Finding E(Y) and Var(Y) with Conditional Expectation

island-boy
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Is it possible to solve for E(Y) and var (Y) when I am only given the distribution f(Y|X)?

I can solve for E(Y|X). But is it possible to find E(Y) and var(Y) given only this info?
 
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No it is not. Let Y be distributed as f(x) = .5 if x = -1 or 1, 0 otherwise. Let X be distributed as f(x) = 1 if x = 1, 0 otherwise. Then f(Y|X) is the distribution of X. You can change the mean and variance of Y to almost whatever you want by moving the other probability mass, and f(Y|X) will not be affected.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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