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How to show that two random variables are independent Watch

1. Hi all,
I was wondering if there's a general way to show that two random variables are independent (as you might have guessed from the title) apart from directly calculating P(X <= k), P(Y <= m), P(X <= k and Y <= m). Is there a way of using PGFs, for instance? (Continuous or discrete variables welcomed.)
It would be nice to have something as simple as "Covariance = 0" - but I know that's only a necessary condition, not a sufficient one.
Thanks!
2. (Original post by Smaug123)
Hi all,
I was wondering if there's a general way to show that two random variables are independent (as you might have guessed from the title) apart from directly calculating P(X <= k), P(Y <= m), P(X <= k and Y <= m). Is there a way of using PGFs, for instance? (Continuous or discrete variables welcomed.)
It would be nice to have something as simple as "Covariance = 0" - but I know that's only a necessary condition, not a sufficient one.
Thanks!
Of course, the general definition is:

Two random variables, X and Y, are independent

is independent of

Other than that, I don't recall any "simple" way of showing independence, although it would be nice to have one!
3. (Original post by Smaug123)
Hi all,
I was wondering if there's a general way to show that two random variables are independent (as you might have guessed from the title) apart from directly calculating P(X <= k), P(Y <= m), P(X <= k and Y <= m). Is there a way of using PGFs, for instance? (Continuous or discrete variables welcomed.)
It would be nice to have something as simple as "Covariance = 0" - but I know that's only a necessary condition, not a sufficient one.
Thanks!
Have you learnt (or are you learning) measure theory?

Another way to check independence is that if both X and Y have densities, then X and Y are independent iff their joint pdf is equal to the product of the marginals.
4. (Original post by Mark13)
Have you learnt (or are you learning) measure theory?
Another way to check independence is that if both X and Y have densities, then X and Y are independent iff their joint pdf is equal to the product of the marginals.
I've heard of measure theory in connection with probability (and also the "measure of the Cantor dust", so something to do with integrals and fractals) but that's as far as it goes so far, I'm afraid. Working out the joint PDF and the marginals is that it requires the same amount of work as working out P(X <= k), doesn't it, or is there a trick I've missed?
5. (Original post by Smaug123)
I've heard of measure theory in connection with probability (and also the "measure of the Cantor dust", so something to do with integrals and fractals) but that's as far as it goes so far, I'm afraid. Working out the joint PDF and the marginals is that it requires the same amount of work as working out P(X <= k), doesn't it, or is there a trick I've missed?
Yeh the pdf method and distribution function method are very similar. There's not really much else you can do which isn't a rephrasing of these ideas in general, without using some measure theory.

For normal random variables, covariance = 0 implies independent, but the result doesn't really generalise much further than that.
6. (Original post by Mark13)
Yeh the pdf method and distribution function method are very similar. There's not really much else you can do which isn't a rephrasing of these ideas in general, without using some measure theory.

For normal random variables, covariance = 0 implies independent, but the result doesn't really generalise much further than that.
Ah, OK - thanks anyway I'm sure I should know why cov=0 implies independent for normal RVs - could you remind me in what general direction the proof goes?
7. (Original post by Smaug123)
Ah, OK - thanks anyway I'm sure I should know why cov=0 implies independent for normal RVs - could you remind me in what general direction the proof goes?
Sorry, my post above isn't quite right - if X and Y are jointly normal then their covariance being zero implies independence, but it's not enough for X and Y just to be normal random variables. Not sure about the proof, proofwiki is probably a good place to look.
8. (Original post by Mark13)
Sorry, my post above isn't quite right - if X and Y are jointly normal then their covariance being zero implies independence, but it's not enough for X and Y just to be normal random variables. Not sure about the proof, proofwiki is probably a good place to look.
Thanks - I suspect it's in my notes somewhere anyway, I just can't remember where straight off.

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