Solution. (a)(i). \(T=T(X_1,\ldots ,X_n)\) is a consistent estimator of \(\tau (\theta )\) if it converges to \(\tau (\theta )\) in probability: for every \(\varepsilon >0\) and every \(\theta \), \[\lim _{n\to \infty }P\left (\left |T-\tau (\theta )\right |>\varepsilon \right )=0.\]

(ii). The power function of a test with critical region \(C\) is \[\pi (\theta )=P\left (\underline {X}\in C\mid \theta \right ),\] the probability of rejecting \(H_0\), regarded as a function of \(\theta \) over the whole parameter space. Its values on \(\Theta _0\) give the size; its values on \(\Theta _1\) give the power.

(b)(i). Here \((k+\alpha -1)^{(k)}\) denotes the falling factorial of \(k\) terms, \((k+\alpha -1)(k+\alpha -2)\cdots \alpha \). Now \[E\left (X^k\right )=\int _0^\infty x^k\, \frac {x^{\alpha -1}e^{-\frac {x}{\beta }}}{\Gamma (\alpha )\beta ^\alpha }\,dx =\frac {1}{\Gamma (\alpha )\beta ^\alpha } \int _0^\infty x^{\alpha +k-1}e^{-\frac {x}{\beta }}\,dx.\] The remaining integral is the gamma function’s own integral, and equals \(\Gamma (\alpha +k)\beta ^{\alpha +k}\). Hence \[E\left (X^k\right )=\frac {\Gamma (\alpha +k)\beta ^{\alpha +k}} {\Gamma (\alpha )\beta ^\alpha } =\beta ^k\,\frac {\Gamma (\alpha +k)}{\Gamma (\alpha )} =\beta ^k(\alpha +k-1)(\alpha +k-2)\cdots \alpha =\beta ^k(k+\alpha -1)^{(k)},\] the middle step repeating \(\Gamma (z+1)=z\Gamma (z)\) exactly \(k\) times.

(ii). Taking \(k=1\) and \(k=2\), \[E(X)=\alpha \beta ,\qquad E\left (X^2\right )=\beta ^2(\alpha +1)\alpha ,\] \[\therefore \quad \text {Var}(X)=\alpha (\alpha +1)\beta ^2-\alpha ^2\beta ^2 =\alpha \beta ^2.\]

(c)(i). First reconcile the form given with the familiar one. Since \[\binom {-k}{x}(-1)^x=\binom {x+k-1}{x},\] and \((\theta -1)^x=(-1)^x(1-\theta )^x\), the stated probability function is \[f(x)=\binom {-k}{x}\theta ^k(\theta -1)^x=\binom {x+k-1}{x}\theta ^k(1-\theta )^x,\] the number of failures before the \(k\)th success, with success probability \(\theta \).

Its moment generating function is \[M_X(t)=\sum _{x=0}^\infty e^{tx}\binom {x+k-1}{x}\theta ^k(1-\theta )^x =\theta ^k\sum _{x=0}^\infty \binom {x+k-1}{x} \left [(1-\theta )e^t\right ]^x =\left [\frac {\theta }{1-(1-\theta )e^t}\right ]^k,\] using the negative binomial series \(\sum _x\binom {x+k-1}{x}z^x=(1-z)^{-k}\), valid for \(\left |(1-\theta )e^t\right |<1\).

The \(X_i\) are independent, so the mgf of the sum is the product of the mgfs: \[M_Y(t)=\prod _{i=1}^nM_{X_i}(t) =\prod _{i=1}^n\left [\frac {\theta }{1-(1-\theta )e^t}\right ]^{k_i} =\left [\frac {\theta }{1-(1-\theta )e^t}\right ]^{\sum k_i}.\] This is the mgf of \(NB\left (\sum k_i,\theta \right )\), and a moment generating function determines a distribution uniquely, so \[Y\sim NB\left (\sum _{i=1}^nk_i,\theta \right ).\qquad \blacksquare \]

(ii). Work with the distribution function first. By independence the joint density is \(f_X(x)f_Y(y)\), so \[F_Z(z)=P(X+Y\leq z) =\int _{-\infty }^{\infty }\int _{-\infty }^{z-x}f_X(x)f_Y(y)\,dy\,dx =\int _{-\infty }^{\infty }f_X(x)F_Y(z-x)\,dx.\] Differentiating with respect to \(z\) under the integral sign, \[f_Z(z)=\frac {d}{dz}F_Z(z) =\int _{-\infty }^{\infty }f_X(x)\,\frac {\partial }{\partial z}F_Y(z-x)\,dx =\int _{-\infty }^{\infty }f_X(x)f_Y(z-x)\,dx.\qquad \blacksquare \]