Nate's blog » pdf http://www.natenewz.com My various projects Tue, 29 Jun 2010 18:00:42 +0000 http://wordpress.org/?v=2.9 en hourly 1 Convolution of Two Uniform Probability Distributions http://www.natenewz.com/2009/11/01/convolution-of-two-uniform-probability-distributions/ http://www.natenewz.com/2009/11/01/convolution-of-two-uniform-probability-distributions/#comments Sun, 01 Nov 2009 10:36:27 +0000 nate http://www.natenewz.com/?p=140 The Problem

Two friends A and B meet every morning at the Grand Central Station around 7 am. Suppose the actual times they arrive are independent and uniformly distributed between 6:55 am and 7:05am. Let Z denote the time between arrivals, i.e. Z = time B arrives – time A arrives (can be negative!).
(a) What is the range of values of Z (i.e Im(Z))?
(b) Find the density of Z (derive a formula similar to the convolution formula derived in class).

The Solution

(a) The times for Z can vary from -10 to 10. Suppose friend A arrives at 6:55 and friend B arrives at 7:05. Z would = -10. The other way around will result in Z = 10
(b)

Define Random Variables

X = # of minutes (continuous) after 6:55 friend A arrives
Y = # of minutes (continuous) after 6:55 friend B arrives

Outline of Problem

  • The probability distribution functions of X and Y are uniform (constant and area = 1)
  • The convolution of the pdf’s of X and the negative of Y (X-Y) equal the pdf of Z
  • The result should look closer to the normal curve because of the Central Limit Theorem

The two input pdf’s and the result of the convolution which is the pdf of Z

pdf's and the result of convolution

Performing the convolution

Note: a=10

  • Z=X-Y
  • f_x(t) = 1/a*(u(t) + u(t-a))
  • f_y(t) = 1/a*(u(t) + u(t-a))
  • \int_{-\infty}^{\infty }f_x(t)dt = 1
  • \int_{-\infty}^{\infty }f_y(t)dt = 1
  • f_z(t)=f_x(t)\star f_y(-t)
  • f(t) \star g(t)=\int_{-\infty}^{\infty }f(\tau )g(t-\tau )d\tau
  • f_x(t) \star f_y(-t)=\frac{1}{a} \int_{-a}^{a}[u(\tau )-u(\tau -a)][u(t-\tau +a)-u(t-\tau )]d\tau
  • =\frac{1}{a} \int_{-a}^{a}u(\tau )u(t-\tau +a) -u(\tau )u(t-\tau )-u(\tau -a)u(t-\tau +a)+u(\tau -a)u(t-\tau)d\tau
  • 0= \int_{-a}^{a}u(\tau -a)u(t-\tau +a)+u(\tau -a)u(t-\tau)d\tau
  • =\frac{1}{a}\int_{-a}^{a}u(\tau )u(t-\tau +a)d\tau = \int_{0}^{a}u(t-\tau +a)d\tau
  • \int_{0}^{a}u(t-\tau +a)d\tau\; x=t-\tau +a\; \frac{\mathrm{d} x}{\mathrm{d} \tau} = -1
  • -\int_{t+10}^{t}u(x)dx
  • -tu(t)+(t+a)u(t+a)
  • \int_{0}^{a}u(t-\tau )d\tau\; x=t-\tau\; \frac{\mathrm{d} x}{\mathrm{d} \tau} = -1
  • \int_{t}^{t-10}u(x)dx
  • (t-a)u(t-a)-tu(t)
  • f(t) \star g(t)=\frac{1}{a}(-2tu(t)+(t+a)u(t+a)+(t-a)u(t-a))

Simulation

  • Random number generator is a uniform distribution
  • Histogram of Randbetween(0,10)-Randbetween(0,10) 2092 times divided by 2092
  • Discrete version of this problem

Excel Simulation

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