# Decomposition of a R.V. into dependent and independent parts

## Main Question or Discussion Point

Given random variables X and Y, which are not independent, is it always possible to find a random variable W which is independent from X, such that Y = f(X,W), for some function f?

Example: let the joint distribution of X and Y be
Code:
 Y  0  1
X+-------
0|1/3 1/6
1|1/6 1/3
Then if we let W have the Bernoulli distribution with p = 1/3, and f(x,w) = x XOR w, we have
y = f(X,W).

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gel
yes. You can prove it as follows.

Define,
g_X(y) = P(Y<=y|X)
which gives the cumulative distribution function of Y conditional on X.
Define the variable
U=g_X(Y)
As long as g_X is continuous (so Y has continuous distribution) then U will be uniformly distributed on [0,1] independently of X.

Setting f(x,u)=g_X^{-1}(u) gives Y=f(X,U).

If g_X is not continuous then you can still set Y=f(X,U) for a r.v. U uniform on [0,1] independently of X, and X,Y will have the correct joint distribution. Note that this can require enlarging the probability space in order to introduce U.

Hopefully you can fill in the gaps in that brief proof.