# Uniformly Distributed Random variables

• Oct 5th 2009, 04:52 PM
Dfowj
Uniformly Distributed Random variables
Hey guys, another one of these crazy probability questions...

Let X and Y be independent, each uniformly distributed on {1, 2, ... , n}.

Find P(X + Y = k) for 2 <= k <= 2n.

This is the last of a series of problems, all of the previous ones which were easily solved by drawing up a grid of possible choices. Is there a way to figure this out WITH OUT a grid?

A gentle shove in the right direction would be helpful! Thanks!
• Oct 6th 2009, 01:14 PM
Moo
Hello,
Quote:

Originally Posted by Dfowj
Hey guys, another one of these crazy probability questions...

Let X and Y be independent, each uniformly distributed on {1, 2, ... , n}.

Find P(X + Y = k) for 2 <= k <= 2n.

This is the last of a series of problems, all of the previous ones which were easily solved by drawing up a grid of possible choices. Is there a way to figure this out WITH OUT a grid?

A gentle shove in the right direction would be helpful! Thanks!

think of it as events.

Let's consider the possible values for Y, in the event {X+Y=k}
Y can take any value between {1,2,...,n}

So you can divide the probability space into events :
$\{X+Y=k\} = \bigcup_{j=1}^n \{X+Y=k ~ and ~ Y=j\}$

So this gives $\mathbb{P}(X+Y=k)=\sum_{j=1}^n \mathbb{P}(X+Y=k ~,~ Y=j)$, since the events are disjoint (Y cannot take two values at a same time)

---> $\mathbb{P}(X+Y=k)=\sum_{j=1}^n \mathbb{P}(X+j=k ~,~ Y=j)=\sum_{j=1}^n \mathbb{P}(X=k-j ~,~ Y=j)$

$=\sum_{j=1}^n \mathbb{P}(X=k-j)\mathbb{P}(Y=j)$ because X and Y are independent.

Does it look clear to you ?