Question

Consider a random sample $X_1, \dots, X_n$ from a normal distribution with mean 0 and variance $\theta$. Then find the UMVUE for $\theta^2$. ($S = \sum_{i=1}^n X_i^2$) (A) $T = \frac{S^2}{n(n+2)}$ is UMVUE for $\theta^2$. (B) $T = \frac{S^2}{(n+2)}$ is UMVUE for $\theta^2$. (C) $T = \frac{S^2}{n}$ is UMVUE for $\theta^2$. (D) $T = S^2$ is UMVUE for $\theta^2$.

          Consider a random sample $X_1, \dots, X_n$ from a normal distribution with mean 0 and variance $\theta$.
Then find the UMVUE for $\theta^2$. ($S = \sum_{i=1}^n X_i^2$)
(A) $T = \frac{S^2}{n(n+2)}$ is UMVUE for $\theta^2$.
(B) $T = \frac{S^2}{(n+2)}$ is UMVUE for $\theta^2$.
(C) $T = \frac{S^2}{n}$ is UMVUE for $\theta^2$.
(D) $T = S^2$ is UMVUE for $\theta^2$.

Consider a random sample X1, …, Xn from a normal distribution with mean 0 and variance θ.
Then find the UMVUE for θ^2. (S = ∑i=1^n Xi^2)
(A) T = (S^2)/(n(n+2)) is UMVUE for θ^2.
(B) T = (S^2)/((n+2)) is UMVUE for θ^2.
(C) T = (S^2)/(n) is UMVUE for θ^2.
(D) T = S^2 is UMVUE for θ^2.

Added by Ismael P.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Madhur L

07/10/2023

Step 1

.., xn) from a normal distribution with mean 0 and variance θ. The likelihood function for a random sample x1, ..., xn from a normal distribution with mean 0 and variance θ is given by: L(θ; x1, ..., xn) = (1/(√(2πθ))^n) * exp(-(1/(2θ)) * Σ(xi^2)) Show more…

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Consider a random sample x_(1),dots,x_(n) from a normal distribution with mean 0 and variance heta . Then find the UMVUE for heta ^(2). (S)=(sum_(i=1)^n x_(i)^(2)) (A) T=(s^(2))/(n(n+2)) is UMVUE for heta ^(2). (B) T=(s^(2))/((n+2)) is UMVUE for heta ^(2). (C) T=(s^(2))/(n) is UMVUE for heta ^(2). (D) T=S^(2) is UMVUE for heta ^(2). Consider a random sample X,, ...,X, from a normal distribution with mean 0 and variance Then find the UMVUE for 02.(S=Ei=1X?) S2 A)T= is UMVUE for e2 n(n+2) 52 BT= is UMVUE for 02. (n+2 S2 := I (b) -is UMVUE for e2 (D) T= S2 is UMVUE for e2 Oa O b O c P O