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bivariate normal distribution |
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| Jun3-12, 10:19 AM | #1 |
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bivariate normal distribution
Quick question on bivariate normal distribution please:
I know for a bivariate normal distribution, the two variables are marginally normal and all the conditional distributions are also normal. Is the reverse true? i.e. if you have two random variables that are marginally normal themselves and all the conditional distributions of one variable given a value of the other are also normal, does this result in the joint distribution has to be "bivariate normal" or it can still be another type of joint distribution? thanks. |
| Jun3-12, 12:59 PM | #2 |
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| Jun3-12, 04:01 PM | #3 |
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Recognitions:
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| Jun3-12, 04:59 PM | #4 |
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bivariate normal distribution |
| Jun3-12, 10:32 PM | #5 |
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thanks guys, just to be clear,
even if they are not independent, is the joint distribution has to be bivariate normal? no need for proof. |
| Jun4-12, 02:29 AM | #6 |
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| Jun4-12, 03:35 AM | #7 |
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on top of X and Y are normally distributed but not independent,
if all the conditional distributions are also normal, does that mean X and Y are definitely jointly normal? |
| Jun4-12, 04:21 AM | #8 |
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| Jun4-12, 04:50 AM | #9 |
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thanks,
so you don't think the fact not only all conditional distributions are normal, X and Y are also marginally normal, this is not enough to restrict the joint distribution to be bivariate normal? |
| Jun4-12, 07:05 AM | #10 |
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You could try to further constraint X and Y to behave in such a way that a particular linear combination of their values would not behave normally even if for every particular value of X and Y they do (which is basically what your restriction does). And if that does not work you could still try to mess with the other conditions. |
| Jun4-12, 10:17 AM | #11 |
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thanks,
i don't have the proof, but I think the fact because X and Y are also marginally normal (on top of all conditional distributions are normal), this extra condition does make every linear combination of X and Y normally distributed hence bivariate normal, you don't think that is the case? any idea how to proof? |
| Jun4-12, 10:17 AM | #12 |
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Mathman do you know who is correct?
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| Jun4-12, 01:35 PM | #13 |
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Imagine that given X=x the variance of the normal conditional distribution of Y is inversely proportional to x, and also imagine that the variance of X condition to Y=y is also inversely proportional to y. Now you have X,Y, X|Y=y and Y|X=x following normal distributions, but you are getting in the bivariate distribution a contour line that looks nothing like an ellipse which is what you would expect if it would follow a bivariate joint normal distribution... yeah? I let for you the fun to do the formal proof though
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| Jun4-12, 03:39 PM | #14 |
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Recognitions:
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| Jun4-12, 05:56 PM | #15 |
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With the bivariate normal PDF, the variables may be correlated. Jointly normal random variables are uncorrelated and are a special case of the bivariate normal PDF.
http://athenasc.com/Bivariate-Normal.pdf |
| Jun4-12, 07:47 PM | #16 |
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I've been reading this thread for a couple of days and I also found the above referenced article. As I recall learning once, the missing condition is that the conditional variances must be constant in addition to the conditions given in the original post. I can't find any reference to this and now I'm wondering if it equivalent to the description given in this article or I'm just wrong. Any ideas?
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| Jun4-12, 09:24 PM | #17 |
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viraltux, your example doesn't work as it breaks one of the restrictions.
In your case, as soon as correlation is not zero (the mean of the normal conditional distributions shift), the resultant marginal distribution is no longer normal. In fact, the more I think about it, it does seem the only way of keeping marginal distribution of Y normal is for the normal conditional distributions of Y for each value of X follow the bivariate normal formula, with the mean shifting slightly and variance the same. As soon as you change the variance of these conditional distributions or shift the mean in a different way, the resultant marginal distribution of Y is no longer normal. Therefore I do still think that given the restriction of both X and Y being marginally normal and all conditional distributions being also normal, it does mean that X and Y has to be jointly normal and the joint distribution can not be anything else. Would be great if someone can confirm this or convince otherwise. Thanks. |
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