# Independence PMF

• March 2nd 2010, 02:38 AM
Mohammed
Independence PMF
Hey guys,

I need a little help with this prove. The question is to prove from both sides; so going left to right was easy but right to left, I couldn't figure it out! any suggestion how can I do this?

Pr[X = x; Y = y] = Pr[X = x]Pr[Y = y] , for all x & y belong to R
• March 2nd 2010, 01:11 PM
Danneedshelp
Quote:

Originally Posted by Mohammed
Hey guys,

I need a little help with this prove. The question is to prove from both sides; so going left to right was easy but right to left, I couldn't figure it out! any suggestion how can I do this?

Pr[X = x; Y = y] = Pr[X = x]Pr[Y = y] , for all x & y belong to R

Notice $P(Y=y)=P(Y=y|X=x)$ or equivalently $p(y)=p(y|x)$, since both X and Y are assumed to independent. So, using the latter notation we have that

$p(x)p(y)=p(x)p(y|x)=p(x,y)$.
• March 3rd 2010, 04:46 PM
matheagle
There's nothing to prove here.
this is a valid definition of independence between two random variables.

you can prove that ....

P(X|Y)=P(X) is equivalent to P(X,Y)=P(X)P(Y)

but you have half of a statement.

Quote:

Originally Posted by Mohammed
Hey guys,

I need a little help with this prove. The question is to prove from both sides; so going left to right was easy but right to left, I couldn't figure it out! any suggestion how can I do this?

Pr[X = x; Y = y] = Pr[X = x]Pr[Y = y] , for all x & y belong to R