Actuarial Outpost Independence of x and y (joint)
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#1
07-24-2018, 05:13 AM
 fjmvasa CAS SOA Join Date: Apr 2018 Posts: 7
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.

#2
07-24-2018, 07:24 AM
 Michael Mastroianni Member SOA Join Date: Jan 2018 Posts: 33

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.
__________________
Michael Mastroianni, ASA
Video Course for Exam 1/P: www.ProbabilityExam.com
#3
07-24-2018, 08:50 AM
 fjmvasa CAS SOA Join Date: Apr 2018 Posts: 7

Thank you very much, Michael! Now it’s clearer to me.
#4
07-24-2018, 09:57 AM
 nonlnear Member Non-Actuary Join Date: May 2010 Favorite beer: Civil Society Fresh IPA Posts: 30,566

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.
#5
07-25-2018, 07:50 AM
 Michael Mastroianni Member SOA Join Date: Jan 2018 Posts: 33

Quote:
 Originally Posted by nonlnear 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.
__________________
Michael Mastroianni, ASA
Video Course for Exam 1/P: www.ProbabilityExam.com
#6
07-27-2018, 06:53 PM
 nonlnear Member Non-Actuary Join Date: May 2010 Favorite beer: Civil Society Fresh IPA Posts: 30,566

Quote:
 Originally Posted by Michael Mastroianni 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.