Proof. If \(g\) is increasing, \(g(X)\leq y\) exactly when \(X\leq g^{-1}(y)\); if decreasing, exactly when \(X\geq g^{-1}(y)\). Differentiate by the chain rule. □

Solution. The function \(x^2\) is not monotone, so apply the method directly. For \(y > 0\), \[F_Y(y) = P(X^2 \leq y) = P(-\sqrt {y} \leq X \leq \sqrt {y}) = F_X(\sqrt {y}) - F_X(-\sqrt {y}) ,\] and \(F_Y(y) = 0\) for \(y\leq 0\). Differentiating by the chain rule, \[f_Y(y) = f_X(\sqrt {y}) \cdot \frac {1}{2\sqrt {y}} + f_X(-\sqrt {y}) \cdot \frac {1}{2\sqrt {y}} = \frac {1}{2\sqrt {y}} \left [ f_X(\sqrt {y}) + f_X(-\sqrt {y}) \right ], \qquad y > 0 .\] For \(X\sim N(0,1)\) this gives \(f_Y(y) = \frac {y^{-1/2}e^{-y/2}}{\sqrt {2\pi }}\), the \(\chi ^2(1)\) density: the square of a standard normal variable is chi-square with one degree of freedom.

Solution. For \(x > 0\), \(F_X(x) = \int _0^x 2e^{-2t}\, dt = 1 - e^{-2x}\). Since \(x > 0\), \(Y = e^X\) takes values in \((1, \infty )\), and \(e^x\) is increasing, so for \(y>1\), \[F_Y(y) = P(e^X \leq y) = P(X \leq \ln y) = 1 - e^{-2\ln y} = 1 - y^{-2} .\] Differentiating, \(f_Y(y) = 2y^{-3}\) for \(y > 1\), a Pareto distribution.

Solution. As \(x\) runs from \(1\) down to \(0^+\), \(y = -2\ln x\) runs from \(0\) up to \(\infty \), so the support of \(Y\) is \(y > 0\). The function is decreasing, and for \(y>0\) \[F_Y(y) = P(-2\ln X \leq y) = P\left (\ln X \geq -\frac {y}{2}\right ) = P\left (X \geq e^{-y/2}\right ) = 1 - e^{-y/2},\] the inequality reversing on division by \(-2\). Hence \(f_Y(y) = \frac {1}{2} e^{-y/2}\) for \(y > 0\).

Solution. For \(0<u<1\), \(P(U\leq u) = P(F(X)\leq u) = P\left (X\leq F^{-1}(u)\right ) = F\left (F^{-1}(u)\right ) = u\), which is the uniform c.d.f.

Proof. \(P(Y = y) = P(X\in \{x : g(x) = y\})\), which is a single term when \(g\) is one to one. □

Solution. The inverse is \(x = y + 1\), so \[f_Y(y) = f_X(y+1) = pq^y, \qquad y = 0, 1, 2, \dots .\] The distribution of \(X\) counts the trial on which the first success occurs; \(Y\) counts the failures before it.

Proof. A one-to-one differentiable function with non-zero derivative is strictly monotone, so this is Theorem 1.3.1; the absolute value covers both the increasing and the decreasing case. □

Solution. As in Example 1.3.4, the support of \(Y\) is \((0,\infty )\). The inverse is \(x = e^{-y}\), with \(J = \frac {dx}{dy} = -e^{-y}\) and \(|J| = e^{-y}\). By Theorem 1.3.9, \[f_Y(y) = \frac {1}{B(\alpha , \beta )} (e^{-y})^{\alpha - 1} (1 - e^{-y})^{\beta - 1} \cdot e^{-y} = \frac {1}{B(\alpha , \beta )} e^{-\alpha y} (1 - e^{-y})^{\beta - 1}, \qquad y > 0 .\] With \(\beta = 1\) this is \(\alpha e^{-\alpha y}\): if \(X\sim \text {BETA}(\alpha ,1)\), then \(-\ln X\sim \text {EXP}\left (\frac 1\alpha \right )\).

Proof. For a region \(C\subseteq B\), \(P((U,V)\in C) = P((X,Y)\in w(C)) = \iint _{w(C)}f_{X,Y}\,dx\,dy\), and by the change of variables formula for double integrals this is \(\iint _C f_{X,Y}(w_1,w_2)|J|\,du\,dv\). □

Solution.

(i)

By independence, the joint p.d.f. is \[ f_{X,Y}(x,y) = \frac {1}{\Gamma (a)\Gamma (b)} x^{a-1} y^{b-1} e^{-(x+y)}, \qquad x, y > 0 .\] The transformation \(u = x + y\), \(v = \frac {x}{x+y}\) has inverse \(x = uv\), \(y = u(1-v)\), and maps the quadrant \(x,y>0\) onto the strip \(u > 0\), \(0 < v < 1\).

The Jacobian is \[ J = \begin {vmatrix} v & u \\ 1-v & -u \end {vmatrix} = -uv - u(1-v) = -u ,\qquad |J| = u ,\] so by Theorem 1.3.11 \begin {align*} f_{U,V}(u,v) &= \frac {(uv)^{a-1} (u(1-v))^{b-1} e^{-u}}{\Gamma (a)\Gamma (b)}\cdot u\\ &= \frac {u^{a+b-1} e^{-u}\, v^{a-1} (1-v)^{b-1}}{\Gamma (a)\Gamma (b)} ,\qquad u>0,\ 0<v<1 . \end {align*}

(ii)

The joint p.d.f. factorises as \[f_{U,V}(u,v) = \frac {u^{a + b - 1} e^{-u}}{\Gamma (a + b)} \times \frac {\Gamma (a + b)\, v^{a - 1}(1 - v)^{b - 1}}{\Gamma (a)\Gamma (b)} ,\] a function of \(u\) times a function of \(v\), on the product set \((0,\infty )\times (0,1)\). By the factorisation criterion, \(U\) and \(V\) are independent.

(iii)

The two factors are themselves densities: integrating out \(v\) uses \(\int _0^1v^{a-1}(1 - v)^{b-1}\,dv = B(a,b) = \frac {\Gamma (a)\Gamma (b)}{\Gamma (a + b)}\), and integrating out \(u\) uses \(\int _0^\infty u^{a+b-1}e^{-u}\,du = \Gamma (a + b)\). So \[f_U(u) = \frac {u^{a + b - 1} \, e^{-u}}{\Gamma (a + b)},\ u > 0,\qquad f_V(v) = \frac {v^{a - 1}\, (1 - v)^{b - 1}}{B(a,b)},\ 0 < v < 1 ,\] that is, \(U\sim \text {GAM}(a + b ,1)\) and \(V \sim \text {BETA}(a,b)\).

Solution. The inverse is \(y = v\), \(x = \frac {u}{v}\), so \[g(u,v) = \frac {1}{36} \cdot \frac {u}{v}\cdot v = \frac {u}{36}\] on the support \(B = \{(u,v) : v \in \{1,2,3\},\ \frac {u}{v} \in \{1,2,3\}\}\): for \(v = 1\), \(u\in \{1,2,3\}\); for \(v = 2\), \(u\in \{2,4,6\}\); for \(v = 3\), \(u\in \{3,6,9\}\). The joint and marginal p.f.s are shown in the table.

Solution.

(i)

The joint p.f. of \(X\) and \(Y\) is \[f(x,y) = \frac {e^{-(\mu + \lambda )}\, \mu ^x\, \lambda ^y}{x!\, y!} ,\qquad x, y = 0, 1, 2, \dots .\] The inverse of \(u = x + y\), \(v = y\) is \(y = v\), \(x = u - v\), and \(x\geq 0\) means \(v\leq u\), so \[g(u,v) = \frac {e^{-(\mu + \lambda )}\, \mu ^{u-v}\, \lambda ^v}{(u - v)!\, v!} ,\qquad u = 0, 1, 2, \dots ;\ v = 0, 1, \dots , u .\]

(ii)

Summing over \(v\) and using the binomial theorem, \[g_U(u) = \frac {e^{-(\mu + \lambda )}}{u!}\sum _{v = 0}^u\binom {u}{v}\mu ^{u - v}\lambda ^v = \frac {e^{-(\mu + \lambda )}\,(\mu + \lambda )^u}{u!} ,\qquad u = 0, 1, 2, \dots ,\] so \(U\sim \text {POI}(\mu + \lambda )\).

Solution. Dividing the joint p.f. by the marginal, \[P(V = v\mid U = u) = \frac {g(u,v)}{g_U(u)} = \binom {u}{v}\left (\frac {\lambda }{\mu + \lambda }\right )^v\left (\frac {\mu }{\mu + \lambda }\right )^{u - v},\qquad v = 0,\dots ,u ,\] the binomial distribution with \(u\) trials and success probability \(\frac {\lambda }{\mu + \lambda }\). Given the total number of events, each one comes from the second source with probability proportional to its rate.

Proof. (a) The c.d.f. is the integral of \(f\) over the half-plane \(x + y \leq z\).

Integrating \(x\) first, \[F_Z(z) = \iint \limits _{x+y \leq z} f(x,y)\,dx\, dy = \int _{-\infty }^\infty \int ^{z-y}_{-\infty }f(x,y)\, dx\,dy .\] Differentiating under the integral sign, by the fundamental theorem of calculus, \[f_Z(z) = \int _{-\infty }^\infty \frac {\partial }{\partial z}\left [ \int ^{z-y}_{-\infty }f(x,y)\, dx\right ]dy = \int _{-\infty }^\infty f(z-y,y)\,dy ,\] and integrating \(y\) first instead gives the other form.

(b) The c.d.f. is the integral over the half-plane \(x - y\leq w\), that is \(y\geq x - w\).

So \(F_W(w) = \int _{-\infty }^\infty \int _{x - w}^\infty f(x,y)\,dy\,dx\), and differentiating with respect to \(w\), the lower limit \(x - w\) contributing with a minus sign twice, \[f_W(w) = \int _{-\infty }^\infty f(x, x - w)\,dx .\] The substitution \(x = w + y\) gives the other form. □

Solution. Here \(f_X(x) = \frac {1}{\lambda }e^{-x/\lambda }\) for \(x>0\) and likewise for \(Y\). The integrand \(f_X(x)f_Y(z - x)\) is non-zero only when \(x>0\) and \(z - x>0\), so for \(z>0\), \[f_Z(z) = \int _0^z \frac {1}{\lambda } e^{-x/\lambda } \cdot \frac {1}{\lambda } e^{-(z-x)/\lambda }\, dx = \int ^z_0 \frac {1}{\lambda ^2} e^{-z/\lambda }\, dx = \frac {z}{\lambda ^2} e^{-z/\lambda } ,\qquad z>0 .\] This is \(\frac {z^{2-1}e^{-z/\lambda }}{\Gamma (2)\lambda ^2}\), so \(X + Y \sim \text {GAM}(2,\lambda )\).

Solution. By Corollary 1.3.19, \[f_W(w) = \frac {1}{2\pi } \int ^{\infty }_{-\infty } e^{-\frac {1}{2}(w + y)^2} e^{-\frac {1}{2}y^2}\, dy = \frac {1}{2\pi } \int ^{\infty }_{-\infty } e^{-\frac {1}{2}(w^2 + 2wy + 2y^2)}\, dy .\] Completing the square, \(w^2 + 2wy + 2y^2 = 2\left (y + \frac {w}{2}\right )^2 + \frac {w^2}{2}\), so \[f_W(w) = \frac {e^{-w^2/4}}{2\pi }\int ^{\infty }_{-\infty }e^{-\left (y + \frac {w}{2}\right )^2}\,dy = \frac {e^{-w^2/4}}{2\pi }\sqrt \pi = \frac {e^{-w^2/4}}{\sqrt {4\pi }} ,\qquad -\infty < w < \infty .\] This is the \(N(0,2)\) density: the difference of two independent standard normals is normal with variance \(2\).

Solution. The integrand \(f_X(x)f_Y(z - x)\) is \(1\) when \(0<x<1\) and \(0<z - x<1\), that is \(\max (0,z - 1)<x<\min (1,z)\), and \(0\) otherwise. So \[f_Z(z) = \begin {cases} z, & 0<z\leq 1,\\ 2 - z, & 1<z<2 , \end {cases}\] the triangular distribution on \((0,2)\). The density of a sum is smoother than those of its terms.

Proof. \(\{Z = z\}\) is the disjoint union over \(x\) of the events \(\{X = x,\ Y = z - x\}\), and similarly for \(W\). □

Solution. By Theorem 1.3.24, each probability is the sum of the entries on a diagonal of the table: for \(Z = z\) the cells with \(x + y = z\), and for \(W = w\) the cells with \(x - y = w\). For example \(P(Z = 4) = f(1,3) + f(2,2) + f(3,1) = 0.02 + 0.20 + 0.02 = 0.24\). The results are

and

Each set of probabilities sums to \(1\).

Proof. For the quotient, take \(U = X\) and \(Z = \frac {Y}{X}\), with inverse \(x = u\), \(y = uz\) and Jacobian \(\begin {vmatrix}1 & 0\\ z & u\end {vmatrix} = u\). By Theorem 1.3.11 the joint p.d.f. of \((U,Z)\) is \(f(u,uz)|u|\), and integrating out \(u\) gives \(f_Z\). For the product, take \(U = X\), \(W = XY\), with inverse \(x = u\), \(y = \frac {w}{u}\) and Jacobian \(\frac 1u\), and integrate out \(u\) again. □

Solution. By Corollary 1.3.27, for \(z>0\), \[f_Z(z) = \int ^{\infty }_0x\,\lambda e^{-\lambda x}\,\lambda e^{-\lambda zx}\, dx = \lambda ^2\int ^{\infty }_0xe^{-\lambda (1 + z)x}\, dx = \frac {\lambda ^2}{\lambda ^2 (1 + z)^2} = \frac {1}{(1+z)^2} ,\] using \(\int _0^\infty xe^{-cx}\,dx = \frac {1}{c^2}\). The parameter \(\lambda \) has disappeared, as it must: the ratio of two lifetimes does not depend on the unit of time.

Solution. By Corollary 1.3.27, using the symmetry of the integrand in \(x\), \[f_Z(z) = 2\int _0^\infty x\frac {e^{-x^2/2}}{\sqrt {2\pi }}\frac {e^{-z^2x^2/2}} {\sqrt {2\pi }}\,dx = \frac {1}{\pi }\int _0^\infty xe^{-(1 + z^2)x^2/2}\,dx = \frac {1}{\pi (1 + z^2)} .\] The Cauchy distribution has no mean: \(\int |z|f_Z(z)\,dz\) diverges.

Proof. By independence the expectation of the product factorises: \[M_Y(t) = E\left (e^{tX_1}e^{tX_2}\cdots e^{tX_n}\right ) = E\left (e^{tX_1}\right ) E\left (e^{tX_2}\right ) \cdots E\left (e^{tX_n}\right ) .\] □

Proof. For 1, \(M_{2X/\beta }(t) = M_X\left (\frac {2t}{\beta }\right ) = (1 - 2t)^{-\alpha }\), the m.g.f. of \(\chi ^2(2\alpha )\) (Theorem 1.3.35 below). For 2, \(\prod _i(1 - \beta t)^{-\alpha _i} = (1 - \beta t)^{-\sum \alpha _i}\), and 3 is the case \(\alpha _i = 1\). For 4, each \(Z_i = \frac {X_i - \mu }{\sigma }\) is standard normal, \(Z_i^2\sim \chi ^2(1) = \text {GAM}\left (\frac 12,2\right )\) by Example 1.3.2, and the sum of \(n\) independent such variables is \(\text {GAM}\left (\frac n2,2\right ) = \chi ^2(n)\) by 2. □

Solution. The m.g.f. of \(\text {POI}(\mu _i)\) is \(M_{X_i}(t) = \sum _x e^{tx}\frac {e^{-\mu _i}\mu _i^x}{x!} = e^{\mu _i(e^t - 1)}\), so by Theorem 1.3.30 \[M_Y(t) = \prod ^n_{i =1} e^{\mu _i(e^t-1)} = e^{\left (\sum _i\mu _i\right )(e^t - 1)},\] which is the m.g.f. of \(\text {POI}\left (\sum _i\mu _i\right )\). By uniqueness, \(Y\) has that distribution. This repeats Example 1.3.16 in two lines.

Proof. With \(M_{X_i}(t) = e^{\mu _i t + \frac {1}{2}\sigma ^2_it^2}\) and \(Y = \sum _i a_i X_i\), \[M_Y(t) = \prod ^n_{i = 1}E\left (e^{(a_it)X_i}\right ) = \prod ^n_{i = 1}e^{\mu _ia_it + \frac {1}{2}\sigma ^2_ia^2_it^2} = e^{\left (\sum _i\mu _ia_i\right )t + \frac {1}{2}\left (\sum _i\sigma ^2_ia^2_i\right )t^2},\] the m.g.f. of the stated normal distribution. □

Proof. Take \(a_i = 1\), and then \(a_i = \frac 1n\), in Theorem 1.3.33. □

Proof. For \(t<\frac 12\), substituting \(y = \left (\frac {1}{2}-t\right )x\), \[M_X(t) = \frac {1}{2^{n/2}\Gamma \left (\frac {n}{2}\right )}\int _0^{\infty }x^{\frac {n} {2}-1}\, e^{-\left (\frac {1}{2}-t\right )x}\, dx = \frac {1}{2^{n/2}\Gamma \left ( \frac {n}{2}\right )}\cdot \frac {\Gamma \left (\frac {n}{2}\right )}{\left (\frac {1}{2}-t \right )^{n/2}} = (1-2t)^{-n/2} .\] Then \(M'_X(t) = n(1 - 2t)^{-\frac {n}{2}-1}\) and \(M''_X(t) = n(n + 2)(1 - 2t)^{-\frac {n}{2}-2}\), so \(E(X) = n\), \(E(X^2) = n^2 + 2n\) and \(\text {Var}(X) = 2n\). □

Solution. By special result 3, \(\sum X_i = n\bar X\sim \text {GAM}(n,\beta )\), and by result 1, \(\frac {2n\bar X}{\beta }\sim \chi ^2(2n)\). With \(n = 10\), \(P(\bar X>c\beta ) = P\left (\chi ^2(20)>20c\right ) = 0.05\) when \(20c = 31.41\), the upper \(5\%\) point of \(\chi ^2(20)\), so \(c = 1.571\).

Proof. The variables \(Z_j = \frac {X_j - \mu _j}{\sigma _j}\) are independent standard normal, so by Theorem 1.3.30 \(M_W(t) = \left [E\left (e^{tZ^2}\right )\right ]^k\) with \(Z\sim N(0,1)\). For \(t<\frac 12\), substituting \(u = \sqrt {1 - 2t}\,z\), \[E\left (e^{tZ^2}\right ) = \int ^{\infty }_{-\infty } \frac {1}{\sqrt {2 \pi }}e^{-\frac {1}{2}(1 - 2t)z^2}\, dz = \frac {1}{\sqrt {1 - 2t}}\int ^{\infty }_{-\infty } \frac {1}{\sqrt {2\pi }}e^{-\frac {1}{2} u^2} \, du = \frac {1}{\sqrt {1 - 2t}} .\] So \(M_W(t) = (1 - 2t)^{-k/2}\), and by uniqueness \(W\sim \chi ^2(k)\). □

Proof. The plan is to introduce a second variable \(Z = V\), find the joint p.d.f. of \((X,Z)\) by Theorem 1.3.11, and integrate out \(z\). The joint p.d.f. of \(U\) and \(V\) is \[f_{U,V}(u,v) = \frac {u^{\frac {m}{2}-1}v^{\frac {n}{2}-1} e^{-(u+v)/2}}{2^{\frac {m + n}{2}}\, \Gamma \left (\frac {m}{2}\right ) \, \Gamma \left (\frac {n}{2}\right )} ,\qquad u,v>0 .\] The inverse of \(x = \frac {u/m}{v/n}\), \(z = v\) is \(u = \frac {m}{n}xz\), \(v = z\), with Jacobian \[J = \begin {vmatrix} \frac {\partial u}{\partial x} & \frac {\partial u}{\partial z}\\[1ex] \frac {\partial v}{\partial x} & \frac {\partial v}{\partial z} \end {vmatrix} = \begin {vmatrix} \frac {m}{n}z & \frac {m}{n}x\\ 0 & 1 \end {vmatrix} = \frac {m}{n}z .\] So for \(x,z>0\), \[f_{X,Z}(x,z) = \frac {\left (\frac {m}{n}xz\right )^{\frac {m}{2}-1} z^{\frac {n}{2}-1} e^{-\frac {z}{2}\left (1 + \frac {m}{n}x\right )}}{2^{\frac {m+n}{2}} \Gamma \left (\frac {m}{2}\right )\Gamma \left (\frac {n}{2}\right )}\cdot \frac {m}{n}z = \frac {\left (\frac {m}{n}\right )^{m/2}x^{\frac {m}{2}-1}\, z^{\frac {m+n}{2}-1}\, e^{-\frac {z}{2}\left (1 + \frac {m}{n}x\right )}}{2^{\frac {m+n}{2}} \Gamma \left (\frac {m}{2}\right )\Gamma \left (\frac {n}{2}\right )} .\] Integrating out \(z\) with \(\int _0^\infty z^{a-1}e^{-cz}\,dz = \frac {\Gamma (a)}{c^a}\), where \(a = \frac {m+n}{2}\) and \(c = \frac 12\left (1 + \frac mnx\right )\), \[f_X(x) = \frac {\left (\frac {m}{n}\right )^{m/2}x^{\frac {m}{2}-1}}{2^{\frac {m+n}{2}} \Gamma \left (\frac {m}{2}\right )\Gamma \left (\frac {n}{2}\right )}\cdot \frac {\Gamma \left (\frac {m + n}{2}\right )2^{\frac {m+n}{2}}}{\left (1 + \frac {m}{n}x\right )^{ \frac {m+n}{2}}},\] which is the \(F(m,n)\) density. □

Proof. The joint p.d.f. of \(Z\) and \(U\) is \[f_{Z,U}(z,u) = \frac {1}{\sqrt {2\pi }}e^{-z^2/2}\cdot \frac {u^{\frac {m}{2}-1} e^{-u/2}}{\Gamma \left (\frac {m}{2}\right ) 2^{m/2}} ,\qquad -\infty <z<\infty ,\ u>0 .\] Transform to \(X = U\) and \(Y = \frac {Z}{\sqrt {U/m}}\), with inverse \(u = x\), \(z = y\sqrt {\frac {x}{m}}\) and Jacobian \[J = \begin {vmatrix} \frac {\partial u}{\partial x} & \frac {\partial u}{\partial y}\\[1ex] \frac {\partial z}{\partial x} & \frac {\partial z}{\partial y} \end {vmatrix} = \begin {vmatrix} 1 & 0 \\ \frac {y}{2\sqrt {mx}} & \sqrt {\frac {x}{m}} \end {vmatrix} = \sqrt {\frac {x}{m}} .\] So, for \(x>0\) and all \(y\), \[f_{X,Y}(x,y) = \frac {x^{\frac {m}{2}-1}\, e^{-\frac {x}{2}\left (1 + \frac {y^2}{m}\right )}}{\sqrt {2\pi }\, 2^{m/2}\, \Gamma \left (\frac {m}{2}\right )} \sqrt {\frac {x}{m}} = \frac {x^{\frac {m+1}{2} - 1}e^{-\frac {x}{2}\left (1 + \frac {y^2}{m}\right )}}{\sqrt {2\pi m}\, 2^{m/2}\, \Gamma \left (\frac {m}{2}\right )} .\] Integrating out \(x\) with the same gamma integral, \(a = \frac {m+1}{2}\) and \(c = \frac 12\left (1 + \frac {y^2}{m}\right )\), \[f_Y(y) = \frac {1}{\sqrt {2\pi m}\, 2^{m/2}\, \Gamma \left (\frac {m}{2}\right )}\cdot \frac {\Gamma \left (\frac {m+1}{2}\right )2^{\frac {m+1}{2}}}{\left (1 + \frac {y^2}{m}\right )^{\frac {m+1}{2}}} = \frac {\Gamma \left (\frac {m + 1}{2}\right )}{\sqrt {m \pi }\, \Gamma \left (\frac {m}{2}\right )\left (1 + \frac {y^2}{m}\right )^{\frac {m+1}{2}}} ,\] since \(\frac {2^{\frac {m+1}{2}}}{\sqrt {2}\,2^{m/2}} = 1\). □

Solution. Write \(T = Z\sqrt {m/U}\) with \(Z\) and \(U\) independent. Then \(E(T) = E(Z)E\left ( \sqrt {m/U}\right ) = 0\) when the second factor is finite, which needs \(m>1\). For the variance, \(E(T^2) = E(Z^2)\,m\,E\left (U^{-1}\right )\), and for \(U\sim \chi ^2(m)\), \[E\left (U^{-1}\right ) = \int _0^\infty \frac {u^{\frac m2 - 2}e^{-u/2}}{\Gamma \left ( \frac m2\right )2^{m/2}}\,du = \frac {\Gamma \left (\frac m2 - 1\right )2^{\frac m2 - 1}}{\Gamma \left (\frac m2\right )2^{m/2}} = \frac {1}{m - 2},\qquad m>2 .\] So \(\text {Var}(T) = \frac {m}{m - 2}\), larger than the normal variance \(1\) and tending to it as \(m\) grows.

Solution. Since \(\frac {(m - 1)S_1^2}{\sigma ^2}\sim \chi ^2(m - 1)\) and \(\frac {(n - 1)S_2^2}{\sigma ^2}\sim \chi ^2(n - 1)\) independently, Theorem 1.3.42 gives \[\frac {S_1^2}{S_2^2} = \frac {\frac {(m-1)S_1^2}{\sigma ^2}\big /(m - 1)}{\frac {(n - 1)S_2^2}{\sigma ^2}\big /(n - 1)}\sim F(m - 1, n - 1) ,\] the common \(\sigma ^2\) cancelling. With \(m = n = 10\) this is \(F(9,9)\), whose upper \(5\%\) point is \(3.18\) and upper \(10\%\) point is \(2.44\), so the probability lies between \(0.05\) and \(0.10\); more precisely it is about \(0.054\). This ratio is the test statistic for equality of two variances.