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But we're talking about "continuous variables", random variables with infinite support for which P(X=a)=0, for all a. It's kind of like just a single point in
by H8crilA 6y ago
But we're talking about "continuous variables", random variables with infinite support for which P(X=a)=0, for all a.
It's kind of like just a single point in the convolution for "X_1 - X_2", corresponding to "X_1 - X_2 = 0" ?
- epistasis 6y agoI'm not sure I understand what you are trying to convey. If we have two continuous random variables in a probability space, we have a mapping P(X, Y): R^2 --> R, such that the integral sums to 1. If X and Y are independent, then P(X, Y) == P(X)P(Y). The examples in the linked article are a bit confusing because some of them have P(X) and P(Y) as different distributions. But the plot only shows the plot of the 2D domain along the 1D line of X==Y, which isn't a terribly common occurrence in actual practice.
- H8crilA 6y agoP(X,Y), when integrated along all possible diagonals given by x+y=z where z goes through all real numbers, is what a convolution is (of P(X) and P(Y) generated by those variables). This gives you the distribution of the variable Z=X+Y. So, we're talking about the same thing, namely that multiplying two probability density functions pointwise roughly corresponds to a single point where X==Y (the same as X-Y==0), but otherwise doesn't seem terribly useful.