Need help with a conditional variance proof.

[Kuma]

need urgent help with a conditional variance proof.

I have been given this problem and I'm pretty stumped.

I want to prove that Y=g(X) if and only if var(YlX) = 0.

so if var(YlX)=0 then

E(Y^2lX) - E(YlX)^2 = 0

E(Y^2lX) =E(YlX)^2

so what should I do now? I tried showing that this equation is true when y =g(X), but in the end I end up with var(g(X)) = 0...unless I did something wrong.

 

[Stephen Tashi]

I want to prove that Y=g(X) if and only if var(YlX) = 0.

In an advanced course on probability, I doubt that statement is true because there are techicalities about "sets of measure zero". If this problem is from an introductory course in probability, I assume it deals with discrete distributions.

Suppose there is a value [itex] X_0 [/itex], such that Y can have two or more different values when [itex] X = X_0 [/itex]. Show that at least one of these possible values for Y contribues a positive term to the sum for computing the variance [itex] E(Y|X_0) [/itex].

 

[Kuma]

thanks for the reply. I kind of see where this is going. So for example if I had y^2 = x, then for each value of x there would be 2 values of y.

there is still a few things I'm a bit lost on though. Why would I want to compute the variance of the conditional expectation? I'm looking for the variance to be zero.

the question let's me assume that these are discrete rvs.

the conditional expectation of (YlX=x0) is calculated by sigma x * p(xly). So if I had two different values for x, am I trying to make the sum 0?

 

[Stephen Tashi]

tWhy would I want to compute the variance of the conditional expectation? I'm looking for the variance to be zero.

My mistake. I should have written [itex] Var(Y|X_0) [/itex] instead of [itex] E(Y|X_0) [/itex].

If the joint density is f(x,y)

[itex] Var(Y|X_0) = \sum (y - \bar{y} )^2 C f(y,x_0) [/itex]
where [itex] C = \frac {1}{ \sum f(y,x_0)} [/itex] and the sums are taken over all possible values of Y.

If Y is not a function of X then there are at least two values [itex] y_1 [/itex] and [itex] y_2 [/itex] with [itex] f(y1,x_0) > 0 [/itex] and [itex] f(y2,x_0) > 0 [/itex]. Only one of these values can be equal to [itex] \bar{y} [/itex].

I evaded the issue of whether [itex] \bar{y} [/itex] is [itex] E(Y)[/itex] or [itex] E(Y|x_0) [/itex] because I don't remember the precise definition of conditional variance!

 

[Kuma]

I'm a bit confused about the part where you mentioned that if y was not a function of x, then there are 2 y values or more. Why is that? I'm not sure what it has to do with y nkt being a function of x.

also you sai that only one of f(yn, x) can be y bar. How js that possible since in the discrete case the joint distribution is a probability.

 

[Stephen Tashi]

I'm a bit confused about the part where you mentioned that if y was not a function of x, then there are 2 y values or more. Why is that? I'm not sure what it has to do with y nkt being a function of x.

If, for each value of x, there is one and only one possible y value possible then y is a function of x. Essentially that is the definition of a function. If y is not a function of x then either there is a y_i not associated with any of the x_i or there is an x_i with two or more possible y_i's.

also you sai that only one of f(yn, x) can be y bar. How js that possible since in the discrete case the joint distribution is a probability.

I didn't say that only one of the f(yn,x) can be y bar. I said only one of y1 and y2 can be y bar.

 

[Kuma]

but what about for multi valued functions that are not 1-1, so in that case like you said, y is not a function of x? I kind of get it because if it was a 1-1 function, where there is only 1 y for every x, then the variance would be 0. Since µ would just be g(X) and the variance of that is 0.

so jf I understand this correctly, that is the proof in itself? Since the variance of a function of x is 0, I'm done? If there are 2 or more y for each x then y js not a function of x.

 

[Stephen Tashi]

but what about for multi valued functions that are not 1-1, so in that case like you said, y is not a function of x?

"Multi-valued functions" are not functions, just like "counterfeit money" isn't money.