Solution. (a)(i). If \(Z\sim N(0,1)\) and \(V\sim \chi ^2(n)\) are independent, then \[T=\frac {Z}{\sqrt {V/n}}\] has the \(t\)-distribution with \(n\) degrees of freedom.

(ii). A test of \(H_0\) against a simple \(H_1\) with critical region \(C\) is most powerful of size \(\alpha \) if its size is \(\alpha \) and, for every other test with critical region \(A\) of size at most \(\alpha \), \[P\left (\underline {X}\in C\mid H_1\right )\geq P\left (\underline {X}\in A\mid H_1\right ).\]

(b)(i). For \(0<x<1\) and \(0<y<1\), \[F(x,y)=\int _0^x\int _0^y\frac {2}{3}(u+1)\,dv\,du =\frac {2}{3}\,y\int _0^x(u+1)\,du =\frac {2}{3}\,y\left [\frac {u^2}{2}+u\right ]_0^x =\frac {y\left (x^2+2x\right )}{3}.\] Completing the definition over the plane, \(F(x,y)=0\) if \(x\leq 0\) or \(y\leq 0\); \(F(x,y)=\frac {x^2+2x}{3}\) for \(0<x<1,\ y\geq 1\); \(F(x,y)=y\) for \(x\geq 1,\ 0<y<1\); and \(F(x,y)=1\) for \(x\geq 1,\ y\geq 1\).

(ii). The marginal c.d.f.s are the limits of \(F\) in the other variable: \[F_X(x)=F(x,1)=\frac {x^2+2x}{3},\qquad F_Y(y)=F(1,y)=\frac {y(1+2)}{3}=y.\] Differentiating, \[f_X(x)=\frac {2x+2}{3}=\frac {2}{3}(x+1),\quad 0<x<1;\qquad f_Y(y)=1,\quad 0<y<1.\] So \(Y\) is uniform on \((0,1)\).

(iii). \[F_X(x)\,F_Y(y)=\frac {x^2+2x}{3}\cdot y=F(x,y)\] for every \((x,y)\), so \(X\) and \(Y\) are independent. The same conclusion follows from the density: \(f(x,y)=\frac {2}{3}(x+1)\cdot 1=f_X(x)f_Y(y)\), and the support is a rectangle.

(c)(i). Here \(E(X)=\frac {1}{\theta }\). Equating the first population moment to the first sample moment, \[\frac {1}{\theta }=\bar X\qquad \implies \qquad \tilde \theta =\frac {1}{\bar X}.\]

(ii). \(L(\theta )=\theta ^ne^{-\theta \sum x_i}\), so \[\ell (\theta )=n\ln \theta -\theta \sum x_i,\qquad \ell '(\theta )=\frac {n}{\theta }-\sum x_i=0 \quad \implies \quad \hat \theta =\frac {n}{\sum x_i}=\frac {1}{\bar X},\] and \(\ell ''(\theta )=-\frac {n}{\theta ^2}<0\) confirms a maximum. Here the two methods agree.

(iii). Write \(S=\sum X_i\sim \text {GAM}\left (n,\frac {1}{\theta }\right )\), for which \(E\left (S^j\right )=\frac {\Gamma (n+j)}{\Gamma (n)\,\theta ^j}\). With \(\bar X=S/n\), \[E\left (\bar X^2\right )=\frac {(n+1)n}{n^2\theta ^2}=\frac {n+1}{n\theta ^2} \ \longrightarrow \ \frac {1}{\theta ^2},\] so \(T\) is asymptotically unbiased, and \[\text {Var}\left (\bar X^2\right ) =\frac {(n+3)(n+2)(n+1)n-n^2(n+1)^2}{n^4\theta ^4} =\frac {(n+1)(4n+6)}{n^3\theta ^4}\ \longrightarrow \ 0.\] Mean square error therefore tends to zero, so by Chebyshev’s inequality \(T\stackrel {p}{\longrightarrow }\frac {1}{\theta ^2}\): \[\therefore \quad T=\bar X^2\ \text {\textbf {is a consistent estimator of}}\ \frac {1}{\theta ^2}.\]