Fix probability space $(\Omega, F, P)$, let $D$ be a nonnegative random variable with $E[D] = 1$. Define a new probability measure $Q$ on $(\Omega, F)$ by $$Q[A] := E[D 1_A]$$I have the following result and I don't see where does it come from : the expectation of a random variable $X$ with respect to $Q$ is given by$$E_Q[X]=E_P[DX] $$
If $X$ and $D$ are discrete random variables, then $$E_Q[X]=\sum x Q(X=x)=\sum x E(D 1_{X=x}) = \sum x \sum y P(D=y \cap X=x) $$which I struggle to make it equal to $E_P[DX]$.Therefore, I tried the other way letting $$E_P[DX]= \sum i P(DX=i) = \sum i \sum P(X=j \cap D=\frac ij ) $$which is also still a bit far from the desired result. Is there a way out ?