Re: Question about computing probablities by conditioning
- From: J W <julianw@xxxxxxxxxxxx>
- Date: Mon, 04 Sep 2006 21:30:15 -0700
smartgelu@xxxxxxx wrote:
hi everyone:
i got a question that puzzled me. we know that
P(E)=sum P(E | Y=y)P(Y=y) if Y is discrete
=integration P(E | Y=y)f(y)dy if Y is
continuous
f(y) is the continuous probablity density fuction of random variable Y
we can compute P(E) via the above way, that's the 1-D case.
how about 2-D case?
could anybody tell me whether the following equation is correct or not
P(E)=integration P(E |X=x,Y=y)f(x)f(y)dxdy X,Y are both continuous
ramdom variables
If it is right,could you provide a simple way to prove it?
thx!
You're right... sometimes. The general expression you're looking for is
P(E) = int [ P(E | X=x, Y=y) f(x,y) dx dy ]
where f(x,y) is the JOINT distribution of the random variables X and Y. In the case where X and Y are independent, f(x,y) = f(x)g(y), where f and g are the marginal (univariate) densities of X and Y; this will give the same expression (modulo notation) that you proposed. (Note to pedants: yes, I know that one shouldn't write P(E | X=x,Y=y) in the continuous case, but I didn't want to stray too far from the original notation).
As for a proof, I believe the "real" way involves some measure theory (in fact, this is the way that conditional probability is rigorously defined... if you think about it, what does it REALLY mean that X has some distribution given Y?). Perhaps some others know an easier, more heuristic argument.
-J
.
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