Solution. Throughout, \(a^{(j)}\) is the falling factorial \(a(a-1)\cdots (a-j+1)\), so \(X^{(j)}=X(X-1)\cdots (X-j+1)\) and \(E\left (X^{(j)}\right )\) is the \(j\)th factorial moment. Note also that \(q=1-p\), so \(-q=p-1\).

Start from the negative binomial series, valid for \(|z|<1\): \[\sum _{x=0}^\infty \binom {-k}{x}z^x=(1+z)^{-k}.\] Differentiating \(j\) times term by term, \[\sum _{x=0}^\infty x^{(j)}\binom {-k}{x}z^{x-j} =\frac {d^j}{dz^j}(1+z)^{-k} =(-k)(-k-1)\cdots (-k-j+1)(1+z)^{-k-j} =(-k)^{(j)}(1+z)^{-k-j},\] and multiplying through by \(z^j\), \[\sum _{x=0}^\infty x^{(j)}\binom {-k}{x}z^x=(-k)^{(j)}z^j(1+z)^{-k-j}.\] Now put \(z=-q\) and multiply by \(p^k\), which is exactly the expectation asked for: \[E\left (X^{(j)}\right )=p^k(-k)^{(j)}(-q)^j(1-q)^{-k-j} =p^k(-k)^{(j)}(-q)^jp^{-k-j} =(-k)^{(j)}\left (\frac {-q}{p}\right )^j =(-k)^{(j)}\left (\frac {p-1}{p}\right )^j.\qquad \blacksquare \] (Taking \(j=0\) gives \(p^kp^{-k}=1\), so the probabilities do sum to one.)

For the variance, read off the first two factorial moments: \[E(X)=(-k)\left (\frac {-q}{p}\right )=\frac {kq}{p},\qquad E\left [X(X-1)\right ]=(-k)(-k-1)\left (\frac {-q}{p}\right )^2 =\frac {k(k+1)q^2}{p^2}.\] Since \(E\left (X^2\right )=E\left [X(X-1)\right ]+E(X)\), \[\text {Var}(X)=\frac {k(k+1)q^2}{p^2}+\frac {kq}{p}-\frac {k^2q^2}{p^2} =\frac {kq^2}{p^2}+\frac {kq}{p} =\frac {kq}{p^2}\left (q+p\right ) \] \[\therefore \quad \text {Var}(X)=\frac {kq}{p^2}.\]

Solution. The support is the triangle \(0\leq x\leq y\leq 1\) — not the unit square. Every integral below has to respect that, and it is where most of the marks are.

(a). Integrating \(x\) from \(0\) to \(y\), then \(y\) from \(0\) to \(1\), \[1=k\int _0^1\!\!\int _0^y(x+y)\,dx\,dy =k\int _0^1\left [\frac {x^2}{2}+xy\right ]_0^y dy =k\int _0^1\frac {3y^2}{2}\,dy =\frac {k}{2},\] \[\therefore \quad k=2.\]

(b). The joint c.d.f. splits into cases according to whether the corner \((x,y)\) lies above or below the line \(u=v\).

For \(0\leq x\leq y\leq 1\) the region \(\{U\leq x,\ V\leq y\}\) meets the support in \(\{0\leq u\leq x,\ u\leq v\leq y\}\), so \[F(x,y)=\int _0^x\!\!\int _u^y2(u+v)\,dv\,du =\int _0^x\left [2uv+v^2\right ]_u^y du =\int _0^x\left (2uy+y^2-3u^2\right )du,\] \[F(x,y)=x^2y+xy^2-x^3.\]

For \(0\leq y\leq x\) with \(y\leq 1\) the constraint \(U\leq x\) is not binding, because the support already forces \(U\leq V\leq y\leq x\). So \(F(x,y)=P(V\leq y)=y^3\). Collecting the cases, \[F(x,y)=\begin {cases} 0, & x<0\ \text {or}\ y<0,\\[2pt] x^2y+xy^2-x^3, & 0\leq x\leq y\leq 1,\\[2pt] y^3, & 0\leq y\leq x,\ y\leq 1,\\[2pt] x^2+x-x^3, & 0\leq x\leq 1,\ y>1,\\[2pt] 1, & x>1,\ y>1. \end {cases}\] The two middle rows agree on \(x=y\), where both give \(y^3\), so \(F\) is continuous across the diagonal — worth checking, because a mistake in the limits almost always shows up as a jump there.

(c). Let \(y\to 1\) and \(x\to 1\) in turn: \[F_X(x)=F(x,1)=x^2+x-x^3,\quad 0\leq x\leq 1;\qquad F_Y(y)=F(1,y)=y^3,\quad 0\leq y\leq 1,\] each \(0\) below its range and \(1\) above.

(d). Differentiating the marginal c.d.f.s, \[f_X(x)=2x+1-3x^2,\quad 0\leq x\leq 1;\qquad f_Y(y)=3y^2,\quad 0\leq y\leq 1.\] The same answers come from integrating the joint density directly, \[f_X(x)=\int _x^12(x+y)\,dy=2x+1-3x^2,\qquad f_Y(y)=\int _0^y2(x+y)\,dx=3y^2,\] and both integrate to \(1\) over \([0,1]\).

Solution. (a). Since \(\theta >0\) the integral converges at infinity: \[1=\int _\alpha ^\infty kx^{-(\theta +1)}dx =k\left [\frac {x^{-\theta }}{-\theta }\right ]_\alpha ^\infty =\frac {k\alpha ^{-\theta }}{\theta } \qquad \implies \qquad k=\theta \alpha ^\theta .\] This is the Pareto distribution with scale \(\alpha \) and shape \(\theta \).

(b). The transformation \(y=\theta \ln \left (\frac {x}{\alpha }\right )\) is strictly increasing, mapping \(x>\alpha \) onto \(y>0\), with inverse and derivative \[x=\alpha e^{\frac {y}{\theta }},\qquad \frac {dx}{dy}=\frac {\alpha }{\theta }e^{\frac {y}{\theta }}.\] Hence \[f_Y(y)=f_X\!\left (\alpha e^{\frac {y}{\theta }}\right )\left |\frac {dx}{dy}\right | =\theta \alpha ^\theta \left (\alpha e^{\frac {y}{\theta }}\right )^{-(\theta +1)} \cdot \frac {\alpha }{\theta }e^{\frac {y}{\theta }} =\alpha ^{\theta -\theta -1+1}\, e^{-\frac {(\theta +1)y}{\theta }+\frac {y}{\theta }} =e^{-y},\] \[\therefore \quad Y\sim \text {EXP}(1),\qquad y>0.\]

Solution. (a). \(y=x^3\) is increasing, taking \(0<x<3\) onto \(0<y<27\), with \(x=y^{\frac 13}\) and \(\frac {dx}{dy}=\frac 13y^{-\frac 23}\). So \[f_Y(y)=\frac {\left (y^{\frac 13}\right )^2}{9}\cdot \frac {1}{3}y^{-\frac 23} =\frac {y^{\frac 23}}{9}\cdot \frac {y^{-\frac 23}}{3}=\frac {1}{27}, \qquad 0<y<27,\] \[\therefore \quad Y\sim \text {UNIF}(0,27).\]

(b)(i). \(y=e^{-x}\) is decreasing, taking \(0<x<1\) onto \(e^{-1}<y<1\), with \(x=-\ln y\) and \(\left |\frac {dx}{dy}\right |=\frac {1}{y}\). Since \(f_X\equiv 1\), \[f_Y(y)=\frac {1}{y},\qquad \frac {1}{e}<y<1,\] and \(\int _{1/e}^1\frac {dy}{y}=\ln 1-\ln e^{-1}=1\), as it must.

(ii). Here \(v=x(1-x)\) is not monotone on \((0,1)\): it rises to \(\frac 14\) at \(x=\frac 12\) and falls back. The change-of-variable formula does not apply directly, so work with the distribution function.

Solving \(x^2-x+v=0\) gives the two roots \(x=\frac {1\pm \sqrt {1-4v}}{2}\), and \(X(1-X)\leq v\) holds outside them: \[F_V(v)=P\left (X\leq \frac {1-\sqrt {1-4v}}{2}\right ) +P\left (X\geq \frac {1+\sqrt {1-4v}}{2}\right ).\] As \(X\) is uniform on \((0,1)\) these probabilities are the lengths themselves: \[F_V(v)=\frac {1-\sqrt {1-4v}}{2}+1-\frac {1+\sqrt {1-4v}}{2}=1-\sqrt {1-4v}.\] Differentiating, \[f_V(v)=\frac {2}{\sqrt {1-4v}},\qquad 0<v<\frac 14,\] which integrates to \(\left [-\sqrt {1-4v}\right ]_0^{1/4}=1\).

(c). Again \(y=x^2\) is not monotone on \((-2,4)\), and this time the two branches cover different ranges. For \(0<y<4\) both \(x=\pm \sqrt y\) lie in \((-2,4)\); for \(4<y<16\) only \(x=+\sqrt y\) does, since \(-\sqrt y<-2\). With \(f_X(x)=\frac {x^2}{24}\) and \(\left |\frac {dx}{dy}\right |=\frac {1}{2\sqrt y}\), \[f_Y(y)=\begin {cases} \dfrac {2\cdot \frac {y}{24}}{2\sqrt y}=\dfrac {\sqrt y}{24}, & 0<y<4,\\[10pt] \dfrac {\frac {y}{24}}{2\sqrt y}=\dfrac {\sqrt y}{48}, & 4<y<16. \end {cases}\] As a check, \[\int _0^4\frac {\sqrt y}{24}\,dy+\int _4^{16}\frac {\sqrt y}{48}\,dy =\frac {16}{72}+\frac {112}{144}=\frac {2}{9}+\frac {7}{9}=1.\]

Solution. For a discrete variable there is no Jacobian. When the map is one-to-one the probabilities simply move with the values: \(f_Y(y)=f_X\left (w(y)\right )\), where \(w\) is the inverse.

(a). On \(x=0,1,2,3\) the map \(y=x^2\) is one-to-one (no negative values to collide), so \(Y\) takes \(0,1,4,9\) with \(x=\sqrt y\): \[f_Y(y)=\binom {3}{\sqrt y}\left (\frac 23\right )^{\sqrt y} \left (\frac 13\right )^{3-\sqrt y},\qquad y=0,1,4,9,\] that is \[f_Y(0)=\frac {1}{27},\quad f_Y(1)=\frac {6}{27},\quad f_Y(4)=\frac {12}{27},\quad f_Y(9)=\frac {8}{27},\] which total \(\frac {27}{27}=1\).

(b). \(y=2x+1\) carries \(x=1,2,3\) to \(y=3,5,7\), and the probabilities go with them: \[f_Y(y)=\frac 13,\qquad y=3,5,7.\]

(c). \(y=x^3\) carries \(x=1,2,3,\ldots \) to the cubes \(y=1,8,27,\ldots \), with \(x=y^{\frac 13}\): \[f_Y(y)=\left (\frac 12\right )^{y^{1/3}},\qquad y=1,8,27,64,\ldots \]

(d). \(Y=n-X\) takes the values \(y=0,1,\ldots ,n\), and \[f_Y(y)=P(X=n-y)=\binom {n}{n-y}p^{n-y}q^{y}=\binom {n}{y}q^{y}p^{n-y},\] using \(\binom {n}{n-y}=\binom {n}{y}\). \[\therefore \quad Y\sim BIN(n,q)=BIN(n,1-p).\] This is obvious once stated: if \(X\) counts the successes in \(n\) trials then \(n-X\) counts the failures, and a failure has probability \(q\).

Solution. (a). Put \(U=\frac {X}{Y}\) and \(V=Y\), so \(x=uv\), \(y=v\) and \[|J|=\left |\begin {matrix}v & u\\0 & 1\end {matrix}\right |=v.\] Then for \(u>0\), \(v>0\), \[f_{U,V}(u,v)=e^{-uv}e^{-v}\cdot v=v\,e^{-v(1+u)},\] \[f_U(u)=\int _0^\infty v\,e^{-v(1+u)}dv=\frac {1}{(1+u)^2},\qquad u>0,\] the integral being a gamma integral with shape \(2\) and scale \(\frac {1}{1+u}\). As a check, \(\int _0^\infty (1+u)^{-2}du=1\).

(b). By symmetry it is enough to work with the density directly. With \(T=\frac {X}{Y}\) and \(V=Y\) the same Jacobian gives \(|J|=|v|\), and \[f_{T,V}(t,v)=\frac {1}{2\pi \sigma ^2} \exp \left [-\frac {(tv)^2+v^2}{2\sigma ^2}\right ]|v|,\] so, using the symmetry in \(v\) to fold the integral onto \(v>0\), \[f_T(t)=2\int _0^\infty \frac {v}{2\pi \sigma ^2} \exp \left [-\frac {v^2\left (1+t^2\right )}{2\sigma ^2}\right ]dv.\] Substituting \(w=v^2\), \(dw=2v\,dv\), \[f_T(t)=\frac {1}{2\pi \sigma ^2}\int _0^\infty \exp \left [-\frac {w\left (1+t^2\right )}{2\sigma ^2}\right ]dw =\frac {1}{2\pi \sigma ^2}\cdot \frac {2\sigma ^2}{1+t^2} =\frac {1}{\pi \left (1+t^2\right )}.\] Every \(\sigma \) has cancelled.

Solution. Put \(U=X\) and \(V=XY\), so \(x=u\), \(y=\frac {v}{u}\) and \(|J|=\frac {1}{u}\). Both original variables lie in \((0,1)\), so the image is \(0<v<u<1\), and there \[f_{U,V}(u,v)=1\cdot \frac {1}{u}=\frac {1}{u}.\] Integrating \(u\) out over \(v<u<1\), \[f_V(v)=\int _v^1\frac {du}{u}=\left [\ln u\right ]_v^1=-\ln v,\qquad 0<v<1.\] As a check, \(\int _0^1(-\ln v)\,dv=\left [v-v\ln v\right ]_0^1=1\).

Solution. Inverting, \(x=uv\) and \(y=u(1-v)\), with \[|J|=\left |\begin {matrix}v & u\\ 1-v & -u\end {matrix}\right | =\left |-uv-u(1-v)\right |=u.\] The image of \(\{x>0,y>0\}\) is \(\{u>0,\ 0<v<1\}\). Substituting, \[f_{U,V}(u,v)=\frac {(uv)^{\alpha -1}e^{-uv}}{\Gamma (\alpha )}\cdot \frac {\left [u(1-v)\right ]^{\beta -1}e^{-u(1-v)}}{\Gamma (\beta )}\cdot u =\frac {u^{\alpha +\beta -1}e^{-u}\,v^{\alpha -1}(1-v)^{\beta -1}} {\Gamma (\alpha )\Gamma (\beta )}.\] Multiply and divide by \(\Gamma (\alpha +\beta )\) to split it: \[f_{U,V}(u,v)= \underbrace {\frac {u^{\alpha +\beta -1}e^{-u}}{\Gamma (\alpha +\beta )}} _{\text {GAM}(\alpha +\beta ,\,1)}\cdot \underbrace {\frac {\Gamma (\alpha +\beta )}{\Gamma (\alpha )\Gamma (\beta )} v^{\alpha -1}(1-v)^{\beta -1}}_{\text {BETA}(\alpha ,\,\beta )}.\] The joint density is a function of \(u\) alone times a function of \(v\) alone, over a rectangular region, so \(U\) and \(V\) are independent with the stated marginals. \(\blacksquare \)

Solution. (a). The map is increasing with \(x=\frac {\beta y}{2}\) and \(\frac {dx}{dy}=\frac {\beta }{2}\), so \[f_Y(y)=\frac {1}{\Gamma (\alpha )\beta ^\alpha } \left (\frac {\beta y}{2}\right )^{\alpha -1}e^{-\frac {y}{2}}\cdot \frac {\beta }{2} =\frac {1}{2^{\alpha }\Gamma (\alpha )}y^{\alpha -1}e^{-\frac {y}{2}},\qquad y>0.\] This is the \(\chi ^2\) density with \(2\alpha \) degrees of freedom, since \(\frac {2\alpha }{2}=\alpha \). \(\blacksquare \)

(b). Each \(X_i\) has \(M(t)=(1-\lambda t)^{-1}\) for \(t<\frac 1\lambda \), so by independence \[M_{\sum X_i}(t)=\prod _{i=1}^n(1-\lambda t)^{-1}=(1-\lambda t)^{-n},\] which is the moment generating function of \(\text {GAM}(n,\lambda )\). Hence \[\sum _{i=1}^nX_i\sim \text {GAM}(n,\lambda ),\qquad E\left (\sum X_i\right )=n\lambda ,\qquad \text {Var}\left (\sum X_i\right )=n\lambda ^2.\]

Solution. A linear combination of independent normal variables is normal, so only the mean and variance are needed: \[E(U)=\frac {\mu -3\mu }{2}=-\mu ,\] \[\text {Var}(U)=\frac {1}{4}\left [\text {Var}(Y_1)+9\,\text {Var}(Y_2)\right ] =\frac {1}{4}\left (\sigma ^2+9\sigma ^2\right )=\frac {5\sigma ^2}{2}.\] \[\therefore \quad U\sim N\!\left (-\mu ,\ \frac {5\sigma ^2}{2}\right ).\] Formally, \(M_U(t)=M_{Y_1}\!\left (\frac {t}{2}\right ) M_{Y_2}\!\left (-\frac {3t}{2}\right )\), and multiplying the two normal moment generating functions gives \(\exp \left (-\mu t+\frac {5\sigma ^2t^2}{4}\right )\), as stated.

Solution. Use the convolution formula, taking care with the support. For \(u>0\) the constraint \(y>0\) and \(x=u+y>0\) reduces to \(y>0\): \[f_U(u)=\int _0^\infty f_X(u+y)f_Y(y)\,dy =\frac {1}{\theta ^2}\int _0^\infty e^{-\frac {u+y}{\theta }}e^{-\frac {y}{\theta }}dy =\frac {e^{-\frac {u}{\theta }}}{\theta ^2}\int _0^\infty e^{-\frac {2y}{\theta }}dy =\frac {e^{-\frac {u}{\theta }}}{2\theta }.\] For \(u<0\) the roles of \(X\) and \(Y\) swap, and since they are identically distributed the answer is the mirror image. Combining, \[f_U(u)=\frac {1}{2\theta }e^{-\frac {|u|}{\theta }},\qquad -\infty <u<\infty ,\] the Laplace (double exponential) distribution. It integrates to \(2\cdot \frac {1}{2\theta }\cdot \theta =1\).

Solution. Standardise first: \(\frac {X}{\sigma }\) and \(\frac {Y}{\sigma }\) are independent \(N(0,1)\), so \[\frac {W}{\sigma ^2}=\left (\frac {X}{\sigma }\right )^2 +\left (\frac {Y}{\sigma }\right )^2\sim \chi ^2(2),\] being a sum of two independent squared standard normals. Since \(\chi ^2(2)=\text {GAM}(1,2)\) is the exponential distribution with mean \(2\), \[W\sim \text {EXP}\left (\text {mean }2\sigma ^2\right ),\qquad f_W(w)=\frac {1}{2\sigma ^2}e^{-\frac {w}{2\sigma ^2}},\quad w>0.\]

Solution. (a). Each \(X_i\) has \(M(t)=\exp \left (\mu t+\frac {\sigma ^2t^2}{2}\right )\). By independence, \[M_{\bar X}(t)=E\left [e^{\frac {t}{n}\sum X_i}\right ] =\left [M\!\left (\frac {t}{n}\right )\right ]^n =\left [\exp \left (\frac {\mu t}{n}+\frac {\sigma ^2t^2}{2n^2}\right )\right ]^n =\exp \left (\mu t+\frac {\left (\frac {\sigma ^2}{n}\right )t^2}{2}\right ),\] which is the moment generating function of a normal distribution with mean \(\mu \) and variance \(\frac {\sigma ^2}{n}\): \[\bar X\sim N\!\left (\mu ,\ \frac {\sigma ^2}{n}\right ),\qquad E\left (\bar X\right )=\mu ,\qquad \text {Var}\left (\bar X\right )=\frac {\sigma ^2}{n}.\]

(b). Standardising a normal variable gives a standard normal, and by (a) the standard deviation of \(\bar X\) is \(\frac {\sigma }{\sqrt n}\), so the first statement is immediate. The square of a standard normal is \(\chi ^2(1)\) by definition, giving the second. For the third, each \(\frac {X_i-\mu }{\sigma }\sim N(0,1)\) and they are independent, so the sum of their squares is a sum of \(n\) independent \(\chi ^2(1)\) variables, which is \(\chi ^2(n)\) because degrees of freedom add.

Solution. Each \(X_i\) has \(M(t)=(1-\beta t)^{-\alpha }\), so \[M_{\bar X}(t)=\left [M\!\left (\frac {t}{n}\right )\right ]^n =\left (1-\frac {\beta t}{n}\right )^{-n\alpha },\] which is the moment generating function of a gamma distribution with shape \(n\alpha \) and scale \(\frac {\beta }{n}\): \[\bar X\sim \text {GAM}\!\left (n\alpha ,\ \frac {\beta }{n}\right ),\qquad f_{\bar X}(w)=\frac {w^{n\alpha -1}e^{-\frac {nw}{\beta }}} {\Gamma (n\alpha )\left (\frac {\beta }{n}\right )^{n\alpha }},\quad w>0.\] Its moments follow from the gamma formulas \(E=\text {shape}\times \text {scale}\) and \(\text {Var}=\text {shape}\times \text {scale}^2\): \[E\left (\bar X\right )=n\alpha \cdot \frac {\beta }{n}=\alpha \beta ,\qquad \text {Var}\left (\bar X\right )=n\alpha \left (\frac {\beta }{n}\right )^2 =\frac {\alpha \beta ^2}{n}.\]

Solution. For \(i<j\) the joint density of \(Y_i\) and \(Y_j\) is \[f(y_i,y_j)=\frac {n!}{(i-1)!\,(j-i-1)!\,(n-j)!} \left [F(y_i)\right ]^{i-1} \left [F(y_j)-F(y_i)\right ]^{j-i-1} \left [1-F(y_j)\right ]^{n-j}f(y_i)f(y_j),\] the multinomial count of the ways to place \(i-1\) observations below \(y_i\), one at \(y_i\), \(j-i-1\) between, one at \(y_j\) and \(n-j\) above.

Here \(n=6\), \(i=2\), \(j=5\), and for the uniform distribution \(F(y)=y\) and \(f(y)=1\) on \((0,1)\). The constant is \[\frac {6!}{1!\,2!\,1!}=\frac {720}{2}=360,\] \[\therefore \quad f(y_2,y_5)=360\,y_2\left (y_5-y_2\right )^2\left (1-y_5\right ), \qquad 0<y_2<y_5<1.\] Integrating over the triangle confirms a total mass of \(1\).

Solution. Here \(F(x)=1-e^{-\frac {x}{\theta }}\) for \(x>0\).

(a). The minimum exceeds \(y\) exactly when all \(n\) observations do: \[P\left (Y_1>y\right )=\left [1-F(y)\right ]^n =\left [e^{-\frac {y}{\theta }}\right ]^n=e^{-\frac {ny}{\theta }},\] which is the survival function of an exponential distribution with mean \(\frac {\theta }{n}\). Differentiating \(F_{Y_1}(y)=1-e^{-\frac {ny}{\theta }}\), \[f_{Y_1}(y)=\frac {n}{\theta }e^{-\frac {ny}{\theta }},\qquad y>0.\]

(b). The maximum is at most \(y\) exactly when all \(n\) are: \[F_{Y_n}(y)=\left [1-e^{-\frac {y}{\theta }}\right ]^n,\qquad f_{Y_n}(y)=\frac {n}{\theta }\left (1-e^{-\frac {y}{\theta }}\right )^{n-1} e^{-\frac {y}{\theta }},\quad y>0.\]

Solution. On \((0,a)\) we have \(f(x)=\frac 1a\) and \(F(x)=\frac {x}{a}\). The joint density of the minimum and maximum is \[f\left (y_1,y_n\right )=n(n-1)\left [F(y_n)-F(y_1)\right ]^{n-2}f(y_1)f(y_n) =\frac {n(n-1)\left (y_n-y_1\right )^{n-2}}{a^n},\qquad 0<y_1<y_n<a.\] Put \(R=Y_n-Y_1\) and \(S=Y_1\); then \(y_1=s\), \(y_n=s+r\), the Jacobian is \(1\), and the region becomes \(0<r<a\), \(0<s<a-r\). So \[f_R(r)=\int _0^{a-r}\frac {n(n-1)r^{n-2}}{a^n}\,ds =\frac {n(n-1)r^{n-2}(a-r)}{a^n},\qquad 0<r<a.\] It integrates to \(1\), since \[\frac {n(n-1)}{a^n}\left [\frac {a\,r^{n-1}}{n-1}-\frac {r^n}{n}\right ]_0^a =n(n-1)\left (\frac {1}{n-1}-\frac {1}{n}\right )=1.\]

Solution. First the distribution function: \[F(x)=\int _0^x12w(1-w)^2dw =12\int _0^x\left (w-2w^2+w^3\right )dw =12\left [\frac {x^2}{2}-\frac {2x^3}{3}+\frac {x^4}{4}\right ],\] \[F(x)=6x^2-8x^3+3x^4=x^2\left (3x^2-8x+6\right ),\qquad 0<x<1,\] and \(F(1)=6-8+3=1\), as it must be.

The joint density of the extremes is the general formula with \(i=1\), \(j=n\): \[f\left (y_1,y_n\right )=n(n-1)\left [F(y_n)-F(y_1)\right ]^{n-2}f(y_1)f(y_n),\] \[\therefore \quad f\left (y_1,y_n\right )= 144\,n(n-1)\,y_1\left (1-y_1\right )^2y_n\left (1-y_n\right )^2 \left [y_n^2\left (3y_n^2-8y_n+6\right )-y_1^2\left (3y_1^2-8y_1+6\right )\right ]^{n-2}\] for \(0<y_1<y_n<1\), the \(144\) being \(12\times 12\) from the two densities.

Solution. The joint density of the order statistics from a sample of size \(2\) is \(2!\,f(y_1)f(y_2)\) on \(y_1<y_2\): \[f\left (y_1,y_2\right )=2\left (2y_1\right )\left (2y_2\right )=8y_1y_2, \qquad 0<y_1<y_2<1.\] With \(u=\frac {y_1}{y_2}\) and \(v=y_2\) the inverse is \(y_1=uv\), \(y_2=v\), so \[|J|=\left |\begin {matrix}v & u\\0 & 1\end {matrix}\right |=v,\] and the region \(0<y_1<y_2<1\) becomes the rectangle \(0<u<1\), \(0<v<1\). Hence \[f_{U,V}(u,v)=8(uv)(v)\cdot v=8uv^3 =\underbrace {\left (2u\right )}_{0<u<1}\cdot \underbrace {\left (4v^3\right )}_{0<v<1}.\] The joint density factorises over a rectangle, so \(U\) and \(V\) are independent, with \[f_U(u)=2u\quad (0<u<1),\qquad f_V(v)=4v^3\quad (0<v<1),\] each integrating to \(1\).