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Utah State University
ECE 6010
Stochastic Processes
Homework # 5 Solutions

  1. Let $ X_{1} \sim \mathcal{U}(0,1)$ and $ X_{2} \sim
\mathcal{U}(0,1)$ (independent). Let $ Y_{1} = \sqrt{-2 \ln X_{1}}
\cos(2 \pi X_{2}) $ and $ Y_{2} = \sqrt{-2 \ln X_{1}}
\sin(2 \pi X_{2}) $ . Show that $ Y_{1} \sim \mathcal{N}(0,1)$ and $ Y_{2} \sim \mathcal{N}(0,1)$

    Here we have,

    $\displaystyle Y_{1}^{2} + Y_{2}^{2} = -2 \ln X_{1}[(\cos(2 \pi X_{2}))^{2} +
(\sin(2 \pi X_{2}))^{2}] = -2 \ln X_{1} $

    Therefore,

    $\displaystyle X_{1} = e^{- \frac{Y_{1}^{2} + Y_{2}^{2}}{2}} $

    Also, $ Y_{2}/Y_{1} = \tan (2 \pi X_{2})$ , therefore

    $\displaystyle X_{2} = \frac{1}{2 \pi} \tan^{-1} \left( \frac{Y_{2}}{Y_{1}}
\right) $

    Jacobian is given by

    $\displaystyle J = \left[
\begin{array}{cc}
\frac{\partial x_{1}}{\partial y_{1...
...}}{\partial y_{1}} & \frac{\partial
x_{2}}{\partial y_{2}}
\end{array} \right]
$

    Therefore,

    $\displaystyle \vert J\vert^{-1} = \ldots = \frac{X_{1}}{2 \pi} $

    Now,

    $\displaystyle f_{Y_{1},Y_{2}}(y_{1},y_{2})$ $\displaystyle =$ $\displaystyle \vert J\vert^{-1}
f_{X_{1},X_{2}}(x_{1},x_{2})$  
      $\displaystyle =$ $\displaystyle \frac{X_{1}}{2 \pi} f_{X_{1}}(x_{1}) f_{X_{2}}(x_{2})$  
      $\displaystyle =$ $\displaystyle \frac{1}{2 \pi} e^{- \frac{Y_{1}^{2} + Y_{2}^{2}}{2}}$  
      $\displaystyle =$ $\displaystyle \frac{1}{ \sqrt{2 \pi}} e^{- \frac{Y_{1}^{2}}{2}} \cdot
\frac{1}{ \sqrt{2 \pi}} e^{- \frac{Y_{2}^{2}}{2}}$  
      $\displaystyle \Rightarrow$ $\displaystyle Y_{1} \sim \mathcal{N}(0,1)$     and $\displaystyle Y_{2} \sim
\mathcal{N}(0,1)$  

  2. If $ X$ and $ Y$ are independent and $ Y \sim \mathcal{U}(0,1)$ , show that $ Z = X+Y$ has density $ f_{Z}(z) = F_{x}(z)
-F_{x}(z-1)$ .

    Here $ X$ and $ Y$ are independent and $ Y \sim \mathcal{U}(0,1)$ , therefore we have,

    $\displaystyle F_{Z}(z) = P(z \leq Z) = \int_{-\infty}^{\infty}
\int_{-\infty}^{...
...) dx dy =
\int_{-\infty}^{\infty}f_{Y}(y)dy \int_{-\infty}^{z-y} f_{X}(z-y) dx
$

    Now,

    $\displaystyle f_{Z}(z) = \frac{d}{dz} F_{Z}(z) = \frac{d}{dz}
\int_{-\infty}^{\...
...{-\infty}^{z-y} f_{X}(z-y) dx
= \int_{-\infty}^{\infty} f_{Y}(y) f_{X}(z-y) dy
$

    As, $ Y \sim \mathcal{U}(0,1)$ the above integral exists from 0 to 1 only, therefore

    $\displaystyle f_{Z}(z) = \int_{0}^{1} f_{X}(z-y) dy = -F_{X}(z-y) \vert _{0}^{1} =
F_{X}(z) - F_{X}(z-1)
$

  3. Let $ X \sim \Nc(0,\sigma^2)$ and $ Y \sim \Nc(0,\sigma^2)$ . Let $ Z = X^2 + Y^2$ . Show that

    $\displaystyle f_Z(z) = \frac{1}{2\sigma^2} e^{-z/2\sigma^2} u(z)
$

    Such a random variable is said to be chi-squared $ (\chi^2)$ distributed with two degrees of freedom.

    $\displaystyle \begin{aligned}
F_Z(z) &= P(Z \leq z) = \int\int_{x^2+y^2 \leq z}...
...rt{z}} \frac{1}{2\pi\sigma^2}
e^{-r^2/2\sigma^2}   r dr d\theta.
\end{aligned}$

    Taking derivatives,

    $\displaystyle f_Z(z) = \frac{\sqrt{z}}{2 \sqrt{z}\sigma^2} e^{-z/2\sigma^2}
$

  4. If $ X \in \mathcal{U}(0,1)$ and $ Y = -(\ln(X))/ \lambda$ , show that $ Y$ is exponentially distributed.

    We have,

    $\displaystyle Y = -\frac{\ln(X)}{\lambda}   \Rightarrow   \lambda Y =
-(\ln(X))   \Rightarrow   X = e^{- \lambda Y}
$

    $\displaystyle \frac{dx}{dy} = - \lambda e^{- \lambda Y}
\hspace{1cm}\mathrm{so,} \hspace{0.5cm} \left\vert \frac{dx}{dy} \right\vert
= \lambda e^{- \lambda Y}
$

    So,

    $\displaystyle f_{Y}(y) = f_{X}(g^{-1}(y)) \left\vert \frac{dx}{dy} \right\vert ...
...mbda e^{-
\lambda Y} \mbox{ (as $f_{X}(e^{- \lambda Y})=1$ for $0<Y<\infty$)}
$

    Therefore, we have

    $\displaystyle f_{Y}(y) = \left\{
\begin{array}{ll}
\lambda e^{- \lambda Y} & \mbox{ for all $Y > 0$} \\
0 & \mbox{ else}
\end{array}\right.
$

  5. Let $ g(x)$ be a monotone increasing function and let $ Y =
g(X)$ . Show that

    $\displaystyle F_{XY}(x,y) = \left\{
\begin{array}{ll}
F_{X}(x) & \mathrm{if}  y > g(x) \\
F_{Y}(y) & \mathrm{if}  y < g(x).
\end{array} \right. $

    Suppose that $ y>g(x)$ . Then

    $\displaystyle \begin{aligned}
F_{XY}(x,y) &= P(X \leq x, Y \leq y) = P(X \leq x, X \leq g^{-1}(y)).
\end{aligned}$

    But when $ y>g(x)$ , the intersection of $ (-\infty,x]$ and $ (-\infty,g^{-1}(y)]$ is $ (-\infty,x]$ . So

    $\displaystyle F_{XY}(x,y) = P(X\leq x) = F_X(x).
$

    It works oppositely when $ y < g(x)$ .

  6. Let $ Z = aX + bY$ . Show that

    $\displaystyle f_Z(z) = \frac{1}{\vert a\vert} \int_{-\infty}^\infty
f_{XY}(\frac{z-by}{a},y)  dy.
$

    Solution: Introduce the auxiliary variable $ W = Y$ . Then

    $\displaystyle J = a
$

    so

    $\displaystyle f_{ZW}(z,w) = \frac{1}{\vert a\vert}f_{XY}(\frac{z-by}{a},y).
$

    Then integrate out $ w$ to get $ f_Z(z)$ .
  7. Let $ X$ and $ Y$ be independent with $ f_{X}(x) = \alpha
e^{-\alpha x} u(x)$ and $ f_{Y}(y) = \beta
e^{-\beta y} u(y)$ . Determine the density of:
    1. $ Z = X/Y$ .

      $\displaystyle F_{Z}(z) = \int_{-\infty}^{\infty} \int_{-\infty}^{zy}
f_{XY}(x,y)dxdy = \int_{-\infty}^{\infty} \int_{-\infty}^{zy}
f_{X}(zy) f_{Y} (y) dxdy
$

      Therefore,

      $\displaystyle f_{Z}(z) = \int_{-\infty}^{\infty} f_{X}(zy) f_{Y} (y) \vert y\ve...
...\alpha z + \beta)} dy =
\ldots = \frac{ \alpha \beta} { ( \beta + \alpha z)^2}
$

    2. $ Z = \max (X,Y)$ .

      $\displaystyle F_{Z}(z) = p(Z \leq z) = P(\max (X,Y) \leq z) = P(X \leq z, Y \leq
z) = P(X \leq z)P(Y \leq z)
$

      Therefore,

      $\displaystyle F_{Z}(z) = F_{X}(z)F_{Y}(z) = \int_{0}^{z} f_{X}(x)dx \
\int_{0}...
... \
\int_{0}^{z} \beta e^{-\beta y} dy = (1 - e^{-\alpha z})(1 - e^{-\beta z})
$

      So,

      $\displaystyle f_{Z}(z) = \frac{d}{dz} F_{Z}(z) = -(\alpha + \beta) e^{
-z(\alpha+\beta)} + \alpha e^{-\alpha z} + \beta e^{-\beta z}
$

  8. Show that if $ X,Y$ are i.i.d. $ \Nc(0,\sigma^2)$ , then $ Z = X/Y$ has a Cauchy distribution,

    $\displaystyle f_Z(z) = \frac{1/\pi}{z^2 + 1}.
$

    We have

    $\displaystyle f_{XY}(x,y) = \frac{1}{2\pi \sigma^2} \exp\left[
-\frac{1}{2\sigma^2}(x^2 + y^2)\right].
$

    Using the result from the notes we have with $ Z = X/Y$

    $\displaystyle f_Z(z) = \int_{-\infty}^\infty \vert y\vert f_{XY}(zy,y) dy.
$

    Then

    $\displaystyle \begin{aligned}
f_Z(z) &= \frac{2}{2\pi \sigma^2} \int_0^\infty y...
... u = -y^2 a, \text{ so }ydy = -du/2a \\
&= \frac{1}{\pi(z^2+1)}.
\end{aligned}$

Copyright 2008, Todd Moon. Cite/attribute Resource . admin. (2006, June 13). Homework Solutions. Retrieved January 07, 2011, from Free Online Course Materials — USU OpenCourseWare Web site: http://ocw.usu.edu/Electrical_and_Computer_Engineering/Stochastic_Processes/hw5sol.html. This work is licensed under a Creative Commons License Creative Commons License