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### Independent, Uniformly Distributed Random Variables

```
Date: 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/
```
Associated Topics:
High School Probability

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