Solution. Note that \(Y\) can only take non-negative values (\(y > 0\)). \begin {align*} F_Y(y) & = P(Y \leq y) = P(X^2 \leq y)\\ &= P(-\sqrt {y} \leq X \leq \sqrt {y})\\ &= F_X(\sqrt {y}) - F_X(-\sqrt {y}) \end {align*}

Applying the chain rule to differentiate with respect to \(y\): \begin {align*} f_Y(y) & = \frac {d}{dy} [F_X(\sqrt {y}) - F_X(-\sqrt {y})]\\ & = f_X(\sqrt {y}) \cdot \frac {1}{2\sqrt {y}} - f_X(-\sqrt {y}) \cdot \left (-\frac {1}{2\sqrt {y}}\right )\\ &= \frac {1}{2\sqrt {y}} \left [ f_X(\sqrt {y}) + f_X(-\sqrt {y}) \right ], \quad y > 0 \end {align*} □

Solution. Step 1: Find the Population CDF. For \(x > 0\): \[F_X(x) = \int _0^x 2e^{-2t} dt = 1 - e^{-2x}.\] Step 2: Transform. Since \(x > 0\), the range of \(Y = e^X\) is \(\,(e^0, \infty )\), so \(y > 1\). \begin {align*} F_Y(y) & = P(e^X \leq y) = P(X \leq \ln y)\\ &= F_X(\ln y)\\ &= 1 - e^{-2(\ln y)}\\ & = 1 - e^{\ln (y^{-2})}\\ & = 1 - y^{-2}, \quad y > 1 \end {align*}

Step 3: Differentiate. \[ f_Y(y) = \frac {d}{dy}(1 - y^{-2}) = 2y^{-3}, \quad y > 1 \] □

Solution. Step 1: Determine the support of \(Y\).Since \(0 < x < 1\), we evaluate the boundaries:

• As \(x \to 1\), \(Y = -2\ln (1) = 0\).
• As \(x \to 0^+\), \(Y = -2\ln (0^+) \to \infty \).

Thus, the support of \(Y\) is \(y > 0\).

Step 2: Find the CDF \(F_Y(y)\). \begin {align*} F_Y(y) &= P(Y \leq y)\\ &= P(-2\ln X \leq y)\\ &= P\left (\ln X \geq -\frac {y}{2}\right ) \quad \text {(Note: Inequality flips when dividing by -2)}\\ & = P\left (X \geq e^{-y/2}\right )\\ &= 1 - P\left (X < e^{-y/2}\right )\\ &= 1 - F_X(e^{-y/2}) \end {align*}

Since \(F_X(x) = x\) for a Uniform \((0,1)\) distribution: \[ F_Y(y) = 1 - e^{-y/2}, \quad y > 0 \] Step 3: Find the PDF \(f_Y(y)\) by differentiation. \begin {align*} f_Y(y) &= \frac {d}{dy} \left [ 1 - e^{-y/2} \right ]\\ &= 0 - \left ( e^{-y/2} \cdot -\frac {1}{2} \right )\\ &= \frac {1}{2} e^{-y/2}, \quad y > 0. \end {align*} □

Solution. The inverse is \(x = y + 1\). Substituting this into the original PF: \[ f_Y(y) = f_X(y+1) = pq^{(y+1)-1} = pq^y, \quad y = 0, 1, 2, \dots \] Expert Note: This shifts the ”Geometric” distribution from counting the trial of the first success to counting the number of failures before the first success. □

Solution. Determine the Support of \(Y\): Since \(X \in (0, 1)\):

• As \(x \to 1\), \(y = -\ln (1) = 0\).
• As \(x \to 0^+\), \(y = -\ln (0^+) \to \infty \).

Thus, \(Y \in (0, \infty )\).

Find the Inverse Transformation: \[ y = -\ln x \implies -y = \ln x \implies x = e^{-y} \] So, \(h^{-1}(y) = e^{-y}\).

Calculate the Jacobian: \[ J = \frac {dx}{dy} = \frac {d}{dy}(e^{-y}) = -e^{-y} \] The absolute value of the Jacobian is \(|J| = e^{-y}\).

Substitute into the Transformation Formula: \(f_Y(y) = f_X(h^{-1}(y)) \cdot |J|\): \begin {align*} 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^{-y\alpha + y} (1 - e^{-y})^{\beta - 1} e^{-y}\\ & = \frac {1}{B(\alpha , \beta )} e^{-\alpha y} (1 - e^{-y})^{\beta - 1}, \quad y > 0 \end {align*} □

Solution. Inverse Transformation: \(v = y \implies y = v\) and \(u = xy \implies x = u/v\).

Substituting: \[ g(u,v) = \frac {1}{36} \left (\frac {u}{v}\right )v = \frac {u}{36} \]

The possible values for \(v\) are \(\{1, 2, 3\}\). For each \(v\), \(x\) must be in \(\{1, 2, 3\}\). Since \(u = xv\), if \(v=1, u \in \{1,2,3\}\); if \(v=2, u \in \{2,4,6\}\); if \(v=3, u \in \{3,6,9\}\). The support is \[B = \{(u,v) : v \in \{1,2,3\}, u/v \in \{1,2,3\}\}.\]

Solution. \(f_{X,Y}(x,y) = f_X(x)\cdot f_Y(y)\)

\[f_X(x) = \frac {1}{\lambda }\, e^{-x/\lambda } \quad \text {and}\quad f_Y(y) = \frac {1}{\lambda }\hspace {0.1cm}e^{-y/\lambda }\] Let \(Z = X + Y \) , then

\begin {align*} f_Z(z) = \int \limits _x f_X(x) f_Y(z-x)\, dx & = \int \limits _x \frac {1}{\lambda } e^{-x/\lambda } \cdot \frac {1}{\lambda } e^{-(z-x)/\lambda } dx\\ & = \int \limits _x \frac {1}{\lambda ^2} e^{-z/\lambda } dx\\ & = \int ^z_0 \frac {1}{\lambda ^2} e^{-z/\lambda } dx.\\ & = \frac {z}{\lambda ^2} e^{-z/\lambda }\, , \quad z>0,\, \lambda > 0. \end {align*}

\[f_Z(z) = \frac {z^{2-1}}{\lambda ^2}e^{-3/\lambda } \thicksim \operatorname {GAM}(z,\lambda )\] Thus, \(X + Y \thicksim \operatorname {GAM}(z,\lambda )\). □

Solution. \(f_X(x) = \frac {1}{\sqrt {2\pi }}e^{-\frac {1}{2}x^2}\) and \(f_Y(y) = \frac {1}{\sqrt {2\pi }}e^{-\frac {1}{2}y^2}.\)

Let \(W = X - Y\), then \begin {align*} f_W(w) & = \int ^{\infty }_{-\infty } f_X(w+y)f_Y(y) dy\\ & = \int ^{\infty }_{-\infty } \frac {1}{\sqrt {2\pi }}\exp \left \{-\frac {1}{2}(w + y)^2\right \}\cdot \frac {1}{\sqrt {2\pi }}\exp \left \{-\frac {1}{2}y^2\right \}\, dy\\ & = \frac {1}{2\pi } \int ^{\infty }_{-\infty } \exp \Big \{ -\frac {1}{2}(w^2 + 2yw + 2y^2)\Big \}\, dy\\ & = \frac {\exp \Big \{-\frac {1}{2}w^2\Big \}}{2\pi } \int ^{\infty }_{-\infty } \exp \Big \{-\frac {2}{2}\Big (y^2 + yw + \Big (\frac {1}{2}w\Big )^2 - \Big (\frac {1}{2}w\Big )^2\Big )\Big \}\, dy \\ & = \frac {\exp \Big \{-\frac {1}{2}w^2\Big \}}{2\pi } \int ^{\infty }_{-\infty } \exp \Big \{-\Big (y + \frac {1}{2}w\Big )^2\Big \}\cdot \exp \Big \{\frac {1}{4}w^2\Big \}\, dy\\ & = \frac {e^{-\frac {1}{4}w^2}}{\sqrt {2\pi }}\cdot \sqrt {2\pi } \underbrace {\int ^{\infty }_{-\infty } \frac {1}{\sqrt {2 \pi \cdot 1/2}}\exp \Big \{-\frac {1}{2}\cdot \frac {(y + \frac {1}{2}w)^2}{1/2}\Big \}\, dy}_1 \end {align*}

Therefore \[f_W(w) = \frac {e^{-\frac {1}{4}w^2}}{\sqrt {4\pi }}\, ,\quad \quad -\infty < w < \infty \quad N(0,2)\] □

Proof. \begin {align*} F_(Z) & = P(Y/X \leq z)\\ & = P(Y/X \leq z\quad \text {and}\quad X\geq 0) + P(Y/X\leq z \quad \text {and}\quad X<0)\\ & = P(Y \leq zX\quad \text {and}\quad X\geq 0) + P(Y\geq zX \quad \text {and}\quad X<0)\\ & = P(Y \leq zX\quad \text {and}\quad X\geq 0) \,+\, 1- P(Y\leq zX \quad \text {and}\quad X<0)\\ & = \int _0^{\infty }\int _{-\infty }^{zx} f_X(x)\, f_Y(y)\, dy dx\, +\, 1 - \int ^0_{-\infty }\int _{-\infty }^{zx} f_X(x)f_Y(y)\, dy dx. \end {align*}

Differentiating \(F_Z(z)\) we obtain \begin {align*} f_Z(z) & = \frac {d}{dz}F_Z(z)\\ & =\frac {d}{dz}\int _0^{\infty }\int _{-\infty }^{zx}f_X(x) f_Y(y)\, dy dx - \frac {d}{dz}\int _{-\infty }^0\int _{-\infty }^{zx}f_X(x)f_Y(y)\, dy dx\\ & = \int ^{\infty }_0f_X(x)\left (\frac {d}{dz}\int _{-\infty }^{zx} f_Y(y)\, dy\right )dx - \int ^0_{-\infty }f_X(x)\left (\frac {d}{dz}\int ^{zx}_{-\infty } f_Y(y) dy\right )dx \end {align*}

Note that by the Fundamental Theorem of Calculus and the chain rule, we find \[\frac {d}{dz}\int _{-\infty }^{zx}f_Y(y)\, dy = f_Y(zx)\, \frac {d}{dz}(zx) = xf_Y(zx).\] Hence \begin {align*} f_Z(z) & = \int ^{\infty }_0xf_X(x)f_Y(zx)\, dx - \int ^0_{-\infty }xf_X(x)f_Y(zx)\, dx\\ & = \int ^{\infty }_0xf_X(x)f_Y(zx)\, dx + \int ^0_{-\infty }(-x)f_X(x)f_Y(zx)\, dx\\ & = \int ^{\infty }_0|x|f_X(x)f_Y(zx)\, dx + \int ^0_{-\infty }|x|f_X(x) f_Y(zx)\, dx\\ & = \int ^{\infty }_{-\infty }|x|\, f_X(x)f_Y(zx)\, dx. \end {align*} □

Solution. We can write \begin {align*} f_Z(z) & = \int ^{\infty }_0x(\lambda e^{-\lambda zx}(\lambda e^{-\lambda zx})\, dx\\ & = \lambda ^2\int ^{\infty }_0xe^{-\lambda (1 + z)x}\, dx\\ & = \frac {\lambda ^2}{\lambda (1 + z)}\int ^{\infty }_0x\lambda (1 + z)e^{-\lambda (1 + z)x}\, dx\\ & = \frac {\lambda ^2}{\lambda (1 + z)}\cdot \frac {1}{\lambda (1+z)}\int ^{\infty }_0t e^{-t}\, dt,\quad \quad \text {letting} \,\, t = x\lambda (1 + z)\\ & = \frac {\lambda ^2}{\lambda ^2 (1 + z)^2}\, 1!\\ & = \frac {1}{(1+z)^2}\, , \quad z\geq 0 \end {align*} □

Solution. The distribution for \(Z\) is

and for \(W\) is

Proof. \begin {align*} M_Y(t) & = E(e^{tY})\\ & = E\left (e^{t(X_1 + X_2 + \cdots + X_n)}\right )\\ & = E\left (e^{tX_1 + tX_2 + \cdots + tX_n}\right )\\ & = E\left (e^{tX_1}\right ) E\left (e^{tX_2}\right ) \cdots E\left (e^{tX_n}\right )\quad \quad \text {since the}\,\,\, X_i's\,\,\, \text {are independent}\\ & = M_{X_1}(t)\, M_{X_2}(t)\, \cdots \, M_{X_n}(t) \end {align*} □

Solution. \[f_{X_i}(x_i) = \frac {e^{-\mu _i}\, \mu _i^{x_i}}{x_i!}\, , \quad x_i = 0, 1, 2, \cdots \] Recall that \[M_{X_i}(t) = e^{\mu _i(e^t - 1)}\] now \begin {align*} M_Y(t) & = \prod ^n_{i =1} M_{X_i}(t)\\ & = \prod ^n_{i =1} e^{\mu _i(e^t-1)}\\ & = e^{\left (\sum ^n_{i =1}\mu _i\right )(e^t - 1)} \end {align*}

Therefore, \(\, Y \thicksim \operatorname {POI}\left (\sum ^n_{i = 1}\mu _i\right )\). □

Proof. Recall \[M_{X_i}(t) = e^{\mu _i t + \frac {1}{2}\sigma ^2_it^2}\] Let \(Y = \sum ^n_{i = 1} a_i X_i\) \begin {align*} M_Y(t) & = E\left (e^{tY}\right )\\ & = E\left (e^{t\sum ^n_{i = 1}a_iX_i}\right )\\ & = \prod ^n_{i = 1}E\left (e^{ta_iX_i}\right )\\ & = \prod ^n_{i = 1}e^{\mu _ia_it + \frac {1}{2}\sigma ^2_ia^2_it^2}\\ & = e^{\left (\sum ^n_{i =1}\mu _ia_i\right )t + \frac {1}{2}\left (\sum ^n_{i = 1}\sigma ^2_ia^2_i\right )t^2} \end {align*}

Therefore \[Y = \sum ^n_{i = 1} a_iX_i \, \thicksim \, \operatorname {N}\left (\sum ^n_{i = 1}a_i \mu _i\, , \, \sum ^n_{i = 1} a_i^2\sigma ^2_i\right ).\] □

Proof. \(M_{X_i}(t) = e^{\mu t + \frac {1}{2}\sigma ^2 t^2}\)

Let \(\, Y = \sum ^n_{i = 1} X_i\) \begin {align*} M_Y(t) & = E(e^{tY})\\ & = E\left (e^{t\sum ^n_{i = 1} X_i}\right )\\ & = \left [M_X(t)\right ]^n\\ & = \left (e^{\mu t + \frac {1}{2}\sigma ^2 t^2}\right )^n\\ & = e^{n\mu t + \frac {1}{2}n\sigma ^2t^2}. \end {align*}

Therefore \[Y = \sum ^n_{i = 1}X_i \thicksim \operatorname {N}(n\mu , n\sigma ^2)).\] □

Proof. \begin {align*} M_X(t) & = E\left (e^{tX}\right )\\ & = \int ^{\infty }_0e^{tx}\, \frac {x^{\frac {n}{2-1}}\, e^{-\frac {x}{2}}}{2^{\frac {n}{2}}\, \Gamma \left (\frac {n}{2}\right )}\, dx\\ & = \frac {1}{2^{\frac {n}{2}}\, \Gamma \left (\frac {n}{2}\right )}\,\int _0^{\infty }x^{\frac {n}{2}-1}\, e^{-(\frac {1}{2}-t)x}\, dx\\ & = \frac {1}{2^{\frac {n}{2}}\, \Gamma \left (\frac {n}{2}\right )}\,\int _0^{\infty }\left (\frac {y}{\frac {1}{2}-t}\right )^{\frac {n}{2}-1}\, e^{-y}\, \frac {dy}{\frac {1}{2}-t}\, ,\quad \text {letting}\,\, y = (\frac {1}{2}-t)x\\ & = \frac {1}{2^{\frac {n}{2}}\Gamma \left (\frac {n}{2}\right )}\, \frac {1}{\left (\frac {1}{2}-t\right )^{\frac {n}{2}}}\int _0^{\infty } y^{\frac {n}{2}-1}\, e^{-y}\, dy \, , \quad t > \frac {1}{2}\\ & = \frac {\Gamma \left (\frac {n}{2}\right )}{\Gamma \left (\frac {n}{2}\right ) (1-2t)^{\frac {n}{2}}}\\ & = \left (\frac {1}{1-2t}\right )^{\frac {n}{2}}\, , \quad \quad t > \frac {1}{2}. \end {align*}

\[M'_X(t) = -\frac {n}{2}(1 - 2t)^{-\frac {n}{2}-1}(-2) = n(1 - 2t)^{-\frac {n}{2}-1}.\]

\[M''_X(t) = n\left (-\frac {n}{2}-1\right )(1-2t)^{-\frac {n}{2}-2}(-2) = 2n\left (\frac {n}{2} + 1\right )(1 - 2t)^{-\frac {n}{2}-2}.\]

\[E(X) = M'_X(0) = n.\]

\[E(X^2) = M''_X(0) = 2n\left (\frac {n}{2} + 1\right ).\]

\[\operatorname {Var}(X) = 2n\left (\frac {n}{2}+1\right ) - n^2 = n^2 + 2n -n^2 = 2n.\] □

Proof. We find the m.g.f of \(W\) \begin {align*} M_W(t) = E\left (e^{tw}\right ) & = E\left ( e^{t\sum ^k_{j=1}\left (\frac {X_j-\mu _j}{\sigma _j}\right )^2}\right )\\ & = E\left ( e^{\sum \limits ^k_{j=1} Z_j^2t}\right )\,\,\quad \text {where}\quad Z_j= \frac {X_j - \mu _j}{\sigma _j}\\ & = E\left ( \prod ^k_{j=1} e^{tZ^2_j}\right )\\ & = \prod ^k_{j=1} E\left (e^{tZ^2_j}\right )\\ & = \left [E\left (e^{tZ^2}\right )\right ]^k\quad , \quad Z \thicksim N(0,1) \end {align*}

now find \begin {align*} E\left (e^{tz^2}\right ) & = \int ^{\infty }_{-\infty } e^{tZ^2}\, f_Z(z)\, dz\\ & = \int ^{\infty }_{-\infty } e^{tz^2}\, \frac {1}{\sqrt {2\pi }}e^{-\frac {1}{2}z^2}\,dz\\ & = \int ^{\infty }_{-\infty } \frac {1}{\sqrt {2 \pi }}e^{-\frac {1}{2}(1 - 2t)z^2}\, dz\\ & = \int ^{\infty }_{-\infty } \frac {1}{\sqrt {2\pi }}e^{-\frac {1}{2} u^2} \, \frac {du}{\sqrt {1 - 2t}}\, ,\quad \quad \text {letting}\,\, u = \sqrt {1 - 2t}\,z\\ & = \frac {1}{\sqrt {1 - 2t}}\underbrace {\int ^{\infty }_{-\infty } \frac {1}{\sqrt {2\pi }}e^{-\frac {1}{2} u^2} \, du}_1\\ & = \frac {1}{\sqrt {1 - 2t}}. \end {align*}

Therefore, \[M_w(t) = \left [E\left (e^{tz^2}\right )\right ]^k = \left [ \frac {1}{\sqrt {1 - 2t}}\right ]^k\] By uniqueness theorem of m.g.f. \(W\) has a chi-square distribution with \(k\) degrees of freedom. □

Proof. We find the joint p.d.f of \(U\) and \(V\) \begin {align*} f_{U,V}(u,v) & = f_U(u) \cdot f_V(v)\\ & = \frac {u^{\frac {m}{2}-1}\, e^{-\frac {u}{2}}}{\Gamma \left (\frac {m}{2}\right )\, 2^{\frac {m}{2}}}\cdot \frac {v^{\frac {n}{2}-1}\, e^{-\frac {v}{2}}}{\Gamma \left (\frac {n}{2}\right )\, 2^{\frac {n}{2}}}\\ & = \frac {u^{\frac {m}{2}-1}v^{\frac {n}{2}-1} e^{\frac {-(u+v)}{2}}} {2^{\frac {m + n}{2}}\, \Gamma \left (\frac {m}{2}\right ) \, \Gamma \left (\frac {n}{2}\right )}. \end {align*}

We make a transformation, \[X = \frac {U/m}{V/n} \quad \text {and} \quad Z = V\]

\[X = \frac {U/m}{Z/n} \quad \implies \quad \frac {ZX}{n} = \frac {u}{m}\]

\[\implies \quad U = \frac {m}{n}ZX\quad ;\quad V = Z\] and the Jacobian \[J = \begin {vmatrix} \frac {\partial U}{\partial x} & \frac {\partial U}{\partial z}\\ \frac {\partial V}{\partial x} & \frac {\partial V}{\partial z}\\ \end {vmatrix} = \begin {vmatrix} \frac {m}{n}z & \frac {m}{n}x\\ 0 & 1\\ \end {vmatrix} = \frac {m}{n}z.\]

We find the joint p.d.f of \(X\) and \(Z\) \begin {align*} f_{X,Z}(x,z) & = f_{U,V}(x,z)\, |J|\\ & = \frac {\left (\frac {m}{n}zx\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 )}\,\begin {vmatrix} \frac {m}{n}z\\ \end {vmatrix}\\ & = \frac {\left (\frac {m}{n}\right )^{\frac {m}{2}}z^{\frac {m+n}{2}-1}\, x^{\frac {m}{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 )}\quad ,\quad \quad x>0\, , \, z>0. \end {align*}

Now we find the marginal p.d.f of \(X\) \begin {align*} f_X(x) & = \int \limits _Z f_{X,Z}(x,z)\, dz\\ & = \frac {\left (\frac {m}{n}\right )^{\frac {m}{2}}\, x^{\frac {m}{2}-1}}{2^{\frac {m+n}{2}}\, \Gamma \left (\frac {m}{2}\right )\, \Gamma \left (\frac {n}{2}\right )}\int ^{\infty }_0 z^{\frac {m+n}{2}-1}\, e^{-\frac {z}{2}\left (1 + \frac {m}{n}x\right )}\, dz. \end {align*}

Let \(y = \frac {z}{2}\left (1 + \frac {m}{n}x\right )\) \[\implies \quad dy = \frac {1}{2}\left (1 + \frac {m}{n}x\right )\, dz\] \[dz = \frac {2}{\left (1 + \frac {m}{n}x\right )}\, dy\] now replacing \begin {align*} f_X(x) & = \frac {\left (\frac {m}{n}\right )^{\frac {m}{2}}\, x^{\frac {m}{2}-1}}{2^{\frac {m+n}{2}}\, \Gamma \left (\frac {m}{2}\right )\,\Gamma \left (\frac {n}{2}\right )}\int ^{\infty }_0 \left (\frac {2y}{1 + \frac {mx}{n}}\right )^{\frac {m+n}{2}-1}\, e^{-y}\, \left (\frac {2}{1 + \frac {m}{n}x}\right )\, dy\\ & = \frac {\left (\frac {m}{n}\right )^{\frac {m}{2}}\, x^{\frac {m}{2}-1}\, 2^{\frac {m+n}{2}}}{2^{\frac {m+n}{2}}\, \Gamma \left (\frac {m}{2}\right )\,\Gamma \left (\frac {n}{2}\right )\, \left (1 + \frac {m}{n}x\right )^{\frac {m + n}{2}}} \int ^{\infty }_0 y^{\frac {m+n}{2}-1}\, e^{-y} \, dy\\ & =\frac {\left (\frac {m}{n}\right )^{\frac {m}{2}}\, x^{\frac {m}{2}-1}\,\Gamma \left (\frac {m + n}{2}\right )}{\Gamma \left (\frac {m}{2}\right )\,\Gamma \left (\frac {n}{2}\right )\, \left (1 + \frac {m}{n}x\right )^{\frac {m + n}{2}}}. \end {align*} □

Proof. Let \(Z \thicksim N(0,1)\) and \(U\thicksim \chi ^2_m\) with \(Z\) and \(U\) being independent.

We first find the joint p.d.f. of \(Z\) and \(U\) \begin {align*} f_{Z,U} (z,u) & = f_Z(z)\, f_U(u)\\ & = \frac {1}{\sqrt {2\pi }}e^{-\frac {1}{2}z^2}\, \frac {u^{\frac {m}{2}} e^{-\frac {u}{2}}}{\Gamma \left (\frac {m}{2}\right )\, 2^{\frac {m}{2}}}\\ & = \frac {u^{\frac {m}{2}-1}\, e^{-\frac {1}{2}(u + z^2)}}{\sqrt {2\pi }\, 2^{\frac {m}{2}}\, \Gamma \left (\frac {m}{2}\right )}\, ,\quad z \in (-\infty , \infty )\, , \quad u\in (0,\infty ). \end {align*}

Make a transformation \[X = U\quad \text {and}\quad Y = \frac {Z}{\sqrt {\frac {U}{m}}}\quad \implies \quad U = x\, , \quad Z = Y\sqrt {\frac {X}{m}}.\] The Jacobian \[J = \begin {vmatrix} \frac {\partial U}{\partial x} & \frac {\partial U}{\partial y}\\ \frac {\partial Z}{\partial x} & \frac {\partial Z}{\partial y}\\ \end {vmatrix} = \begin {vmatrix} 1 & 0 \\ ? & \sqrt {\frac {x}{m}}\ \end {vmatrix} = \sqrt {\frac {x}{m}}.\]

Then the joint p.d.f. of \(X\) and \(Y\) \[f_{X,Y}(x,y) = \frac {x^{\frac {m}{2}-1}\, e^{-\frac {1}{2}\left (x + y^2 \frac {x}{m}\right )}}{\sqrt {2\pi }\, 2^{\frac {m}{2}}\, \Gamma \left (\frac {m}{2}\right )} \, \sqrt {\frac {x}{m}}\, ,\quad x>0\, , \quad y \in (-\infty , \infty ).\]

We now find the marginal of \(Y\) \[f_Y(y) = \int \limits _X f_{X,Y}(x,y)\, dy\\ = \frac {1}{\sqrt {2\pi }\, 2^{\frac {m}{2}}\, \Gamma \left (\frac {m}{2}\right )\, \sqrt {m}}\int ^{\infty }_0 x^{\frac {m-1}{2}}\, e^{-\frac {1}{2}\left (x + y^2 \frac {x}{m}\right )}\, dx.\]

Let \(\, V = \frac {1}{2}\left (1 + \frac {y^2}{m}\right )\quad x \implies \quad dV = \frac {1}{2}\left (1 + \frac {y^2}{m}\right )\, dx\)

\[dx = \frac {2 dv}{\left (1 + \frac {y^2}{m}\right )}\quad , \quad X = \frac {2 v}{1 + \frac {y^2}{m}}.\] Thus \begin {align*} f_Y(y) & = \frac {1}{\sqrt {2\pi }\, 2^{\frac {m}{2}}\, \Gamma \left (\frac {m}{2}\right )\, \sqrt {m}} \int ^{\infty }_0 \left (\frac {2 v}{1 +\frac {y^2}{m}}\right )^{\frac {m - 1}{2}}\, e^{-v}\, \frac {2dv}{\left (1 + \frac {y^2}{m}\right )}\\ & = \frac {2^{\frac {m+1}{2}}}{\sqrt {2\pi }\, 2^{\frac {m}{2}}\, \Gamma \left (\frac {m}{2}\right )\, \sqrt {m}\, \left ( 1 + \frac {y^2}{m}\right )^{\frac {m+1}{2}}}\, \underbrace {\int ^{\infty }_0 v^{\frac {m - 1}{2}}\, e^{-1}\, dv}_{\Gamma \left (\frac {m+1}{2}\right )}\\ & = \frac {\Gamma \left (\frac {m + 1}{2}\right )}{\sqrt {\pi }\, \Gamma \left (\frac {m}{2}\right )\, \left (1 + \frac {y^2}{m}\right )^{\frac {m+1}{2}}\, \sqrt {m}}\\ & = \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}}}\, ,\quad y \in (-\infty , \infty ). \end {align*} □