# Thread: Density of two discrete uniform random variables

1. ## Density of two discrete uniform random variables

Let X and Y be independent random variables having the uniform density on {0, 1,..,N}. How do you find the density of min(X,Y)?

Here's what I did.
First i wrote out
P(min(X,Y) >= z) = P(X >= z, Y >= z)
= P(X >=z)*P(Y>=z)
= (N+1 - z + 1)/(N+1) * (N+1 - z + 1)/(N+1)
= (N+1-z+1)^2/(N+1)^2

I'm not whether that is the correct way to start. And I don't know what to do after that. The solution provided by the book is [2(N-z)+1]/(N+1)^2.

Can someone explain how to solve this problem?
Thanks!

2. Originally Posted by feiyingx
Let X and Y be independent random variables having the uniform density on {0, 1,..,N}. How do you find the density of min(X,Y)?

Here's what I did.
First i wrote out
P(min(X,Y) >= z) = P(X >= z, Y >= z)
= P(X >=z)*P(Y>=z)
= (N+1 - z + 1)/(N+1) * (N+1 - z + 1)/(N+1)
= (N+1-z+1)^2/(N+1)^2

I'm not whether that is the correct way to start. And I don't know what to do after that. The solution provided by the book is [2(N-z)+1]/(N+1)^2.

Can someone explain how to solve this problem?
Thanks!
You can't have a density on a discrete sample space, you need a probability
mass function.

RonL

3. How can we compute the mass function of min(X,Y)?

4. Originally Posted by feiyingx
How can we compute the mass function of min(X,Y)?
p(x=c)=1/N, p(y=c)=1/N, p(x>c)=1-c/N, p(y>c)=1-c/N

p(min(x,y)=c) = p(x=c)*p(y>c) + p(y=c)*p(x>c) + p(x=c)p(y=c)

RonL

5. thanks