3.2 Homework Problems (Winter 2020)

Exercise 3.12 (Homework 1, Problem 1) Let $X\sim exp(\lambda)$, where $E(X)=\frac{1}{\lambda}$. What is the p.m.f. of $Y=\lfloor X\rfloor$ (the floor of $X$)? Do you recognize it as a distribution that you have studied in the past?

Proof. For nonnegative integer $a$, and becasue $X$ is a continuous random variable, we have $$\begin{equation} \begin{split} Pr(Y=a)&=Pr(X<a+1)-Pr(X<a)\\ &=(1-e^{-\lambda(a+1)})-(1-e^{-\lambda a})\\ &=e^{-\lambda a}(1-e^{-\lambda}) \end{split} \tag{3.43} \end{equation}$$ and if $a$ is negative integer, then $Pr(Y=a)=0$. Thus, we have the pmf of Y as $$\begin{equation} p_Y(y)=\left\{ \begin{aligned} &e^{-\lambda y}(1-e^{-\lambda}) & y\in\{0,1,2,\cdots\} \\ & 0 & otherwise \end{aligned} \right. \tag{3.44} \end{equation}$$ Form (3.44) we can also recognize that Y has a geometric distribution with parameter $p=e^{-\lambda}$.

Exercise 3.13 (Homework 1, Problem 2) Let $X_1$ and $X_2$ be two independent random variables such that $X_i\sim Gamma(a_i,b)$ for any $a_1,a_2,b>0$. Define $Y=\frac{X_1}{X_1+X_2}$ and $Z=X_1+X_2$.

1. Find the joint pdf for $Y$ and $Z$ and show that these two random variables are independent.

2. Find the marginal pdf of $Z$. Do you recognize this pdf as belonging to some family that you know?

3. Find the marginal pdf of $Y$ . Do you recognize this pdf as belonging to some family that you know?

4. Compute $E(Y^k)$ for any $k>0$.

5. What does this result imply if $a_1=a_2=b=1$?

Proof. (a) Since $X_1$ and $X_2$ are independent $Ga(a_i,b)$, the joint pdf is then $$\begin{equation} \begin{split} f(x_1,x_2)&=\frac{b^{a_1}}{\Gamma(a_1)}x_1^{a_1-1}exp(-bx_1)\times\frac{b^{a_2}}{\Gamma(a_2)}x_2^{a_2-1}exp(-bx_2)\\ &=\frac{b^{a_1+a_2}}{\Gamma(a_1)\Gamma(a_2)}x_1^{a_1-1}exp(-bx_1)x_2^{a_2-1}exp(-bx_2) \end{split} \tag{3.45} \end{equation}$$ for $x_1>0$ and $x_2>0$. Define variable transformation as (3.46), $$\begin{equation} \left\{ \begin{aligned} & y=\frac{x_1}{x_1+x_2} \\ & z=x_1+x_2 \end{aligned} \right. \tag{3.46} \end{equation}$$ Then we have $$\begin{equation} \left\{ \begin{aligned} & x_1=yz \\ & x_2=(1-y)z \end{aligned} \right. \tag{3.47} \end{equation}$$ The Jacobian corresponding to (3.47) has determinant $$\begin{equation} |J|=\begin{vmatrix} z & y\\ -z & 1-y \end{vmatrix}=z \tag{3.48} \end{equation}$$ Thus, the joint distribution of $Y$ and $Z$ can be written as $$\begin{equation} \begin{split} f(y,z)&=\frac{b^{a_1+a_2}}{\Gamma(a_1)\Gamma(a_2)}(yz)^{a_1-1}exp(-byz)((1-y)z)^{a_2-1}exp(-b(1-y)z)z\\ &=y^{a_1-1}(1-y)^{a_2-1}\frac{b^{a_1+a_2}}{\Gamma(a_1)\Gamma(a_2)}z^{a_1+a_2-1}exp(-bz)\\ &=f(y)\cdot f(z) \end{split} \tag{3.49} \end{equation}$$ for $0<y<1$ and $z>0$.Since the joint pdf can be factorized as a function of y only and a function of z only, random variables Y and Z are independent.

2. To get the marginal pdf of Z, we need to integrate out Y from the joint pdf. Therefore $$\begin{equation} \begin{split} f_Z(z)&=\int_0^1y^{a_1-1}(1-y)^{a_2-1}\frac{b^{a_1+a_2}}{\Gamma(a_1)\Gamma(a_2)}z^{a_1+a_2-1}exp(-bz)dy\\ &=\frac{b^{a_1+a_2}}{\Gamma(a_1)\Gamma(a_2)}z^{a_1+a_2-1}exp(-bz) \frac{\Gamma{a_1}\Gamma(a_2)}{\Gamma(a_1+a_2)}\\ &=\frac{b^{a_1+a_2}}{\Gamma(a_1+a_2)}z^{a_1+a_2-1}exp(-bz) \end{split} \tag{3.50} \end{equation}$$ for $z>0$. The second equation in (3.50) uses the fact that this integral is the kernel of a beta distributed random variabel with parameters $a_1$ and $a_2$. From (3.50) it is obvious that the marginal distribution of $Z$ is $Ga(a_1+a_2,b)$.

3. Similarily, integrate out Z from the joint distribution will give us the marginal distribution of Y. $$\begin{equation} \begin{split} f_Z(z)&=\int_0^{\infty}y^{a_1-1}(1-y)^{a_2-1}\frac{b^{a_1+a_2}}{\Gamma(a_1)\Gamma(a_2)}\\ &z^{a_1+a_2-1}exp(-bz)dz\\ &=y^{a_1-1}(1-y)^{a_2-1}\frac{b^{a_1+a_2}}{\Gamma(a_1)\Gamma(a_2)} \frac{\Gamma(a_1+a_2)}{b^{a_1+a_2}}\\ &=\frac{y^{a_1-1}(1-y)^{a_2-1}}{B(a_1,a_2)} \end{split} \tag{3.51} \end{equation}$$ for $y\in(0,1)$. The second equation in (3.51) uses the fact that this integral is the kernel of a gamma distributed random variabel with parameters $a_1+a_2$ and $b$. From (3.51) it is obvious that the marginal distribution of $Y$ is $Beta(a_1,a_2)$.

4. By definition $$\begin{equation} \begin{split} E(Y^k)&=\int_0^1y^k\frac{y^{a_1-1}(1-y)^{a_2-1}}{B(a_1,a_2)}dy\\ &=\frac{1}{B(a_1,a_2)}\int_0^1y^{k+a_1-1}(1-y)^{a_2-1}dy\\ &=\frac{B(a_1+k,a_2)}{B(a_1,a_2)} \end{split} \tag{3.52} \end{equation}$$ for $k>0$. The second equation uses the fact that the integral is the kernel of a beta distributed random variable with parameters $a_1+k$ and $a_2$.

5. If $a_1=a_2=b=1$, then for each $X_i$, the gamma distribution collapse to a exponential distribution with parameter $\lambda=1$. Thus, the marginal distribution of $Y$ is $Beta(1,1)$ which is just a uniform distribution on $(0,1)$. The marginal distribution of $Z$ is the sum of two independent exponential random variables, which is an Erlang distributed random variable with parameter $k=2$ and $\lambda=1$. The result can be generate to $n$ independent random variables $X_i$, $X_i\sim Ga(1,1)=Exp(1)$. The random variable that denote the proportion of random variable $X_i$ to the sum of all random variables $\sum_{i=1}^nX_i$ is a $Beta(1,n-1)$ and the sum $\sum_{i=1}^nX_i$ has an Erlang distribution with parameters $k=n$ and $\lambda=1$ (same as $Gamma(n,1)$).

Exercise 3.14 (Homework 1, Problem 3) Consider three independent random variables $X_1,X_2$, and $X_3$ such that $X_i\stackrel{ind}{\sim} Gamma(a_i,b)$, $i=1,2,3$. Let $$\begin{equation} \mathbf{Y}=(Y_1,Y_2,Y_3)=(\frac{X_1}{X_1+X_2+X_3},\frac{X_2}{X_1+X_2+X_3},\frac{X_3}{X_1+X_2+X_3}) \tag{3.53} \end{equation}$$

1. Show that $\mathbf{Y}\sim Dirichlet(a_1,a_2,a_3)$, a Dirichlet distribution.

2. How can this result be used to generate random variables according to a Dirichlet distribution? Write a simple function in R that takes as inputs $n$, the number of trivariate vectors to be generated, and $\mathbf{a}=(a_1,a_2,a_3)$ generates a matrix of size $n\times 3$ whose rows correspond to independent samples from a Dirichlet distribution with parameter $(a_1,a_2,a_3)$.

Use each of $\mathbf{a}=(0.01,0.01,0.01), (100,100,100)$ and $(3,5,10)$ and comment how the density of $\mathbf{Y}$ changes over $\mathbf{a}$.

Proof. (a) Since $X_i\stackrel{ind.}{\sim}Ga(a_i,b)$, the joint pdf is then $$\begin{equation} f(x_1,x_2,x_3)=\prod_{i=1}^3\frac{b^{a_i}}{\Gamma(a_i)}x_i^{a_i-1}exp(-bx_i) \end{equation}$$ Define variable transformation in (3.54) $$\begin{equation} \left\{ \begin{aligned} & y_1=\frac{x_1}{x_1+x_2+x_3} \\ & y_2=\frac{x_2}{x_1+x_2+x_3} \\ & z=x_1+x_2+x_3 \end{aligned} \right. \tag{3.54} \end{equation}$$ Then we have $$\begin{equation} \left\{ \begin{aligned} & x_1=y_1z \\ & x_2=y_2z \\ & x_3=(1-y_1-y_2)z \end{aligned} \right. \tag{3.55} \end{equation}$$ The Jacobian corresponding to (3.55) has determinant $$\begin{equation} \begin{split} |J|&=\begin{vmatrix} z & 0 & y_1\\ 0 & z & y_2\\ -z & -z & 1-y_1-y_2 \end{vmatrix}\\ &=z^2(1-y_1-y_2)+z^2y_1+z^2y_2=z^2 \end{split} \tag{3.56} \end{equation}$$ Thus, the joint distribution of $Y_1$, $Y_2$ and $Z$ can be written as $$\begin{equation} \begin{split} f(y_1,y_2,z)&=\frac{b^{\sum_{i=1}^3a_i}}{\prod_{i=1}^3\Gamma(a_i)}[(y_1z)^{a_1-1}exp(-by_1z)(y_2z)^{a_2-1}exp(-by_2z)\cdot\\ &((1-y_1-y_2)z)^{a_3-1}exp(-b(1-y_1-y_2)z)]z^2 \end{split} \tag{3.57} \end{equation}$$ for $0<y_1<1,0<y_2<1,0<y_1+y_2<1$ and $z>0$.Denote $y_3=1-y_1-y_2$ and integrate w.r.t. $Z$, then we have $$\begin{equation} \begin{split} f(y_1,y_2,y_3)&=\int_{z=0}^{\infty}\frac{b^{\sum_{i=1}^3a_i}}{\prod_{i=1}^3\Gamma(a_i)}[(y_1)^{a_1-1}(y_2)^{a_2-1}(y_3)^{a_3-1}z^{a_1+a_2+a_3-1}exp(-bz)]dz\\ &=\frac{b^{\sum_{i=1}^3a_i}}{\prod_{i=1}^3\Gamma(a_i)}\frac{\Gamma(\sum_{i=1}^3a_i)}{b^{\sum_{i=1}^3a_i}}(y_1)^{a_1-1}(y_2)^{a_2-1}(y_3)^{a_3-1}\\ &=\frac{\Gamma(\sum_{i=1}^3a_i)}{\prod_{i=1}^3\Gamma(a_i)}(y_1)^{a_1-1}(y_2)^{a_2-1}(y_3)^{a_3-1} \end{split} \tag{3.58} \end{equation}$$ with $0<y_i<1$ and $\sum_{i=1}^3y_i=1$. Therefore, $$\begin{equation} \mathbf{Y}=(y_1,y_2,y_3)\sim Dirichlet(a_1,a_2,a_3) \tag{3.59} \end{equation}$$

2. The R function to generate such samples is shown below.

dirichlet_gen=function(n,a.vec){
  sample=matrix(0,nrow=n,ncol=length(a.vec))
  for (i in 1:n) {
    set.seed(i)
    sample.cur=rep(0,length(a.vec))
    for (j in 1:length(a.vec)) {
      sample.cur[j]=rgamma(1,a.vec[j],1)
    }
    sample[i,]=sample.cur/sum(sample.cur)
  }
  return(sample)
}

To compare the density of samples generated using this function with different choice of $\mathbf{a}$, we plot 500 samples using this function and $\mathbf{a}$ using $(0.01,0.01,0.01)$, $(100,100,100)$ and $(3,5,10)$ in Figure 3.1, Figure 3.2, and Figure 3.3, respectively. From the plot, we notice that when $\mathbf{a}=(0.01,0.01,0.01)$, all samples are on the axises, meaning that in each sample, either 1 or two $Y_i$ are really close to 0. When $\mathbf{a}=(100,100,100)$, the variance of samples are quite small, which makes them centered at the theoretical mean $(1/3,1/3,1/3)$. As for $\mathbf{a}=(3,5,10)$, samples are around their theoretical mean $(1/6,5/18,5/9)$ with larger variance comparing to the previous case.

FIGURE 3.1: Samples using a=(0.01,0.01,0.01)

FIGURE 3.2: Samples using a=(100,100,100)

FIGURE 3.3: Samples using a=(3,5,10)

Exercise 3.15 (Homework 1, Problem 4) $Y$ follows an inverse Gamma distribution with shape parameter $a$ and scale parameter $b$ ($Y\sim IG(a,b)$) if $Y=1/X$ with $X\sim Gamma(a,b)$, (assume the Gamma distribution is parameterized so that $E(X)=ab$).

1. Find the density of $Y$.

2. Compute $E(Y^k)$. Do you need to impose any constrain on the problem for this expectation to exists?

3. Compare $E(Y^k)$ to $1/E(X^k)$.

Proof. (a) Since $X\sim Gamma(a,b)$, the pdf is $$\begin{equation} f(x)=\frac{x^{a-1}e^{-\frac{x}{b}}}{b^a\Gamma(a)} \tag{3.60} \end{equation}$$ with $x>0$. For $Y=\frac{1}{X}$, define variable transformation $y=\frac{1}{x}$ then $x=\frac{1}{y}$ with corresponding Jacobian satisfies $|J|=\frac{1}{y^2}$. Thus, the pdf of y is given by $$\begin{equation} f(y)=\frac{y^{-(a+1)}exp(-\frac{1}{by})}{b^a\Gamma(a)} \tag{3.61} \end{equation}$$ for $y>0$.

2. By definition, we have $$\begin{equation} \begin{split} E(y^k)&=\int_0^{\infty}y^k\frac{y^{-(a+1)}exp(-\frac{1}{by})}{b^a\Gamma(a)}dy\\ &=\frac{1}{b^a\Gamma(a)}\int_0^{\infty}y^{-(a+1-k)}exp(-\frac{1}{by})dy\\ &=\frac{1}{b^a\Gamma(a)}\int_0^{\infty}t^{a-k-1}exp(-\frac{t}{b})dt\\ &=\frac{b^{a-k}\Gamma(a-k)}{b^a\Gamma(a)}=\frac{\Gamma(a-k)}{b^k\Gamma(a)} \end{split} \tag{3.62} \end{equation}$$ The forth equation in (3.62) uses the fact that the intergral, by transforming variable $y=\frac{1}{t}$, is the kernel of a gamma distributed random variabel with parameters $a-k$ and $b$. Therefore, the constrain for $E(Y^k)$ to exist is $a-k>0$.

3. For $X\sim Gamma(a,b)$, by definition we have $$\begin{equation} \begin{split} E(X^k)&=\int_0^{\infty}x^k\frac{x^{a-1}e^{-\frac{x}{b}}}{b^a\Gamma(a)}dx\\ &=\frac{\Gamma(a+k)b^{a+k}}{b^a\Gamma(a)}=\frac{b^k\Gamma(a+k)}{\Gamma(a)} \end{split} \tag{3.63} \end{equation}$$ Therefore, consider the ratio $$\begin{equation} \gamma=\frac{1/E(X^k)}{E(Y^k)}=\frac{\Gamma(a)\Gamma(a)}{\Gamma(a+k)\Gamma(a-k)} \tag{3.64} \end{equation}$$ Since from (3.64) $\gamma=1\Leftrightarrow k=0$, we know that expectation is not invariant to non-linear transformation such as $y=\frac{1}{x}$ since every moment of $Y$ and $X$ are different.

Exercise 3.16 (Homework 1, Problem 5) $Y$ follows a log normal distribution with parameters $\mu$ and $\sigma^2$ (denotes as $Y\sim Log-N(\mu,\sigma^2)$) if $Y=\exp(X)$ and $X\sim N(\mu,\sigma^2)$.

1. Find the denisty of $Y$.

2. Compute the mean and the variance of $Y$.

Proof. (a) Since $X\sim N(\mu,\sigma^2)$, the pdf of $X$ is $$\begin{equation} f(x)=\frac{1}{\sigma\sqrt{2\pi}}exp(-\frac{(x-\mu)^2}{2\sigma^2}) \tag{3.65} \end{equation}$$ with $x\in(-\infty,+\infty)$ For $Y=exp(X)$, consider variable transformation $y=exp(x)$ then $x=\log(y)$ with corresponding Jacobian $|J|=\frac{1}{y}$. Therefore, the pdf of $Y$ is $$\begin{equation} f(y)=\frac{1}{y\sigma\sqrt{2\pi}}exp(-\frac{(\log(y)-\mu)^2}{2\sigma^2}) \tag{3.65} \end{equation}$$ for $y>0$.

2. We have $$\begin{equation} \begin{split} E(Y^k)&=E(e^{kX})=M_{X}(k)\\ &=exp(\mu k+\frac{\sigma^2k^2}{2}) \end{split} \tag{3.66} \end{equation}$$ Therefore, $$\begin{equation} \begin{split} &E(Y)=exp(\mu+\frac{\sigma^2}{2})\\ &Var(Y)=E(Y^2)-(E(Y))^2=e^{\sigma^2+2\mu}(e^{\sigma^2}-1) \end{split} \tag{3.67} \end{equation}$$

Exercise 3.17 (Homework 1, Problem 6) Let $\mathbf{X}=(X_1,\cdots,X_p)$ with $\mathbf{X}\sim N_p(\boldsymbol{\mu},\Sigma)$ and set $\mathbf{Z}_1=(X_1,\cdots,X_q)$ and $\mathbf{Z}_2=(X_{q+1},\cdots,X_p)$ with $1<q<p$. Show that $$\begin{equation} \mathbf{Z}_1|\mathbf{Z}_2\sim N_q(\boldsymbol{\mu}_1+\Sigma_{12}\Sigma_{22}^{-1}(\mathbf{Z}_2-\boldsymbol{\mu}_2),\Sigma_{11}-\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}) \tag{3.68} \end{equation}$$ where $\boldsymbol{\mu}_k$ and $\Sigma_{k\ell}$ denote the block of $\boldsymbol{\mu}$ and $\Sigma$ where the rows correspond to the variables in $Z_k$ and the columns correspond to the variables in $Z_{\ell}$.

Proof. Using the inverse of block matrix formula from Wiki we have $$\begin{equation} \begin{split} \Sigma^{-1}&=\begin{pmatrix} \Sigma_{11} & \Sigma_{12} \\ \Sigma_{21} & \Sigma_{22} \end{pmatrix}^{-1}\\ &=\begin{pmatrix} \Sigma_1^{-1} & -\Sigma_{11}^{-1}\Sigma_{12}\Sigma_{2}^{-1}\\ -\Sigma_2^{-1}\Sigma_{21}\Sigma_{11}^{-1}& \Sigma_2^{-1} \end{pmatrix} \end{split} \tag{3.69} \end{equation}$$ with $\Sigma_1=\Sigma_{11}-\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}$ and $\Sigma_2=\Sigma_{22}-\Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12}$. We have $$\begin{equation} \begin{split} f(\mathbf{z}_1|\mathbf{z}_2)&\propto f(\mathbf{z}_1,\mathbf{z}_2)\\ &\propto exp\{-\frac{1}{2}\begin{pmatrix} \mathbf{z}_1-\boldsymbol{\mu}_1\\ \mathbf{z}_2-\boldsymbol{\mu}_2 \end{pmatrix}^T\begin{pmatrix} \Sigma_{11} & \Sigma_{12} \\ \Sigma_{21} & \Sigma_{22} \end{pmatrix}^{-1}\begin{pmatrix} \mathbf{z}_1-\boldsymbol{\mu}_1\\ \mathbf{z}_2-\boldsymbol{\mu}_2 \end{pmatrix} \}\\ &\propto exp\{-\frac{1}{2}[(\mathbf{z}_1-\boldsymbol{\mu}_1)^T\Sigma_{1}^{-1}(\mathbf{z}_1-\boldsymbol{\mu}_1)-(\mathbf{z}_2-\boldsymbol{\mu}_2)^T\Sigma_2^{-1}\Sigma_{21}\Sigma_{11}^{-1}(\mathbf{z}_1-\boldsymbol{\mu}_1)\\ &-(\mathbf{z}_1-\boldsymbol{\mu}_1)^T\Sigma_{11}^{-1}\Sigma_{12}\Sigma_{2}^{-1}(\mathbf{z}_2-\boldsymbol{\mu}_2)]\}\\ &\propto exp\{-\frac{1}{2}[\mathbf{z}_1^T\Sigma_1^{-1}\mathbf{z}_1-\mathbf{z}_1^T(\Sigma_{1}^{-1}\boldsymbol{\mu}_1+\Sigma_{11}^{-1}\Sigma_{12}\Sigma_{2}^{-1}(\mathbf{z}_2-\boldsymbol{\mu}_2)\\ &-(\boldsymbol{\mu}_1^T\Sigma_{1}^{-1}+(\mathbf{z}_2-\boldsymbol{\mu}_2)^T\Sigma_2^{-1}\Sigma_{21}\Sigma_{11}^{-1})\mathbf{z}_1]\}\\ &=exp\{-\frac{1}{2}[(\mathbf{z}_1-(\boldsymbol{\mu}_1+\Sigma_{12}\Sigma_{22}^{-1}(\mathbf{z}_2-\boldsymbol{\mu}_2)))^T\Sigma_1^{-1}\\ &(\mathbf{z}_1-(\boldsymbol{\mu}_1+\Sigma_{12}\Sigma_{22}^{-1}(\mathbf{z}_2-\boldsymbol{\mu}_2)))]\} \end{split} \tag{3.70} \end{equation}$$ Therefore, by recoginzing the kernel we have $\mathbf{z}_1|\mathbf{z}_2\sim N(\boldsymbol{\mu}_1+\Sigma_{12}\Sigma_{22}^{-1}(\mathbf{z}_2-\boldsymbol{\mu}_2),\Sigma_{11}-\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21})$ as we desired.

Exercise 3.18 (Homework 1, Problem 7) Show that if $X\sim exp(\beta)$, then

1. $Y=X^{1/\gamma}$ has a Weibull distribution with parameters $\gamma$ and $\beta$ with $\gamma>0$ a constant.

2. $Y=(2X/\beta)^{1/2}$ has the Rayleigh distribution.

For both parts, derive the form of the p.d.f., verify that is a p.d.f., and calculate the mean and the variance.

Proof. (a) Since $X\sim Exp(\beta)$, the pdf of $X$ is $$\begin{equation} f(x)=\beta exp(-\beta x) \tag{3.71} \end{equation}$$ for $x>0$. Consider variable transformation $y=x^{1/\gamma}$ then $x=y^{\gamma}$ and hence $|J|=\gamma y^{\gamma-1}$, we have the pdf of y as $$\begin{equation} f(y)=\beta\gamma y^{\gamma-1}exp(-\beta y^{\gamma}) \tag{3.72} \end{equation}$$ for $y>0$, which is a Weibull distribution with parameters $\gamma>0$ and $\beta>0$.

To verify the pdf, consider $$\begin{equation} \begin{split} \int_0^{\infty}f(y)dy&=\int_0^{\infty}\beta\gamma y^{\gamma-1}exp(-\beta y^{\gamma})dy\\ &=\int_0^{\infty}\beta exp(-\beta y^{\gamma})dy^{\gamma}=1 \end{split} \tag{3.73} \end{equation}$$ Thus, it is a proper pdf.

To compute mean and variance, we have $$\begin{equation} \begin{split} E(Y^k)&=E(X^{k/\gamma})=\beta\int_0^{\infty}x^{k/\gamma}exp(-\beta x)dx\\ &=\frac{\Gamma(\frac{k}{\gamma}+1)}{\beta^{k/\gamma}} \end{split} \tag{3.74} \end{equation}$$ where the second equation uses the fact that the integral is the kernel of a gamma distributed random variable with parameter $\frac{k}{\gamma}+1$ and $\beta$. Therefore $$\begin{equation} \begin{split} &E(Y)=\frac{\Gamma(\frac{1}{\gamma}+1)}{\beta^{1/\gamma}}\\ &Var(Y)=E(Y^2)-(E(Y))^2=\frac{\Gamma(\frac{2}{\gamma}+1)-(\Gamma(\frac{1}{\gamma}+1))^2}{\beta^{2/\gamma}} \end{split} \tag{3.75} \end{equation}$$

2. Similarily, consider variable transformation $y=(2x/\beta)^{1/2}$, then $x=\frac{\beta y^2}{2}$ with Jacobian $|J|=\beta y$ and thus $Y$ has the pdf in (3.76). $$\begin{equation} f(y)=\beta^2yexp(-\frac{\beta^2y^2}{2}) \tag{3.76} \end{equation}$$ for $y>0$. It has the form as Rayleigh distribution with parameter $\beta>0$.

To see it is a legal pdf, consider $$\begin{equation} \begin{split} \int_0^{\infty}f(y)dy&=\int_0^{\infty}\beta^2yexp(-\frac{\beta^2y^2}{2})dy\\ &=\int_0^{\infty}\frac{1}{2}exp(-\frac{\beta^2y^2}{2})d\beta^2y^2=1 \end{split} \tag{3.77} \end{equation}$$

Finally, for mean and variance, we have $$\begin{equation} \begin{split} E(Y^k)&=E((\frac{2X}{\beta})^{k/2})=\frac{2^{k/2}}{\beta^{\frac{k-2}{2}}}\int_0^{\infty}x^{k/2}exp(-\beta x)dx\\ &=\frac{2^{k/2}\Gamma(\frac{k}{2}+1)}{\beta^k} \end{split} \tag{3.78} \end{equation}$$ Therefore $$\begin{equation} \begin{split} &E(Y)=\frac{\sqrt{2}\Gamma(1.5)}{\beta}=\frac{\sqrt{2\pi}}{2\beta}\\ &Var(Y)=E(Y^2)-(E(Y))^2=\frac{2\Gamma(2)}{\beta^2}-(\frac{\sqrt{2\pi}}{2\beta})^2=\frac{4-\pi}{2\beta^2} \end{split} \tag{3.79} \end{equation}$$