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Old 07-24-2018, 06:13 AM
fjmvasa fjmvasa is offline
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Default Independence of x and y (joint)

Hi guys,

Apologies if this may just be a simple question, but I prefer thinking out loud, so here's my question:

For the independence of x and y, do the following conditions have to be BOTH met?

1. The probability space is rectangular
2. f(x,y) = f(x) * f(y)

I'm quite confused.

Thanks in advance!
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Old 07-24-2018, 08:24 AM
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Michael Mastroianni Michael Mastroianni is offline
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The rectangle can be unbounded, but yes. If the support is not rectangular, then the possible values of one variable depend on the value of the other. This is why. If both have a marginal pdf or pmf, then the product of these is their joint pdf or pmf if and only if they are independent.
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Old 07-24-2018, 09:50 AM
fjmvasa fjmvasa is offline
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Thank you very much, Michael! Now it’s clearer to me.
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Old 07-24-2018, 10:57 AM
nonlnear nonlnear is offline
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To be clear, "rectangular" in this case includes the degenerate rectangles: the point and the line.

And the rectangles in question must have edges parallel to the axes. No crooked rectangles.
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Old 07-25-2018, 08:50 AM
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Michael Mastroianni Michael Mastroianni is offline
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Quote:
Originally Posted by nonlnear View Post
To be clear, "rectangular" in this case includes the degenerate rectangles: the point and the line.

And the rectangles in question must have edges parallel to the axes. No crooked rectangles.
I suppose if we’re going to be completely precise, we should be clear that “rectangular” refers to any cartesian product of random variable supports (imagine the white piece of Finland’s flag). Usually the random variables we work with have a support that is a single interval (bounded or unbounded) so this doesn’t really come up. In this case your conditional distributions are undefined in the gaps so there’s no dependence.

You can easily prove that your (2) implies (1) here which is why (2) is equivalent to independence.
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Old 07-27-2018, 07:53 PM
nonlnear nonlnear is offline
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Quote:
Originally Posted by Michael Mastroianni View Post
I suppose if we’re going to be completely precise, we should be clear that “rectangular” refers to any cartesian product of random variable supports (imagine the white piece of Finland’s flag). Usually the random variables we work with have a support that is a single interval (bounded or unbounded) so this doesn’t really come up. In this case your conditional distributions are undefined in the gaps so there’s no dependence.

You can easily prove that your (2) implies (1) here which is why (2) is equivalent to independence.
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