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2024 PEA Q19

Let X1,X2,…,X5X_1,X_2,\dots,X_5 be i.i.d.\ random variables with E(Xi)=0E(X_i)=0 and V(Xi)=1V(X_i)=1. What is the value of

E ⁣[(∑i=15Xi)25]? E\!\left[\frac{(\sum_{i=1}^{5} X_i)^{2}}{5}\right]?
Reveal answer and solution

Answer

B

Solution

  1. 1

    Let S=∑i=15XiS=\sum_{i=1}^{5} X_i. Then E(S)=0E(S)=0 and Var(S)=∑i=15Var(Xi)=5\mathrm{Var}(S)=\sum_{i=1}^{5}\mathrm{Var}(X_i)=5. Hence

  2. 2
    E(S2)=Var(S)+(E(S))2=5. E(S^{2})=\mathrm{Var}(S)+\bigl(E(S)\bigr)^{2}=5.
  3. 3

    Therefore E[S2/5]=1E[S^{2}/5]=1.

Answer structure / marking notes

Independence is used to add variances.

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Content note

Imported from public/resources/isi/msqe/solutions/pea/2024/ISI_MSQE_PEA_2024_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.