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 June 7th, 2013, 09:14 AM #1 Senior Member   Joined: Jul 2012 Posts: 225 Thanks: 0 What am I doing wrong? simple probability Here's the question: X and Y are discrete random variables that get values from 1 to N with equal probability. In other words, X and Y have a uniform distribution with values between 1 and N. We defined another discrete random variable Z. Z=X-Y. Find the probability function of Z. My answer is incorrect, but I don't know why and I'd like an explanation. Here is my answer: Z can get these values: $Z= 1-N,2-N,3-N, ... , N-2,N-1$ $P(Z=k)= P(X-Y=k) = P((X=k+1,Y=1) , (X=k+2,Y=2) , (X=k+3,Y=3), ... , (X=N, Y=N-k))$ We can see that there are N-k events that yield X-Y=k, out of overall $N^{2}$ possible events. This gives us that $P(Z=k) = \frac{N-k}{N^{2}}$ But that is an incorrect answer. Why?
June 7th, 2013, 01:11 PM   #2
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Re: What am I doing wrong? simple probability

Quote:
 Originally Posted by OriaG Here's the question: X and Y are discrete random variables that get values from 1 to N with equal probability. In other words, X and Y have a uniform distribution with values between 1 and N. We defined another discrete random variable Z. Z=X-Y. Find the probability function of Z. My answer is incorrect, but I don't know why and I'd like an explanation. Here is my answer: Z can get these values: $Z= 1-N,2-N,3-N, ... , N-2,N-1$ $P(Z=k)= P(X-Y=k) = P((X=k+1,Y=1) , (X=k+2,Y=2) , (X=k+3,Y=3), ... , (X=N, Y=N-k))$ We can see that there are N-k events that yield X-Y=k, out of overall $N^{2}$ possible events. This gives us that $P(Z=k) = \frac{N-k}{N^{2}}$ But that is an incorrect answer. Why?
When k < 0, you need to check your count.
Should be $P(Z=k) = \frac{N-|k|}{N^{2}}$

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