## Problem 4.79 (2 Points)

If $$\lambda > 0$$ and $$\alpha$$ is a positive integer, the relationship between incomplete gamma integrals and sums of Poisson probabilities is given by $\frac{1}{\Gamma(\alpha)} \int_{\lambda}^{\infty} y^{\alpha-1}e^{-y}\,dy = \sum_{x=0}^{\alpha-1}\frac{\lambda^{x}e^{-\lambda}}{x!}.$ If $$Y$$ has a gamma distribution with $$\alpha = 2$$ and $$\beta = 1$$, find $$P(Y > 1)$$ by using the preceding equality and Table $$3$$ of Appendix III.

##### Solution 4.79

Let $$X$$ have a Poisson distribution with expected value $$\lambda = 1$$. Using the above equation, we have $P(Y > 1) = \frac{1}{\Gamma(2)} \int_{1}^{\infty} y^{2-1}e^{-y}\,dy = \sum_{x=0}^{2-1}\frac{1^{x}e^{-1}}{x!} = P(X\le 1).$ Using Table $$3$$ on page $$787$$, we find that $$P(X\le 1)\doteq 0.736$$. We can find the same value with the R command ppois(1,lambda = 1)

[1] 0.7357589

## Problem 4.89 (10 Points)

Let the random variable $$Y$$ have the Gamma probability distribution with parameters $$\alpha>0$$ and $$\beta>0$$. Then the probability density function for $$Y$$ is $f(y;\alpha,\beta) = \begin{cases} \frac{y^{\alpha-1}e^{-y/\beta}}{\beta^{\alpha}\Gamma(\alpha)} &, y>0 \\ 0 &, \text{otherwise.} \end{cases}$
1. Let $$a$$ be any real number for which $$\alpha+a>0$$. \begin{align*} \beta^\alpha\Gamma(\alpha)\, E[Y^a] &= \int_0^\infty y^a y^{\alpha-1} e^{-y/\beta}\,dy \\ &= \int_0^\infty y^{\alpha+a-1} e^{-y/\beta}\,dy \\ &= \beta^{\alpha+a}\Gamma(\alpha+a). \end{align*} Therefore, $$E[Y^a]=\beta^a \Gamma(\alpha+a) / \Gamma(\alpha)$$.
2. If $$\alpha+a\le 0$$ then $$\alpha+a-1\le -1$$ and the improper integral in part (a) will diverge at 0.
3. Let $$a=1$$. Then by part (a) $\mu = E[Y] = \beta\frac{\Gamma(\alpha+1)}{\Gamma(\alpha)} = \beta\frac{\alpha \Gamma(\alpha)}{\Gamma(\alpha)} = \alpha \beta.$ Therefore $$\mu = \alpha \beta$$.
4. By part (a), $$E[\sqrt{Y}] = E[Y^{1/2}] = \beta^{1/2}\Gamma(\alpha+1/2)/\Gamma(\alpha)$$. This holds for all $$\alpha>0$$.
5. By part (a) \begin{align*} E[1/Y] &= \beta^{-1}\frac{\Gamma(\alpha-1)}{\Gamma(\alpha)} = \frac{1}{\beta (\alpha-1)}, & \alpha > 1. \\ E[1/\sqrt{Y}] &= \beta^{-1/2}\frac{\Gamma(\alpha-\frac{1}{2})}{\Gamma(\alpha)}, & \alpha > \frac{1}{2}. \\ E[1/Y^2] &= \beta^{-2}\frac{\Gamma(\alpha-2)}{\Gamma(\alpha)} = \frac{1}{\beta^2 (\alpha-1) (\alpha-2)}, & \alpha > 2. \end{align*}

## Problem 4.90 (8 Points)

Let the random variable $$Y$$ have the chi-square distribution with $$\nu$$ degrees of freedom, where $$\nu$$ can be any positive integer. Then $$Y$$ has a Gamma distribution with $$\alpha=\nu/2$$ and $$\beta=2$$.
1. By 4.89(a), $$E[Y^a]=2^a\Gamma(\frac{1}{2}\nu+a)/\Gamma(\frac{1}{2}\nu)$$.
2. $$\alpha+a > 0$$ if and only if $$2\alpha = \nu > -2a$$.
3. By part (a), $$E[\sqrt{Y}]=\sqrt{2}\,\Gamma(\frac{1}{2}\nu+\frac{1}{2})/\Gamma(\frac{1}{2}\nu)$$.
4. By part (a) \begin{align*} E[1/Y] &= \frac{1}{2(\frac{1}{2}\nu-1)} = \frac{1}{\nu-2}, &\nu > 2. \\ E[1/\sqrt{Y}] &= \frac{\Gamma(\frac{1}{2}\nu-\frac{1}{2})}{\sqrt{2}\, \Gamma(\frac{1}{2}\nu)}, & \nu > 1. \\ E[1/Y^2] &= \frac{1}{2^2(\frac{1}{2}\nu -1)(\frac{1}{2}\nu - 2)} = \frac{1}{(\nu - 2)(\nu - 4)}, & \nu > 4. \end{align*}

## Problem 6.14 (6 Points)

A member of the Pareto family of distributions (often used in economics to model income distributions) has a distribution function given by $F(y) = \begin{cases} 0, & y < \beta \\ 1 - \left(\frac{\beta}{y}\right)^{\alpha}, & y \ge \beta, \end{cases}$ where $$\alpha > 0$$ and $$\beta > 0$$.

1. Find the density function.

The density function, $$f(y)$$, is the derivative of the distribution function, $$F(y)$$. Therefore, $f(y) = \begin{cases} 0, & y \le \beta \\ \frac{\alpha \beta^{\alpha}}{y^{\alpha+1}}, & y > \beta. \end{cases}$
2. For fixed values of $$\beta$$ and $$\alpha$$, find a transformation $$G(U)$$ so that $$G(U)$$ has the distribution function of $$F$$ when $$U$$ has a uniform distribution on the interval $$(0,1)$$.

For $$0 < u < 1$$, the distribution function for $$U$$ is $$F_U(u) = u$$. Thus, we need to find a function $$y = G(u)$$ such that $F_U(u) = F(y).$ So, we need to solve the equation $$u = 1 - (\beta / y)^{\alpha}$$ for $$y$$. After some routine algebra, we find that $$y = \beta / (1- u)^{1/\alpha}$$. Therefore $Y = \beta / (1 - U)^{1/\alpha}$ will transform the uniform distribution on $$(0,1)$$ to $$F$$.
3. Given that a random sample of size $$5$$ from a uniform distribution on the interval $$(0,1)$$ yielded the values $$0.0058, 0.2048, 0.7692, 0.2475$$ and $$0.6078$$, use the transformation derived in (b) to give values associated with a random variable with a Pareto distribution with $$\alpha=2$$, $$\beta=3$$.

We will use R to do these tedius calculations.

alpha <- 2  # Set alpha to 2
beta <- 3   # Set beta to 3
# Define the transformation G
G <- function(u) {beta / (1 - u)^(1/alpha)}
# Set u to a vector that contains the uniform random sample.
# We use the function "c()" to create a vector of reals.
u <- c(0.0058, 0.2048, 0.7692, 0.2475, 0.6078)
# Apply the transformation G to the values in u.
G(u)
[1] 3.008738 3.364210 6.244582 3.458343 4.790352
# As a sanity check, confirm that all 5 values
# are greater than 3, i.e. greater than beta.

## Problem 6.40 (2 Points)

Let $$n$$ be a positive integer and let $$Y$$ have a Gamma distribution with parameters $$\alpha=n/2$$ and $$\beta>0$$. Using Table A2.2 in Appendix Two $m_Y(t) = (1 - \beta\, t)^{-n/2} = E\left[e^{t Y}\right].$ Let $$W=2Y/\beta$$, then \begin{align*} m_W(t) &= E\left[e^{t W}\right] \\ &= E\left[e^{(2t/\beta) Y}\right] \\ &= m_Y(2t/\beta) \\ &= \left[1 - \beta\left(\frac{2t}{\beta}\right)\right]^{-n/2} \\ &= (1 - 2t)^{-n/2}, \end{align*}

which, by Table A2.2, is the moment generating function for a chi-square distributed random variable with $$\nu=n$$ degrees of freedom. Therefore, by Theorem 6.1 (page 302), $$W$$ has a chi-square distribution with $$n$$ degrees of freedom.

## Problem (2 Points)

Let $$Y_1$$ and $$Y_2$$ be independent random variables with respective probability distributions Poisson($$\lambda_1$$) and Poisson($$\lambda_2$$). By Table A2.1 in Appendix Two, the respective moment generating functions are \begin{align*} m_{Y_1}(t) &= \exp[\lambda_1 (e^t - 1)] \\ m_{Y_2}(t) &= \exp[\lambda_2 (e^t - 1)]. \end{align*}

Let $$Y = Y_1 + Y_2$$. Since $$Y_1$$ and $$Y_2$$ are independent random variables, Theorem 6.2 (page 304) implies that the moment generating function for $$Y$$ is $m_Y(t) = m_{Y_1}(t)\cdot m_{Y_2}(t) = \exp[(\lambda_1 + \lambda_2) (e^t - 1)],$ which, by Table A2.1, is the moment generating function for a random variable that has a Poisson($$\lambda_1 + \lambda_2$$) distribution. Therefore, by Theorem 6.1 (page 302), the random variable $$Y$$ has a Poisson($$\lambda_1+\lambda_2$$) probability distribution. In other words, the sum of independent Poisson random variables with means $$\lambda_1$$ and $$\lambda_2$$ is a Poisson random variable with mean $$\lambda_1 + \lambda_2$$.

## Problem (4 Points)

Let $$Y_1$$ and $$Y_2$$ be independent and identically distributed Exponential random variables with common mean $$E[Y_1] = E[Y_2] = \beta$$. By Table A2.2 in Appendix Two, the (identical) moment generating functions are $m(t) = (1 - \beta\,t)^{-1}.$ Let $$Y = Y_1 + Y_2$$. Since $$Y_1$$ and $$Y_2$$ are independent random variables, Theorem 6.2 (page 304) implies that the moment generating function for $$Y$$ is $m_Y(t) = m(t)\cdot m(t) = m(t)^2 = (1 - \beta\,t)^{-2},$ which, by Table A2.2, is the moment generating function for a random variable that has a Gamma probability distribution with parameters $$\alpha=2$$ and $$\beta$$. Therefore, by Theorem 6.1 (page 302), the random variable $$Y$$ has a Gamma($$2,\beta$$) probability distribution.

Now, let $$Y_1,\dotsc,Y_n$$ be independent and identically distributed Exponential($$\beta$$) random variables. As an induction hypothesis, assume that the random variable $$Y_1 + \dotsb + Y_{n-1}$$ has a Gamma($$n-1,\beta$$) distribution. Let $Y = Y_1 + \dotsb + Y_{n-1} + Y_n = (Y_1 + \dotsb + Y_{n-1}) + Y_n.$ Then, by the induction hypothesis, $$Y$$ is the sum of a Gamma($$n-1,\beta$$) distributed random variable and an Exponential($$\beta$$) distributed random variable. The independence of $$Y_1,\dotsc,Y_n$$ implies that $$(Y_1 + \dotsb + Y_{n-1})$$ and $$Y_n$$ are independent random variables. Therefore, using Table A2.2 and Theorem 6.2, the moment generating function for $$Y$$ is $m(t) = (1-\beta\,t)^{-n+1}\cdot (1-\beta\,t)^{-1} = (1 - \beta\,t)^{-n},$ which, by Table A2.2, is the moment generating function for a random variable that has a Gamma probability distribution with parameters $$\alpha=n$$ and $$\beta$$. Therefore, by Theorem 6.1 (page 302), the random variable $$Y$$ has a Gamma($$n,\beta$$) probability distribution. In other words, the sum of $$n$$ independent and identically distributed Exponential($$\beta$$) random variables is a Gamma($$n,\beta$$) distributed random variable.