Probability of a curve in joint density plane

[CantorSet]

Suppose X and Y are continuous random variables with the joint pdf f_{xy}(x,y) on the [0,1] \times [0,1] square.

Is the probability P(X = Y) then equal to zero since probability here is a volume, and the set that satisfies P(X = Y) is a plane?

Supposing it's not zero, when I tried to evaluate it with the integral

\int_{0}^{1} f_{xy}(t,t)\sqrt2dt

(basically, just a line integral) the answer seems way too big.

Any thoughts from the folks out there?

 

[statdad]

Whenever random variables are continuous the probability of their equality is zero.

 

[CantorSet]

thanks.

 

[mathman]

Whenever random variables are continuous the probability of their equality is zero.

This is not always true. If there is some dependency between X and Y, it may not be true. An extreme example is X=Y, so P(X=Y)=1.

 

[statdad]

This is not always true. If there is some dependency between X and Y, it may not be true. An extreme example is X=Y, so P(X=Y)=1.

Hmmm. I would argue that if X=Y then there is only one random variable. (As stated I'm looking at your comment as a different situation than the case where we construct independent - identically distributed copies of random variables in probability theory.)

 

[mathman]

Hmmm. I would argue that if X=Y then there is only one random variable. (As stated I'm looking at your comment as a different situation than the case where we construct independent - identically distributed copies of random variables in probability theory.)

Fussy! Another example: X=max(Y,1/2). Then P(X=Y)=P(Y≥1/2). You can make up any more as you wish. The main point is the obvious dependency between X and Y.