda}{n} \times \frac {2} {\lambda}\\\\ \boxed{= \frac{2}{n}}[/tex]
