# Uniform probability

I'm confused on how exactly they used independence to get that line. Could someone break down the steps so I can see more clearly

What's wrong with my thought process aswell, P(XY<=z|X=x) is by definition P(xY<=z) which is equal to P(Y<=z/x)
(edited 10 months ago)
Simply the joint being independent means that when you condition on X say, the marginal pdf of Y is independent of X so
p(Y|X) = p(Y,X)/p(X) = p(Y)p(X)/p(X) = p(Y)

As an extreme counter example, you could have Y=X and in this case
p(Y|X) is not p(Y) as the given information about X tells you everything about the value of Y. So saying
P(XY<=z|X=x) is by definition P(xY<=z)
if we had this total dependence, then P(XY<=z|X=x) would reduce to a deterministic x^2<z and not the marginal distribution P(xY<=z)
(edited 10 months ago)
Original post by mqb2766
Simply the joint being independent means that when you condition on X say, the marginal pdf of Y is independent of X so
p(Y|X) = p(Y,X)/p(X) = p(Y)p(X)/p(X) = p(Y)

As an extreme counter example, you could have Y=X and in this case
p(Y|X) is not p(Y) as the given information about X tells you everything about the value of Y. So saying
P(XY<=z|X=x) is by definition P(xY<=z)
if we had this total dependence, then P(XY<=z|X=x) would reduce to a deterministic x^2<z and not the marginal distribution P(xY<=z)

Apologies but I'm a bit confused on your last paragraph, are you saying that it's correct to get P(XY<=z|X=x) = P(xY<=z) due to independence of X,Y. Whereas if X,Y were not independent for example X=Y, then the above would reduce to P(x^2<z) which isn't particularly useful
Original post by Student 999
Apologies but I'm a bit confused on your last paragraph, are you saying that it's correct to get P(XY<=z|X=x) = P(xY<=z) due to independence of X,Y. Whereas if X,Y were not independent for example X=Y, then the above would reduce to P(x^2<z) which isn't particularly useful

Agree with the bold. However, you know that the random variable X has the value of x, then if Y and X are completely dependent (for example), then
P(XY<=z|X=x)
is no longer probabilistic, nevermind a distribution on Y.