Multivariate Distributions, Moments, and Correlationsby shaiguy6 Tags: correlations, distributions, moments, multivariate 

#1
Mar1011, 12:29 PM

P: 13

So if I start with a multivariate distribution f(x,y), I can find the marginal distributions, the conditional probability distributions, all conditional moments, and by the law of iterated expectations, the moments of both X and Y.
It seems to me that I should be able to relate the conditional moments in x to the conditional moments in y. Right? This is mainly coming from intuition. To be a bit more clear. If I have a function V(x,t) and it has all the properties of a joint probability distribution, I can begin to describe its shape by finding the conditional moments in X and in T. But it seems like all of the conditional moments in X should be able to recreate the original function V(x,t) just as well as all the conditional moments in T. I was wondering if this makes any sense at all. I'm not all too familiar with statistics, and feel like a huge dilletent Relatedly, if i want to define a covariance or correlation between my two random variables, x and t, but I only know their joint distribution, V(x,t), then is the way to go about it to comput the first two moments of X and T using the law of iterated expectations, and then find the covariance and correlation that way? Sorry, one final thing. Having a probability distribution p(x) is equivalent to having the infinity of moments of that distribution. My question is, how can you rebuild the probability distribution given all the moments? Sorry for my ramblings :) Any help is appreciated. 



#2
Mar1211, 03:35 AM

P: 2,490

I'm not exactly sure what you're after, but if you know f(x,y) and f(x) you can find the conditional expectation of Y from: [tex] E(YX=x)=\int y \frac{f(x,y)}{f(x)} dy[/tex] In general, for a random variable X conditional on an event B: E(XB) is the sum or integral of the products of all possible values of the RV (or function) and the respective conditional probabilities of each. EDIT: Note the nth moment is [tex] E(Xx)^n[/tex], n>1 and the covariance E[(Xx)(Yy)]. 



#3
Mar1211, 12:15 PM

P: 13

yea thanks, so I've been messing around a bit, and I think I've almost got it figured out (just need to make everything pretty for my specific example). But in general, this is what I have: Given the moments of either conditional distribution functions, you can recreate the characteristic function. Then, to get back the conditional distribution function, you take the inverse fourier transform of the characteristic function. Since we can easily relate each of the conditional distribution functions to eachother, we can then get the conditional distribution function of the other variable, and then we can get the moments from there. This is how the moments are related to eachother. Going from either of the conditional distributions back to the joint distribution is straightfoward enough. If we keep everything in terms of the original moments, we will derive a relationship between the conditional moments and the joint moments (though in general it doesn't look pretty). 



#4
Mar1211, 01:49 PM

P: 2,490

Multivariate Distributions, Moments, and Correlations 



#5
Mar1211, 02:02 PM

P: 13

Because for instance if I have f(x,y) given, I can get the conditional by taking [tex]f(xy)=\frac{f(x,y)}{\int_{\infty}^{\infty} f(x,y)dy}[/tex] THen we can find moments of that in the normal fashion In other words, the marginal distributions f(y) and f(x) can be directly found from f(x,y), unless I am misinterpreting something. So we have: [tex] E[f(YX=x)]=\int_{\infty}^{\infty} y \frac{f(x,y)}{\int_{\infty}^{\infty} f(x,y)dy} dy [/tex] 


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