If $X$ and $Y$ are independent r.v. then $E[Y \mid \sigma(X)] = E[Y]$ using $\sigma$-algebra's

Question

I have seen the following definition of a conditional expectency:

Let $X$ be a r.v. in a probability space $(\Omega, \mathcal{F}, \mathcal{P})$ and let $\mathcal{G}\subseteq\mathcal{F}$ be a sub-$\sigma$-algebra, then $E[X\mid \mathcal{G}]$ is an integrable random variable such that:

1. $E[X\mid\mathcal{G}]$ is $\mathcal{G}$-measurable
2. $\int_G X \mathrm{d}\mathcal{P} = \int_G E[X\mid \mathcal{G}] \mathrm{d}\mathcal{P}$ for each $G\in \mathcal{G}$

Now I would like to prove the following:

If $X$ and $Y$ are independent and Y is integrable, then $E[Y\mid \sigma(X)] = E[Y]$

This implies that I need to prove two things,

1. Is $E[Y]$ $\sigma(X)$-measurable? (where $\sigma(X) = \{X^{-1}(A):A\in \mathcal{B}(\mathbb{R})\}$)

   This looks ok, since $E[Y]$ can be seen as a function which maps $\Omega$ on $E[Y]\in \mathbb{R}$. This implies $E[Y]^{-1}=\Omega \in \sigma(X)$.

2. Is $\int_G E[Y] \mathrm{d}\mathcal{P} \stackrel{?}{=} \int_G Y \mathrm{d}\mathcal{P}\qquad $ ($\forall G\in \sigma(X))$

   I'm don't see the implications of the independence here. The left hand side is equal to $E[Y]\mathcal{P}(G)$, but how about that right hand side? I know that $X$ independent from $Y$ implies $\sigma(X)$ and $\sigma(Y)$ independent, which implies $\forall A\in \sigma(X), \forall B\in \sigma(Y)$, $\mathcal{P}(A\cap B)= \mathcal{P}(A)\cdot \mathcal{P}(B)$ but how can I use this here? And is $\int_G Y \mathrm{d}\mathcal{P}$ where $G\in \sigma(X)$ even properly defined?

Answer 1

1. Is $E[Y]$ $\sigma(X)$-measurable? (where $\sigma(X) = \{X^{-1}(A):A\in \mathcal{B}(\mathbb{R})\}$)

This looks ok, since $E[Y]$ can be seen as a function which maps $\Omega$ on $E[Y]\in \mathbb{R}$. This implies $E[Y]^{-1}=\Omega \in \sigma(X)$.

Actually, $E[Y]$ is a number. Any number $c$ is $\mathscr H$-measurable where $H$ is any $\sigma$-algebra. Why?

We have to show that for any Borel set, i.e. for any $A\in \mathcal{B}(\mathbb{R})\}$ that $c^{-1}(A) \in \mathscr H$.

Here, we only care about whether or not $c \in A$.

• If yes: $c^{-1}(A) = \Omega \in \mathscr H$
• If no: $c^{-1}(A) = \emptyset \in \mathscr H$

Actually, $\{c^{-1}(A)| A \in \mathcal{B}(\mathbb{R})\} = \{\emptyset, \Omega\}$, i.e. any constant random variable's $\sigma$-algebra is the trivial $\sigma$-algebra.

So instead of $E[Y]^{-1}=\Omega \in \sigma(X)$, you should have said:

Let $A \in \mathcal{B}(\mathbb{R})$.

Here, we only care about whether or not $E[Y] \in A$.

• If yes: $E[Y]^{-1}(A) = \Omega \in \sigma(X)$
• If no: $E[Y]^{-1}(A) = \emptyset \in \sigma(X)$

If it was proven in your class that any number $c$ is $\mathscr H$-measurable where $\mathscr H$ is any $\sigma$-algebra, then you may just point it out in addition to saying that $E[Y]$ is a number to conclude $E[Y]$ is $\sigma(X)$-measurable.

2. Is $\int_G E[Y] \mathrm{d}\mathcal{P} \stackrel{?}{=} \int_G Y \mathrm{d}\mathcal{P}\qquad $ ($\forall G\in \sigma(X))$

Well, why is LHS = $E[Y]\mathcal{P}(G)$?

$$LHS = \int_G E[Y] \mathrm{d}\mathcal{P}$$

$$ = \int_{\Omega} E[Y]1_G \mathrm{d}\mathcal{P}$$ $$ = E[Y] \int_{\Omega} 1_G \mathrm{d}\mathcal{P}$$ $$ = E[Y] E[1_G]$$ $$ = E[Y] \mathcal{P}(G)$$

Note that we made use of indicator functions and did not care much what $\sigma$-algebra is represented by $G$. Let's try similarly for RHS.

$$RHS = \int_G Y \mathrm{d}\mathcal{P}$$

$$ = \int_{\Omega} Y1_G \mathrm{d}\mathcal{P}$$

$$ = \int_{\Omega} Y \mathrm{d}\mathcal{P} \int_{\Omega} 1_G \mathrm{d}\mathcal{P} \tag{*}$$

$$ = E[Y] \int_{\Omega} 1_G \mathrm{d}\mathcal{P}$$ $$ = E[Y] E[1_G]$$ $$ = E[Y] \mathcal{P}(G)$$

$(*)$ Actually, $1_G$ and $Y$ are independent because $X$ and $Y$ are independent and $G \in \sigma(X)$. Why?

Hint: What's $\sigma(1_G)$ ?

$\sigma(1_G)$ can be shown to be equal to $\sigma(G) = \{\emptyset, G, G^C, \Omega\}$. Since $G \in \sigma(X)$, $G^C \in \sigma(X)$ and thus $\sigma(G) \subseteq \sigma(X)$.

Now, have you proved the following?

Let $\mathscr H \subseteq \sigma(X)$ where $\mathscr H$ is a $\sigma$-algebra If $\sigma(X)$ and $\sigma(Y)$ are independent, then $\mathscr H$ and $\sigma(Y)$ are independent.

You can use that to say that

$\sigma(1_G) = \sigma(G)$ and $\sigma(Y)$ are independent.

Thus, $1_G$ and $Y$ are independent.

And is $\int_G Y \mathrm{d}\mathcal{P}$ where $G\in \sigma(X)$ even properly defined?

Actually, it's more of the RHS that we should ask that I think.

Anyway, for LHS, try to write like this:

Since $\sigma(X) = \{X^{-1}(A)|A \in \mathscr{B} \}, \ \forall \ G \in \sigma(X)$, we can find a corresponding Borel set $\exists A_G \in \mathscr{B}$ s.t. $G = X^{-1}(A_G)$. Then

$$RHS = \int_G Y \mathrm{d}\mathcal{P}$$

$$\int_{X^{-1}(A_G)} Y \mathrm{d}\mathcal{P}$$

$$\int_{\Omega} Y1_{X^{-1}(A_G)} \mathrm{d}\mathcal{P}$$

$$\int_{\Omega} Y \mathrm{d}\mathcal{P} \int_{\Omega} 1_{X^{-1}(A_G)} \mathrm{d}\mathcal{P}$$

$$E[Y] E[1_{X^{-1}(A_G)}]$$

$$E[Y] \mathcal{P}({X^{-1}(A_G)})$$

$$E[Y] \mathcal L_X(A_G)$$

Thus, LHS is defined if

1. $E[|Y|] < \infty$, which I guess is an omitted assumption. (I mean, $E[Y]$ is not well-defined otherwise.)
2. ${X^{-1}(A_G)} = G \in \mathscr F$ s.t. $\mathcal L_X(A_G) = \mathcal{P}({X^{-1}(A_G)})$ is well-defined. Now $G \in \mathscr G \subseteq \mathscr F$ where $\mathscr G$ is a $\sigma$-algebra that can be $\sigma(X)$ because $X$ is given to be a random variable in $(\Omega, \mathcal F, \mathcal P)$, i.e. $\sigma(X) \subseteq \mathcal F$

Answer 2

1. We just use the fact that constants are measurable.

2. If two independent random variables $Z_1$ and $Z_2$ are integrable, then $$ \mathbb E\left[Z_1Z_2\right]=\mathbb E\left[Z_1 \right]\mathbb E\left[Z_2\right].$$ Use this with $Z_1=Y$ and $Z_2$ the indicator function of $G$.

Answer 3

For any $A \in \mathcal{B}(\mathbb{R})$, \begin{align*} \int_{X^{-1}(A)} E(Y \mid \sigma(X)) d\mathcal{P} &= \int_{X^{-1}(A)} Y d\mathcal{P}\\ &=\int_{\Omega} \pmb{1}_{X^{-1}(A)} Y d\mathcal{P}\\ &=E\left(\pmb{1}_{X^{-1}(A)} Y\right)\\ &=E\left(\pmb{1}_{X^{-1}(A)}\right) E(Y) \hspace{0.35in} \text{(given the independence of $X$ and $Y$)}\\ &=E(Y) \int_{X^{-1}(A)}d\mathcal{P}\\ &=\int_{X^{-1}(A)} E(Y)d\mathcal{P}. \end{align*} That is, \begin{align*} E(Y \mid \sigma(X)) = E(Y). \end{align*}