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

The answer is "
`\bold{(2)/(n)}`".

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 = (Y_1+Y_2)/(2)`

multiply the above value by Var on both sides:

`Var (\hat \lambda_1 )= Var((Y_1+Y_2)/(2) )`

`=(1)/(4)(Var (Y_1)+Var (Y_2))\\\\=(1)/(4)(\lambda+\lambda)\\\\=(1)/(4)( 2\lambda)\\\\=(\lambda)/(2)\\`

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

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

`=Var \{ (\sum Y_i)/(n)\}`

`=(1)/(n^2)\{\sum_(i)^{}Var(Y_i)\}\\\\=(1)/(n^2)\{ n \lambda \}\\\\=(\lambda )/(n)\\`

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

Formula:

`\bold{Efficiency = (Var(\lambda_2))/(Var(\lambda_1))}`

`=((\lambda)/(n))/((\lambda)/(2))\\\\= (\lambda)/(n) * \frac {2} {\lambda}\\\\ \boxed{= (2)/(n)}`