Independent, Uniformly Distributed Random VariablesDate: 01/10/2002 at 11:08:50 From: Damian Subject: Independent uniformly distributed random variables Dear Dr. Math, There are two independent, uniformly distributed random variables X and Y in {0,1,...,n}. I have to calculate the probability of (X = j I X+Y = k) for 0 <= j <= k <= 2*n. I have no idea how to do it. Can you help me? Yours, Damian from Germany Date: 01/10/2002 at 15:50:22 From: Doctor Anthony Subject: Re: Independent uniformly distributed random variables An easy way to do this is geometrically with x taking integer values 0, 1, 2, ..., n along the x axis, and y also taking integer values 0, 1, 2, ..., n along the y axis. The sample space is then the square of lattice points on the n x n square. For x+y = k we have lattice points lying on the straight line joining (k,0) to (0,k). We need to consider two situations (1) 0 < k <= n (2) n < k <= 2n In situation (1) the number of available lattice points is (0,k), (1,k-1), (2,k-2), (3,k-3),..... (k,0) giving a total of k+1 points and the probability that x = j will be 1/(k+1) In situation (2) with k > n we get the same probability by reflecting in the line x+y = n So we have the points (0,2n-k), (1,2n-k-1), (2,2n-k-2),...., (2n-k,0) giving a total of 2n-k+1 point and the probability that x=j will be 1/(2n-k+1) To summarize: For k <=n P(x=j|x+y=k) = 1/(k+1) For n<k<2n P(x=j|x+y=k) = 1/(2n-k+1) - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/