Problems: 2.6, 2.7, 2.9, 2.11, 2.23, 2.24, 2.25, 2.33

• 2.6
  • (a)
  • (b)
  • (c)
• 2.7
  • (a)
• 2.9
• 2.11
  • (a)
  • (b)
• 2.23
  • (a)
  • (b)
• 2.24
  • (a)
  • (b)
  • (c)
• 2.25
  • (a)
  • (b)
• 2.33
  • (a)
  • (b)
  • (c)

2.6

In each of the following find the pdf of $Y$ and show that the pdf integrates to 1.

(a)

\[f_X(x) = \frac{ 1 }{ 2 } e^{-|x|}, \ -\infty < x < \infty; \ Y=|X|^3\]

We can define our sets and functions

\[\begin{array}{l l l l} A_0 = \{ 0 \} & & & \\ A_1 = (-\infty, 0) & g_1(x) = -x^3 & g_1^{-1}(y) = - y^{1/3} & \frac{d}{dy} g_1^{-1}(y) = -\frac{ 1 }{ 3 } y ^{-2/3} \\ A_2 = ( 0, \infty ) & g_2(x) = x^3 & g_2^{-1}(y) = y^{1/3} & \frac{d}{dy} g_2^{-1}(y) = \frac{ 1 }{ 3 } y ^{-2/3} . \end{array}\]

Then,

\[\begin{align} f_Y(y) & = \frac{ 1 }{ 2 } e^{- | - y^{1/3} |} \cdot \Big| -\frac{ 1 }{ 3 } y ^{-2/3} \Big| + \frac{ 1 }{ 2 } e^{- |y^{1/3} |} \cdot \Big| \frac{ 1 }{ 3 } y ^{-2/3} \Big| \\ & = \frac{ 1 }{ 2 } e^{- y^{1/3}} \cdot\frac{ 1 }{ 3 } y ^{-2/3} + \frac{ 1 }{ 2 } e^{- y^{1/3}} \cdot\frac{ 1 }{ 3 } y ^{-2/3} \\ & = \frac{ 1 }{ 3 } e^{- y^{1/3}} \cdot y ^{-2/3} & y \in \mathcal{Y}\\ & = 0 & \text{otherwise.} \end{align}\]

We can then integrate to make sure that we get 1.

\[\begin{align} \int_{-\infty}^{\infty} f_Y(y) dy & = \int_{-\infty}^{\infty} \frac{ 1 }{ 3 } e^{- y^{1/3}} \cdot y ^{-2/3} dy \\ & = \int_{-\infty}^{\infty} \frac{ 1 }{ 3 } e^{-u} y^{-2/3} \cdot 3 y^{2/3} du & u=y^{1/3}, \ du = \frac{ 1 }{ 3 }y^{-2/3}dy \\ & = 1 \end{align}\]

(b)

\[f_X(x) = \frac{ 3 }{ 8 }(x+1)^2, \ -1 < x < 1; \ Y = 1 - X^2\]

We can define our sets and functions

\[\begin{array}{l l l l} A_0 = \{ 0 \} & & & \\ A_1 = (-1, 0) & g_1(x) = 1-x^2 & g_1^{-1}(y) = (1-y)^{1/2} & \frac{d}{dy} g_1^{-1}(y) = -\frac{ 1 }{ 2 }(1-y)^{-1/2} \\ A_2 = ( 0, 1 ) & g_2(x) = 1-x^2 & g_2^{-1}(y) = - (1-y)^{1/2} & \frac{d}{dy} g_2^{-1}(y) = \frac{ 1 }{ 2 }(1-y)^{-1/2} . \end{array}\]

Then,

\[\begin{align} f_Y(y) & = \frac{ 3 }{ 8 }((\sqrt{1-y}) + 1)^2 \cdot \Big| -\frac{ 1 }{ 2 }(1-y)^{-1/2} \Big|+ \frac{ 3 }{ 8 }((-\sqrt{1-y}) + 1)^2 \cdot \Big| \frac{ 1 }{ 2 }(1-y)^{-1/2} \Big| \\ & = \frac{ 3 }{ 8 } \cdot \frac{ 1 }{ 2 }(1-y)^{-1/2} \Big[ \Big( (1-y) + 2 \sqrt{1-y} + 1 \Big) + \Big( (1-y) - 2 \sqrt{1-y} + 1 \Big) \Big] \\ & = \frac{ 3 }{ 16 } (1-y)^{-1/2} (4-2y) & y \in \mathcal{Y}\\ & = 0 & \text{otherwise.} \end{align}\]

We can then integrate to make sure that we get 1.

\[\begin{align} \int_{0}^{1} f_Y(y) dy & = \int_{0}^{1} \frac{ 3 }{ 16 } (1-y)^{-1/2} (4-2y) dy \\ & = \int_{1}^{0}\frac{ 3 }{ 16 }(u)^{-1/2}\cdot 2(1+u)\cdot - du & u=1-y, \ du=-dy \\ & = -\frac{ 6 }{ 16 } \int_{1}^{0} u^{-1/2} + u^{1/2} du \\ & = 1 \end{align}\]

(c)

\[f_X(x) = \frac{ 3 }{ 8 }(x+1)^2, \ -1 < x < 1; \ Y = 1 - X^2 \text{ if } X \leq 0 \text{ and } Y = 1 - X \text{ if } X > 0\] \[\begin{array}{l l l l} A_0 = \{ 0 \} & & & \\ A_1 = (-1, 0) & g_1(x) = 1-x^2 & g_1^{-1}(y) = -(1-y)^{1/2} & \frac{d}{dy} g_1^{-1}(y) = \frac{ 1 }{ 2 }(1-y)^{-1/2} \\ A_2 = ( 0, 1 ) & g_2(x) = 1-x & g_2^{-1}(y) = 1-y & \frac{d}{dy} g_2^{-1}(y) = -1 . \end{array}\]

Thus,

\[\begin{align} f_Y(y) & = \frac{ 3 }{ 8 }(-\sqrt{1-y} + 1)^2 \cdot \frac{ 1 }{ 2\sqrt{1-y} }+\frac{ 3 }{ 8 }((1-y)+1)^2 |-1| & y \in \mathcal{Y}\\ & = 0 & \text{otherwise.} \end{align}\]

We can then integrate to make sure that we get 1.

\[\begin{align} \int_{0}^{1} f_Y(y) dy & = \int_{0}^{1} \frac{ 3 }{ 8 }(-\sqrt{1-y} + 1)^2 \cdot \frac{ 1 }{ 2\sqrt{1-y} }+\frac{ 3 }{ 8 }((1-y)+1)^2 dy \\ & = \frac{ 3 }{ 8 } \Big[ \int_{0}^{1} \frac{ (1-y)^2 - 2(1-y) + 1 }{ 2 \sqrt{1-y} }dy + \int_{0}^{1} (1-y)^2 + 2(y-1) + 1 \Big]\\ & = 1 \end{align}\]

2.7

Let $X$ have pdf $f_X(x) = \frac{ 2 }{ 9 }(x+1)$ for $-1 \leq x \leq 2$.

(a)

Find the pdf of $Y=X^2$. Note that Theorem 2.1.8 is not directly applicable to this problem.

We will start by finding the PDF of $Y$. We can use equation 2.1.11 from the book. Notice, because of the square root and the support of $X$, we will split this into two cases, the first being $x\in [-1, 1], \ y \in [0,1)$ and the second $x \in (1, 2],\ y\in (1, 4]$. Let’s start with the first case.

\[\begin{align} f_Y(y) & = \frac{ 1 }{ 2 \sqrt{y} } (f_X(\sqrt{y}) + f_X(-\sqrt{y})) \\ & = \frac{ 1 }{ 2 \sqrt{y} } \Big( \frac{ 2 }{ 9 }(\sqrt{y} +1) + \frac{ 2 }{ 9 }(-\sqrt{y} +1) \Big)\\ & = \frac{ 1 }{ 2 \sqrt{y} } (\frac{ 4 }{ 9 })\\ & = \frac{ 2 }{ 9 \sqrt{y} } & y \in (0,1] \end{align}\]

Now we will look at the second case.

\[\begin{align} f_Y(y) & = \frac{ 1 }{ 2 \sqrt{y} } (f_X(\sqrt{y}) + f_X(-\sqrt{1})) \\ & = \frac{ 1 }{ 2 \sqrt{y} } ( \frac{ 2 }{ 9 }(\sqrt{y} +1) + \frac{ 2 }{ 9 }(-1 +1)) \\ & = \frac{ 1 }{ 2 \sqrt{y} } (\frac{ 2 }{ 9 } \sqrt{y} + \frac{ 2 }{ 9 })\\ & = \frac{ 1 }{ 9 } + \frac{ 1 }{ 9 \sqrt{y} } & y \in (1, 4] \end{align}\]

Putting these together gives:

\[f_Y(y) = \begin{cases} \begin{align} & \frac{ 2 }{ 9 \sqrt{y} } & y \in (0,1] \\ & \frac{ 1 }{ 9 } + \frac{ 1 }{ 9 \sqrt{y} } & y \in (1, 4] \\ & 0 & \text{otherwise} \end{align} \end{cases}\]

2.9

If the random variable $X$ has pdf

\[f(x)= \begin{cases} \begin{align} & \frac{ x-1 }{ 2 } & 1 < x < 3\\ & 0 & \text{otherwise,} \end{align} \end{cases}\]

find a monotone function $u(x)$ such that the random variable $Y=u(X)$ has a uniform(0,1) distribution.

Take $Y = F_X(x)$, then by the Probability Integral Transformation, $Y \sim U(0,1)$.

\[\int_1^x \frac{ z-1 }{ 2} dz = \frac{ 1 }{ 4 }x^2 - \frac{ 1 }{ 2 }x - \frac{ 1 }{ 4 }\] \[Y = F_X(x)= \begin{cases} \begin{align} & 0 & x \leq 1\\ & \frac{ 1 }{ 4 }x^2 - \frac{ 1 }{ 2 }x - \frac{ 1 }{ 4 } & 1 < x < 3 \\ & 1 & x \geq 3 \end{align} \end{cases}\]

2.11

Let $X$ have the standard normal pdf, $f_X(x) = (1/\sqrt{2\pi}) e^{-x^2/2}$.

(a)

Find $E(X^2)$ directly, and then by using the pdf of $Y=X^2$ from Example 2.1.7 and calculate $E(Y)$.

\[\begin{align} E(X^2) & = \int_{-\infty}^{\infty} x^2 \cdot f_X(x) dx \\ & = \int_{-\infty}^{\infty} x^2 \cdot (1/\sqrt{2\pi}) e^{-x^2/2} dx \\ & u = x & v = -e^{x^2/2}\\ & du = dx & dv = x e^{x^2/2}dx\\ & = \frac{ 1 }{ \sqrt{2 \pi} }\Big( x \cdot -e^{x^2/2} |_{-\infty}^{\infty} - \int_{-\infty}^{\infty} -e^{x^2/2} dx \Big) \\ & = \frac{ 1 }{ \sqrt{2 \pi} } - (-\sqrt{2 \pi})\\ & = 1 \end{align}\]

We should get the same thing for $E(Y)$. We will start by getting its PDF. We can use example 2.1.7 for this.

\[\begin{align} f_Y(y) & = \frac{ 1 }{ 2 \sqrt{y} } (f_X(\sqrt{y}) + f_X(-\sqrt{y})) \\ & = \frac{ 1 }{ 2 \sqrt{y} } \Big( (1/\sqrt{2\pi}) e^{-(\sqrt{y})^2/2} + (1/\sqrt{2\pi}) e^{-(-\sqrt{y})^2/2} \Big) \\ & = \frac{ 1 }{ \sqrt{2 \pi y} } (e^{-y/2}) \end{align}\]

Then,

\[\begin{align} E(Y) & = \int_{0}^{\infty} y \cdot f_Y(y) dy \\ & = \int_{0}^{\infty} y \cdot \frac{ 1 }{ \sqrt{2 \pi y} } (e^{-y/2}) dy\\ & = \frac{ 1 }{ \sqrt{2 \pi} } \int_{-\infty}^{\infty} \sqrt{y} \cdot (e^{-y/2}) dy\\ \\ & u = \sqrt{y} & v = -2 e^{-y/2} \\ & du = \frac{ 1 }{ 2\sqrt{y} } dy & dv = e^{-y/2} dy \\ \\ & = \frac{ 1 }{ \sqrt{2 \pi} } \cdot \sqrt{y} \cdot -2 e^{-y/2} \Big|_{0}^{\infty} - \int_0^{\infty} \frac{ 1 }{ \sqrt{2 \pi} } \cdot -2 e^{-y/2} \cdot \frac{ 1 }{ 2\sqrt{y} } dy \\ & = 0 + \int_0^{\infty} \frac{ 1 }{ \sqrt{2 y \pi} } \cdot e^{-y/2} dy \\ & = 1. \end{align}\]

(b)

Find the pdf of $Y = | X |$, and find its mean and variance.

We will start by finding the pdf of $Y$.

\[\begin{align} F_Y(y) & = P(Y \leq y) \\ & = P( | X | \leq y) \\ & = P(-y \leq X \leq y) \\ & = F_X(y) - F_X(-y)\\ f_Y(y) & = \frac{ d }{ dy } F_Y(y) \\ & = \frac{ d }{ dy } \Big[ F_X(y) - F_X(-y) \Big] \\ & = f_X(y) (1) - f_X(-y)(-1) \\ & = (1/\sqrt{2\pi}) e^{-(y)^2/2} - (1/\sqrt{2\pi}) e^{-(-y)^2/2} \\ & = \frac{ 2 }{ \sqrt{2\pi} } e^{-y^2/2} & y \in [0, \infty) \\ & = 0 & \text{otherwise} \end{align}\]

Now we can find the expectation.

\[\begin{align} E(Y) & = \int_{0}^{\infty} y \cdot f_Y(y) dy \\ & = \int_{0}^{\infty} y \cdot \frac{ 2 }{ \sqrt{2\pi} } e^{-y^2/2} dy \\ & = \frac{ 2 }{ \sqrt{2\pi} } \int_{0}^{\infty} y \cdot e^{-y^2/2} dy \\\\ & u = -y^2 /2 & du = -y dy \\ \\ & = \sqrt{\frac{ 2 }{ \pi }} \cdot \int y e^u \frac{ -1 }{ y }du\\ & = -\sqrt{\frac{ 2 }{ \pi }} e^u \Big| \\ & = -\sqrt{\frac{ 2 }{ \pi }} e^{-y^2/2} \Big|_{0}^{\infty} \\ & = -\sqrt{\frac{ 2 }{ \pi }} (0 - 1)\\ & = \sqrt{\frac{ 2 }{ \pi }} \end{align}\]

Notice that we need to find $E(Y^2)$ in order to find the variance.

\[\begin{align} E(Y^2) & = \int_{0}^{\infty} y^2 \cdot f_Y(y) dy \\ & = \int_{0}^{\infty} y^2 \cdot \frac{ 2 }{ \sqrt{2\pi} } e^{-y^2/2} dy \\ \\ & u = y & v = -e^{y^2/2}\\ & du = dy & dv = y e^{y^2/2}dy \\ \\ & = \frac{ 1 }{ \sqrt{2 \pi} }\Big( y \cdot -e^{y^2/2} |_{-\infty}^{\infty} - \int_{-\infty}^{\infty} -e^{y^2/2} dy \Big) \\ & = \sqrt{\frac{ 2 }{ \pi }} ( 0 + \sqrt{\frac{ \pi }{ 2 }}) \\ & = 1\\ Var(Y) & = E(Y^2) - E(Y)^2 \\ & = 1 - \Big( \sqrt{\frac{ 2 }{ \pi }} \Big)^2 \\ & = 1 - \frac{ 2 }{ \pi } \end{align}\]

2.23

Let $X$ have the pdf

\[\begin{align} f(x) & = \frac{ 1 }{2 } (1+x), & -1 < x< 1. \end{align}\]

(a)

Find the pdf of $Y = X^2$.

\[\begin{align} F_Y(y) & = P(Y \leq y) \\ & = P(X^2 \leq y)\\ & = P(-\sqrt{y} \leq X \leq \sqrt{y}) \\ & = F_X(\sqrt{y}) - F_X(\sqrt{-y})\\ f_Y(y) & = \frac{ d }{ dy } (F_X(\sqrt{y}) - F_X(\sqrt{-y})) \\ & = f_X(\sqrt{y}) \cdot \frac{ d }{ dy } \sqrt{y} - f_X(\sqrt{-y})\cdot \frac{ d }{ dy } \sqrt{-y}\\ & = f_X(\sqrt{y}) \cdot \frac{ 1 }{ 2 \sqrt{y} } - f_X(\sqrt{-y})\cdot \frac{ - 1 }{ 2 \sqrt{y} }\\ & = \frac{ 1 }{ 2 \sqrt{y} } \Big( \frac{ 1 }{ 2 } ( 1 + \sqrt{y}) + \frac{ 1 }{ 2 } ( 1 - \sqrt{y}) \Big) \\ & = \frac{ 1 }{ 2 \sqrt{y} } & 0 < y< 1 \\ & = 0 & \text{otherwise} \end{align}\]

(b)

Find $E(Y)$ and $Var(Y)$.

\[\begin{align} E(Y) & = \int_0^1 y \cdot \frac{ 1 }{ 2\sqrt{y} } dy\\ & = \frac{ 1 }{ 2 } \int_0^1 \sqrt{y} dy \\ & = \frac{ 1 }{ 2 } (\frac{ 2 }{ 3 } \cdot y^{3/2})\Big|_0^1 \\ & = \frac{ 1 }{ 3 }\\ \\ E(Y^2) & = \int_0^1 y^2 \cdot \frac{ 1 }{ 2\sqrt{y} } dy\\ & = \frac{ 1 }{ 2 } \int_0^1 y^{3/2} dy \\ & = \frac{ 1 }{ 2 } \frac{ 2 }{ 5 } (y^{5/2}) \Big|_0^1 \\ & = \frac{ 1 }{ 5 } \\ \\ Var(Y) & = E(Y^2) - E(Y)^2 \\ & = \frac{ 1 }{ 5 } - \frac{ 1 }{ 9 } \\ & = \frac{ 4 }{ 45 } \end{align}\]

2.24

Compute $E(X)$ and $Var(X)$ for each of the following probability distributions.

(a)

\[f_X(x) = ax^{a-1},\ 0 < x< 1,\ a > 0\] \[\begin{align} E(X) & = \int_{0}^{1} x \cdot a x^{a-1} dx \\ & = \int_{0}^{1} a x^{a} dx\\ & = \frac{ a }{ a+1 } x^{a+1} \Big|_{0}^{1} \\ & = \frac{ a }{ a+1 } \\ \\ E(X^2) & = \int_{0}^{1} x^2 \cdot a x^{a-1} dx \\ & = \int_{0}^{1} a x^{a+1} dx\\ & = \frac{ a }{ a+2 } x^{a+2} \Big|_{0}^{1} \\ & = \frac{ a }{ a+2} \\ \\ Var(X) & = \frac{ a }{ a+2} - (\frac{ a }{ a+1 })^2 \end{align}\]

(b)

\[f_X(x) = \frac{ 1 }{ n }, \ x = 1, 2,\dots , n, \ n > 0 \text{ an integer}\] \[\begin{align} E(X) & = \sum_0^1 \frac{ 1 }{ n }x \\ & = \frac{ 1 }{ n } \sum_0^1 x \\ & = \frac{ 1 }{ n } \cdot \frac{ n (n+1) }{ 2 } \\ & = \frac{ n+1 }{ 2 }\\ \\ E(X^2) & = \sum_0^1 \frac{ 1 }{ n }x^2 \\ & = \frac{ 1 }{ n } \sum_0^1 x^2 \\ & = \frac{ 1 }{ n } \cdot \frac{ (2n+1)(n+1)n }{ 6 } \\ & = \frac{ (2n+1)(n+1) }{ 6 } \\ \\ Var(X) & = \frac{ (2n+1)(n+1) }{ 6 } - (\frac{ n+1 }{ 2 })^2 \\ & = \frac{ 1 }{ 12 }(n^2 - 1) \end{align}\]

(c)

\[f_X(x) = \frac{ 3 }{ 2 } (x-1)^2, \ 0 < x < 2\] \[\begin{align} E(X) & = \int_0^2 x \cdot \Big( \frac{ 3 }{ 2 }x^2 - 3x + \frac{ 3 }{ 2 } dx \Big) \\ & = \Big( \frac{ 3 }{ 8 }x^4-x^3+\frac{ 3 }{ 4}x^2 \Big) \Big|_0^2 \\ & = 1 \\ \\ E(X^2) & = \int_0^2 x^2 \cdot \Big( \frac{ 3 }{ 2 }x^2 - 3x + \frac{ 3 }{ 2 } dx \Big) \\ & = \Big( \frac{ 3 }{ 10 }x^5-\frac{ 3 }{ 4 }x^4+\frac{ 1 }{ 2}x^3 \Big) \Big|_0^2 \\ & = \frac{ 8 }{ 5 } \\ \\ Var(X) & = \frac{ 8 }{ 5 } - 1^2 \\ & = \frac{ 3 }{ 5 } \end{align}\]

2.25

Suppose the pdf $f_X(x)$ of a random variable $X$ is an even function. ($f_X(x)$ is an even function if $f_X(x) = f_X(-x)$ for every $x$.) Show that

(a)

$X$ and $-X$ are identically distributed.

To show that two random variables are identically distributed, we need to show that $F_X(x) = F_{-X}(x)$. We will start by taking $Y=-X$ and finding $F_Y(y)$.

\[\begin{align} F_Y(y) & = P(Y \leq y)\\ & = P(-X \leq y) \\ & = P(X > y)\\ & = 1 - P(X \leq y)\\ & = 1 - F_X(y) \end{align}\]

Now we will find the PDF of $Y$.

\[\begin{align} f_Y(y) & = \frac{ d }{ dy }(1 - F_X(-y)) \\ & = - f_X(-y) \cdot -1 \\ & = f_X(-y) \\ & = f_X(y) & \text{since } f_X(x) \text{ is an even function} \end{align}\]

Then,

\[\begin{align} F_Y(y) & = \int_{-\infty}^{y} f_X(x)dx \\ & = F_X(x). \end{align}\]

Thus, they are identically distributed.

(b)

$M_X(t)$ is symmetric about 0.

For a function to be symmetric about 0, we must have $M_X(t-0) = M_X(0-t)$.

\[\begin{align} M_X(t-0) & = E(e^{(t-0)x})\\ & = E(e^{tx})\\ & = \int e^{tx} f_X(x) dx \\ M_X(0-t) & = E(e^{(0-t)x})\\ & = E(e^{-tx})\\ & = \int e^{-tx} f_X(x) dx \\ \end{align}\]

Since $F_X$ and $F_{-X}$ are identically distributed,

\[\begin{align} M_X(t-0) & = M_{-X}(t-0)\\ & = \int e^{(t-0) -x} f_{-X}(x) dx \\ & = \int e^{-tx} f_{-X}(x) dx \\ & = \int e^{-tx} f_{X}(x) dx & \text{by identical distribution}\\ & = M_X(0-t). \end{align}\]

Thus, $M_X(t)$ is symmetric about 0.

2.33

In each of the following cases verify the expression given for the moment generating function, and in each case use the mgf to calculate $E(X)$ and $Var(X)$.

(a)

\[P(X=x) = \frac{ e^{-\lambda} \lambda^x }{ x! },\ M_X(t) = e^{\lambda (e^t - 1)},\ x = 0, 1, \dots;\ \lambda > 0\] \[\begin{align} M_X(t) & = E(e^{tx}) \\ & = \sum e^{tx} \cdot \frac{ e^{-\lambda} \lambda^x }{ x! }\\ & = \sum e^{t^x} \cdot \frac{ e^{-\lambda} \lambda^x }{ x! }\\ & = \frac{ e^{-\lambda} (e^t \cdot \lambda)^x }{ x! }\\ & = \frac{ e^{-\lambda} }{ e^{-e^{t \lambda}} } \cdot \sum \frac{ e^{-e^{t \lambda}}\cdot (e^t \cdot \lambda)^x }{ x! } \\ & = \frac{ e^{-\lambda} }{ e^{-e^{t \lambda}} } \cdot 1 \\ & = e^{\lambda(e^t - 1)} \\ \\ E(X) & = M^{(1)}_X(0)\\ & = e^{\lambda(e^t-1)} \cdot (\lambda e^t) \Big|_{t=0} \\ & = \lambda \\ \\ E(X^2) & = M^{(2)}_X(0)\\ & = e^{\lambda (e^t - 1)}(\lambda e^t)(\lambda e^t) + e^{\lambda (e^t-1)}\cdot \lambda e^t \Big|_{t=0} \\ & = \lambda^2 + \lambda \\ \\ Var(X) & = \lambda^2 + \lambda - (\lambda)^2 \\ & = \lambda \end{align}\]

(b)

\[P(X=x) = p(1-p)^x,\ M_X(t) = \frac{ p }{ 1-(1-p)e^t },\ x=0,1,\dots ;\ 0<p<1\] \[\begin{align} M_X(t) & = E(e^{tx}) \\ & = \sum e^{tx} p (1-p)^x \\ & = \sum e^{t^x} p (1-p)^x \\ & = p \sum (e^t(1-p))^x \\ & = \frac{ p }{ 1 - e^t(1-p) } & \text{by geometric series} \end{align}\]

Notice that our geometric series argument requires

\[\begin{align} e^t(1-p) & < 1 \\ e^t & < \frac{ 1 }{ 1-p }\\ t & < - \log(1-p). \end{align}\] \[\begin{align} E(X) & = M_X^{(1)}(0) \\ & = \frac{ (1-e^t(1-p))\cdot 0 - p(-e^t(1-p)) }{ (1-e^t(1-p))^2 }\Big|_{t=0} \\ & = \frac{ p(1-p) }{ p^2 }\\ & = \frac{ 1-p }{ p } \\ \\ E(X^2) & = M_X^{(2)}(0) \\ & = p e^t(1-p)\cdot (-2)(1-e^t(1-p))^{-3}(-e^t(1-p))+(1-e^t(1-p))^{-2}\cdot pe^t(1-p) \Big|_{t=0}\\ & = \frac{ -2(1-p)(p-1) }{ p^2 } + \frac{ p(1-p) }{ p^2 } \\ & = \frac{ (p-2)(p-1) }{ p^2 } \\ \\ Var(X) & = \frac{ (p-2)(p-1) }{ p^2 } - (\frac{ 1-p }{ p })^2 \\ & = \frac{ 1-p }{ p^2 } \end{align}\]

(c)

\[f_X(x) = \frac{ e^{-(x-\mu)^2/(2\sigma^2)} }{ \sqrt{2\pi} \sigma },\ M_X(t) = e^{\mu t+ \sigma^2 t^2/2},\ -\infty < x< \infty,\ -\infty < \mu < \infty, \ \sigma > 0\] \[\begin{align} M_X(t) & = E(e^{tx}) \\ & = \int e^{tx} \cdot \frac{ e^{-(x-\mu)^2/(2\sigma^2)} }{ \sqrt{2\pi} \sigma } dx \\ & = \int \frac{ 1 }{ \sqrt{2\pi} \sigma} \cdot e^{tx-(x-\mu)^2/(2\sigma^2)}dx \\ & = \int \frac{ 1 }{ \sqrt{2\pi} \sigma} \cdot e^{(2 \sigma^2 tx-(x-\mu)^2)/(2\sigma^2)} dx \end{align}\]

We can now focus on the numerator of the exponential.

\[\begin{align} 2 \sigma^2 tx-(x-\mu)^2 & = 2 \sigma^2 tx - x^2 - \mu^2 + 2\mu x \\ & = - ( x^2 - 2x (\sigma^2 t + \mu) + \mu^2 )\\ & = -(x-(\mu + \sigma^2 t))^2 + (2 \mu \sigma^2 t + (\sigma^2 t )^2) \end{align}\]

Thus,

\[\begin{align} M_X(t) & = e^{(2 \mu \sigma^2 t + (\sigma^2 t)^2)/2 \sigma^2 } \cdot \int \frac{ 1 }{ \sqrt{2\pi} \sigma} \cdot e^{-(x-(\mu + \sigma^2 t))^2 / 2\sigma^2} dc \\ & = e^{(2 \mu \sigma^2 t + (\sigma^2 t)^2)/2 \sigma^2 } \cdot 1 \\ & = e^{\mu t + \frac{ \sigma^2 t^2 }{ 2 }} \\ \\ E(X) & = M_X^{(1)}(0) \\ & = e^{\mu t} \sigma^2 t e^{\frac{ \sigma^2 t }{ 2} } + e^{\frac{ \sigma^2 t }{ 2} } \cdot \mu e^{\mu t} \Big|_{t=0}\\ & = \mu \\ \\ E(X^2) & = M_X^{(2)}(0)) \\ & = e^{\mu t} \cdot (\sigma^2 e^{\sigma^2 t^2 /2}+ \sigma^2 t \cdot \sigma^2 t \cdot e^{\sigma^2 t^2 /2}) + \mu (\sigma^2 t + \mu e^t\cdot e^{\sigma^2 t /2}) \Big|_{t=0} \\ & = \sigma^2 + \mu^2 \\ \\ Var(X) & = \sigma^2 + \mu^2 - (\mu)^2 \\ & = \sigma^2 \end{align}\]