# Computing expectation under a change of measure [duplicate]

Let $X$ be a random variable on a probability space $(\Omega,\mathscr F, P)$. Define a new probability measure $$\tilde P(A) = E[1_A X]$$ for all $A\in\mathscr F$. Let $\tilde E$ be expectation taken with respect to the new measure $\tilde{P}$.

Suppose now that $Y$ is also a random variable $(\Omega,\mathscr F)$. Then intuitively the expectation should be computed as $$\tilde E [1_A Y] = E[1_A YX],$$ but I'm not sure how to prove this rigorously using the definition.

This is a standard exercise in proving identities with the Lebesgue integral. Prove it first with a simple function, i.e. if $\phi(\omega) = \sum a_i \chi_{A_i}(\omega)$ then $$\int_A \phi(\omega) \tilde{P}(d\omega) = \int_A \phi(\omega)X(\omega) P(d\omega).$$ Now prove it for non-negative random variables. To do this, it should be known that there is a sequence of increasing simple functions $\phi_n \to Y$, and apply the previous result and monotone convergence theorem. Then the result follows for random variables follows.

I'm changing $\tilde{\mathbb P}$ to $\mathbb Q$.

We're given that

$$\mathbb Q(A) = E^{\mathbb P}[X1_A] \tag{1}$$

$$LHS(1) := \mathbb Q(A) = \int_{\Omega}1_A d\mathbb Q$$

$$RHS(1) := \int_{\Omega} X1_A d\mathbb P$$

Thus, $$X1_Ad\mathbb P = 1_Ad\mathbb Q$$

which I guess is equivalent to

$$Xd\mathbb P = d\mathbb Q \tag{*}$$

Now, we want to prove that

$$E^{\mathbb Q}[1_AY] = E^{\mathbb P}[1_AXY] \tag{2}$$

$$LHS(2) := \int_{\Omega} Y1_A d\mathbb Q$$

$$RHS(2) := \int_{\Omega} XY1_A d\mathbb P$$

Now, $LHS(2) = RHS(2)$ if $XY1_Ad\mathbb P = Y1_Ad\mathbb Q$

which I guess is equivalent to $(*)$.

• $X 1_A dP = 1_A dQ$ does not strictly mean anything, and the implication to $XdP = dQ$ is dubious. It seems to me that you are implicitly using the Radon-Nikodym thm, saying that $X$ is the derivative of $Q$ w.r.t. $P$, which is a bit of a stretch. Commented Mar 17, 2018 at 10:18
• @Sisyphus Thanks. Well I did say 'I guess'. Could it be that we instead have, for $(*)$, $P(X d\mathbb P = d\mathbb Q)=1$, so, now, we have $(1) \to (*) \to (2)$? Also is $(*) \to (2)$ wrong or dubious?
– BCLC
Commented Mar 19, 2018 at 5:51
• Once again you are using Radon-Nikodym implicitly. It is not 'wrong', but it requires a lot more machinery than necessary. $XdP = dQ$ doesn't mean anything algebraically, just as $f'(x)dx = dy$ doesn't mean anything algebraically. Commented Mar 19, 2018 at 8:18
• @Sisyphus, about #1, do you mean you're not sure if I'm right or wrong, but the answer is likely related to R-N theorem? About #2, okay I was unclear. I meant to ask if you have issues with the second implication. If so, are they the same as the first implication (either with or without almost surely)?
– BCLC
Commented Mar 19, 2018 at 8:37
• @Sisyphus, actually is $P(XdP=dQ)=1$ dubious? I just realised that even though RN is unique a.s., I don't know which probability measure to use for the a.s. Is it P? Q? Other?
– BCLC
Commented Mar 19, 2018 at 8:39