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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 $0<Y<\infty$)} $<\infty$)} $" align="middle" border="0" height="54" width="618" />

    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. $ 0$} \\ 0 & \mbox{ else} \end{array}\right. $" align="middle" border="0" height="55" width="248" />

  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. $ g(x) \\ F_{Y}(y) & \mathrm{if} y < g(x). \end{array} \right. $" align="middle" border="0" height="54" width="263" />

    Suppose that $ y>g(x)$g(x)$" align="middle" border="0" height="32" width="62" /> . 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)$g(x)$" align="middle" border="0" height="32" width="62" /> , 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)$< g(x)$" align="middle" border="0" height="32" width="62" /> .

  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 November 23, 2009, 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
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