StatstriveISI intelligencePractice
Back to MSQE practice
PEAMCQModerate

2026 PEA Q13

For X1,…,Xn∼Poisson(λ)X_1, \dots, X_n \sim \text{Poisson}(\lambda) (i.i.d.), which is unbiased for λ2\lambda^2?

Reveal answer and solution

Answer

D

Solution

  1. 1

    For Poisson, E(Xˉ)=λE(\bar{X}) = \lambda and Var⁡(Xˉ)=λ/n\operatorname{Var}(\bar{X}) = \lambda/n. Therefore

  2. 2
    E(Xˉ2)=Var⁡(Xˉ)+(EXˉ)2=λn+λ2. E(\bar{X}^2) = \operatorname{Var}(\bar{X}) + (E\bar{X})^2 = \frac{\lambda}{n} + \lambda^2.
  3. 3

    Consequently

  4. 4
    E ⁣(Xˉ2−Xˉn)=λn+λ2−λn=λ2. E\!\left(\bar{X}^2 - \frac{\bar{X}}{n}\right) = \frac{\lambda}{n} + \lambda^2 - \frac{\lambda}{n} = \lambda^2.
  5. 5

    Hence Xˉ2−Xˉ/n\bar{X}^2 - \bar{X}/n is unbiased for λ2\lambda^2.

Answer structure / marking notes

Option (D)-correction subtracts Xˉ/n\bar X/n, not Xˉ\bar X. The bias of Xˉ2\bar X^2 is λ/n\lambda/n, not λ\lambda.

%--------------------------------------------------

Content note

Imported from public/resources/isi/msqe/solutions/pea/2026/ISI_MSQE_PEA_2026_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.