# Prove $E^{\mathbb Q}[Y]=E^{\mathbb P}[XY]$ if $E^{\mathbb P}[X]=\mathbb P(X>0)=1$ and $\mathbb Q(A)=E^{\mathbb Q}[X1_A]$ [closed]

How do I prove the following? I don't know where to start.

If $X$ is a random variable with $E^{\mathbb P}[X] = \mathbb P(X>0)=1$ and $\mathbb Q$ is the probability measure defined by $\mathbb Q(A)=E^{\mathbb Q}[X1_A]$ then $E^{\mathbb Q}[Y]=E^{\mathbb P}[XY]$

• For $Y=1_A$ the statement holds by definition of $Q$. So it also holds for step functions. Then there's a unique way to extend it to arbitrary measurable function $Y$. Commented Mar 14, 2018 at 1:52
• @Berci Do we need standard machine ? We deduce that $X$ is the radon-nikodym derivative from the definition of $Q$ which is precisely what (and all?) we need to prove the result. Or is it that R-N derivative is merely a part of the proof, and we really need to go through the standard machine?
– BCLC
Commented Mar 14, 2018 at 7:27
• @Berci Did you mean 'simple' rather than 'step'? Or you meant 'step' and then the next parts are 'simple', 'nonnegative' and then 'measurable' ?
– BCLC
Commented Mar 14, 2018 at 8:27

Use what David Williams calls the standard machine:

I guess $Y,X,XY \in \mathscr L ^{1}(\Omega, \mathscr F, \mathbb P)$, $Y,X,XY \in \mathscr L ^{1}(\Omega, \mathscr F, \mathbb Q)$ and $A \in \mathscr F$.

We want to show that $$E^{\mathbb Q}[Y]=E^{\mathbb P}[XY] \tag{*}$$

Doing this consists of four parts:

1. Show $(*)$ is true for $Y=1_A$

I think this is $$LHS(*) = \int_{\Omega} Y d \mathbb Q = \mathbb Q(A)$$

1. Show $(*)$ is true for nonnegative simple functions, $Y \in SF^{+}$, that is, $Y=a_11_{A_1} + \cdots + a_n1_{A_n}$.

I think this is $$LHS(*) = \int_{\Omega} Y d \mathbb Q = a_1 \mathbb Q(A_1) + \cdots + a_n\mathbb Q(A_n)$$

1. Show $(*)$ is true for nonnegative measurable functions, $Y \in (m \mathscr F)^{+}$.

I think this $$LHS(*) = \int_{\Omega} Y d \mathbb Q = \sup_{h \in SF^{+}, h \le Y}\{\int_{\Omega} h d \mathbb Q\}$$

1. Show $(*)$ is true for measurable functions $Y = Y^{+} - Y^{-}$

I think this $$LHS(*) = \int_{\Omega} Y d \mathbb Q = \int_{\Omega} Y^{+} d \mathbb Q - \int_{\Omega} Y^{-} d \mathbb Q$$

Wait I don't think we need the standard machine. 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

$$E^{\mathbb Q}[Y] = E^{\mathbb P}[XY] \tag{2}$$

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

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

Now, $LHS(2) = RHS(2)$ if $XYd\mathbb P = Yd\mathbb Q$

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

So um, QED? Or maybe the last part is where standard machine comes in.