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I have difficulty visualizing what a joint pdf of two independent random variables might look like. The one that I can think of is a cylindrical extension of a univariate Gaussian (let's say extent from x to y). However, in this case, only X is independent of Y (in the sense that the choice of Y does not affect the distribution of X) but Y is entirely dependent on X (in the same sense).

Sam
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1 Answers1

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Independence is symmetric, as you can see both from the definition $P(X\cap Y)=P(X)P(Y)$ and from the symmetric form of Bayes’ theorem, $P(A\mid B)P(B)=P(B\mid A)P(A)$.

An example of independent variables that might help to visualize the concept is afforded by two univariate Gaussians with different variances. The product of their probability density functions has an elliptically shaped bulge; along each line parallel to an axis it yields a scaled version of the marginal density for that axis.

joriki
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  • But if say the choice of x will scale the distribution of y (given fixed x), it feels like y is dependent on x. – Sam Mar 26 '20 at 09:37
  • @Zzy1130: I'm not sure what that means. Could you elaborate? – joriki Mar 26 '20 at 09:38
  • For example in that elliptically shaped bulge, if I fix x=10, I will get a y distribution, I fix x=20, I will get another y distribution. If x and y are independent, this two independent should look identical. But I think it's not the case for an elliptically shaped bulge. – Sam Mar 27 '20 at 12:51
  • @Zzy1130: Well, it depends on the elliptically shaped bulge. If the eliptically shaped bulge arises, as in my example, from multiplying the two marginal probability density functions, then the two sections will be similar by construction. For example, look at this eliptically shaped bulge. – joriki Mar 27 '20 at 12:58
  • yeah, but if you want them to be exactly identical (with exactly the same amplitude), so far the only solution I think think of (I spent another bit of time and conclude on this) is f(x,y)= C where C is just an arbitrary constant. Besides, if the two sections just differ by scaling, they cannot represent pdf at the same time as one of them cannot be normalized. – Sam Mar 30 '20 at 03:13
  • @Zzy1130: That last sentence seems to be due to a misconception. Sections of a joint probability distribution are not themselves probability density functions. – joriki Apr 03 '20 at 17:03
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    Oh yes let me think through again. thanks for pointing that out. – Sam Apr 10 '20 at 09:06
  • wait but isn't that slice just the conditional probability when fixing one random variable? And since conditional probability are also normalized (given a fixed condition), it can be regarded as probability density function right? – Sam Apr 12 '20 at 09:10
  • @Zzy1130: I'm not sure I understand. Yes, if you normalize it, it's a conditional probability. But normalizing it rescales it; I thought you didn't want to do that. But maybe I just don't get what you're trying to achieve. – joriki Apr 12 '20 at 09:15
  • Yes you are right. Need to rescale. I forgot that. – Sam Apr 13 '20 at 05:06