Question

Consider the following method of estimating λ for a Poisson distribution. Observe that

p₀ = P(X = 0) = e(^-λ)

Letting Y denote the number of zeros from an i.i.d. sample of size n, λ might be estimated by

λ˜ = − log(Y/n)

Use the method of propagation of error to obtain approximate expressions for the variance and the bias of this estimate. Compare the variance of this estimate to the variance of the mle, computing relative efficiencies for various values of λ. Note that Y ∼ bin(n, p₀).

Answer 1

Let Y denote the number of zeros from an i.i.d. sample of size n, then

$Y\sim Bin(n,p_0),~p_0=e^{-\lambda}$

$\hat{p}_0=e^{-\hat{\lambda}}=Y/n,~since~E(Y)=np_0$

Therefore

$\hat{\lambda}=-log(Y/n)$

Let

$f(X_1,X_2,....,X_n)=-log(Y/n),~where~X_i\sim P(\lambda),i=1,2,..,n,~independently$

The approximate expression for the variance of f(X₁,X₂,...,X_n) is

$\sigma_f^2\approx \sum_{i=1}^{n}\begin{vmatrix} \frac{\partial f }{\partial x_i} \end{vmatrix}^2\sigma_i^2,~\sigma_i^2=Var(X_i)$

since X_i's are independent.

Therefore

$\sigma_f^2\approx \sum_{i=1}^{n}\frac{1}{y^2}\lambda=\frac{n\lambda}{y^2}$, where $\frac{\partial f}{\partial x_i}=-\frac{1}{y},~\sigma_i^2=\lambda ~\forall ~i$

where y is the observed value of Y.

Bias=

$\sum_{i=1}^{n} \frac{\partial f}{\partial x_i} E(X_i)-\lambda=-\frac{n\lambda}{y}-\lambda$

$\hat{\lambda}_{MLE}=\frac{1}{n}\sum_{i=1}^nX_i=\bar{X}$

Therefore,

$Var(\hat{\lambda}_{MLE})=Var(\bar{X})=\frac{\lambda}{n}$.

$\frac{Var(\hat{\lambda}_{MLE})}{Var(\hat{\lambda}_{MPE})} =\frac{y^2}{n^2}\leq 1$

= holds if n=y i.e. all n samples take the value zero.