Solution. (a)(i). The \(k\)th sample moment about the mean is \[M_k=\frac {1}{n}\sum _{i=1}^n\left (X_i-\bar X\right )^k.\]

(ii). \(T\) is an unbiased estimator of \(\tau (\theta )\) if \(E_\theta (T)=\tau (\theta )\) for every \(\theta \) in the parameter space.

(b)(i). The transformation is one-to-one: \(X=V\) and \(Y=U-V\). Substituting, \[f_{U,V}(u,v)=f\left (v,\,u-v\right )=\frac {1}{21}\left [v+(u-v)\right ] =\frac {u}{21},\] on the image of the original support, namely \(v=1,2,3\) with \(u=v+1,v+2\): \[\begin {array}{c|cccc} & u=2 & u=3 & u=4 & u=5\\\hline v=1 & \frac {2}{21} & \frac {3}{21} & & \\ v=2 & & \frac {3}{21} & \frac {4}{21} & \\ v=3 & & & \frac {4}{21} & \frac {5}{21} \end {array}\] The six probabilities total \(\frac {2+3+3+4+4+5}{21}=1\), as they must.

(ii). Summing each column, \[f_U(2)=\frac {2}{21},\quad f_U(3)=\frac {6}{21},\quad f_U(4)=\frac {8}{21},\quad f_U(5)=\frac {5}{21},\] which again totals \(1\).

(iii). Summing each row, \[f_V(1)=\frac {5}{21},\quad f_V(2)=\frac {7}{21},\quad f_V(3)=\frac {9}{21}.\] Since \(V=X\) this must be the marginal of \(X\), and reading it straight off the original table gives \(\frac {2+3}{21},\frac {3+4}{21},\frac {4+5}{21}\) — the same thing, which is a free check on the work.

(c)(i). The likelihood is \[L\left (\sigma ^2\right )=\left (2\pi \sigma ^2\right )^{-\frac {n}{2}} \exp \left (-\frac {\sum x_i^2}{2\sigma ^2}\right ),\] and maximising over all \(\sigma ^2>0\) gives \(\hat \sigma ^2=\frac {1}{n}\sum x_i^2\). Hence \[\lambda =\frac {L\left (\sigma _0^2\right )}{L\left (\hat \sigma ^2\right )} =\left (\frac {\hat \sigma ^2}{\sigma _0^2}\right )^{\frac {n}{2}} \exp \left (-\frac {\sum x_i^2}{2\sigma _0^2}+\frac {n}{2}\right ).\] Writing \(T=\frac {\sum x_i^2}{\sigma _0^2}\), we have \(\frac {\hat \sigma ^2}{\sigma _0^2}=\frac {T}{n}\) and therefore \[\lambda =\left (\frac {T}{n}\right )^{\frac {n}{2}} e^{\frac {n}{2}-\frac {T}{2}} =\left [\frac {T}{n}\,e^{1-\frac {T}{n}}\right ]^{\frac {n}{2}},\] which depends on the sample only through \(T\). \(\blacksquare \)

(ii). Under \(H_0\) each \(\frac {X_i}{\sigma _0}\sim N(0,1)\) and the \(X_i\) are independent, so \(T\) is a sum of \(n\) independent squared standard normals: \[T=\sum _{i=1}^n\left (\frac {X_i}{\sigma _0}\right )^2\sim \chi ^2(n).\]