Conditional expectation with respect to a sigma algebra

sum · **Joined:** Tue, 3 May 2011 06:53:24 UTC **Posts:** 8

I'm going to bump this topic because I am also having trouble getting an intuitive grasp, but this seems intuitive to the contributing posters.

Basically, if you have something like E(X|Y=y)

, where

is a random variable and Y=y

is an event, then you get a scalar. However, $E(X|\mathcal{G})$ is not a scalar, but in fact, a random variable, but I don't know what this random variable represents. In the first case, the intuition is: if I know this event Y=y

happens, what is the new expected value of

. However, in the second case, I'm confused by the intuition. It doesn't make sense to say: what is the expected value of

if $\mathcal{G}$ happens (because what does it mean for $\mathcal{G}$ to happen? This is a collection of events, not just one.)

Can anyone explain this?

outermeasure · **Posted:** Fri, 2 Dec 2011 16:20:04 UTC

Split topic from here.

sum wrote:

I'm going to bump this topic because I am also having trouble getting an intuitive grasp, but this seems intuitive to the contributing posters.

Basically, if you have something like E(X|Y=y)

, where

is a random variable and Y=y

is an event, then you get a scalar. However, $E(X|\mathcal{G})$ is not a scalar, but in fact, a random variable, but I don't know what this random variable represents. In the first case, the intuition is: if I know this event Y=y

happens, what is the new expected value of

. However, in the second case, I'm confused by the intuition. It doesn't make sense to say: what is the expected value of

if $\mathcal{G}$ happens (because what does it mean for $\mathcal{G}$ to happen? This is a collection of events, not just one.)

Can anyone explain this?

It may be useful to think of it this in the other way:

Conditional expectation is about the "best prediction" you can make with incomplete information (especially if you are thinking about stochastic processes --- we know about what happened in the past (well, sort of), but we in general don't know what will happen in the future, but to write $(X_t)_{t\in\mathbb{R}}$ means you know what happens at time t for all t). So $\mathbb{E}(X\mid\mathcal{G})$ represents the maximal information contained in $\mathcal{G}$ about

(again, in stochastic processes your $\mathcal{G}$ will typically be something like $\mathcal{F}_T=\sigma((X_t)_{t\leq T})$ and you want $\mathbb{E}(X_{T+1}\mid\mathcal{F}_T)$ ). Since

itself is a random variable, we should still get a random variable back, but this time it has to be $\mathcal{G}$ -measurable since all we know is $\mathcal{G}$ (so, for example, if $\mathcal{G}$ doesn't distinguish between two points, then there is no reason we can make different prediction about

at these two points).

Then $\mathbb{E}(X\mid A)$ as a scalar quantity, where A is a nonnull event, becomes a special case. It is the value of $\mathbb{E}(X\mid\sigma(A))$ at points of

(recall $\sigma(A)=\{\varnothing,A,A^c,\Omega\}$ is the sigma algebra generated by

), since all you know is whether A happened or not (it has). And of course, the "best" prediction we got about

in this case is its expected value "when A happens", which is your scalar (if

is scalar-valued) $\mathbb{E}(X\mid A)=\dfrac{\mathbb{E}(X1_A)}{\mathbb{P}(A)}$ .

sum · **Joined:** Tue, 3 May 2011 06:53:24 UTC **Posts:** 8

Thanks for feeding some intuition in this. I still can't get a good explicit picture of it, but I'm understanding little tidbits. Let me make some statements and do an example problem to see if they make sense.

Say we have $E[X|Y]:\Omega\rightarrow\mathbb{R}$ , where X,Y

are random variables. Then, I'm assuming the random variable represents $E[X|Y](\omega)=E[X|Y=Y(\omega)]$ ? If not, then disregard the rest.

Now, let $\Omega=[-1,1]$ with the typical Borel sets, so $P(d\omega)=\frac{1}{2}d\omega$ . Define $X(\omega)=\omega^2$ and $Y(\omega)=e^\omega$ . Then, I want to see if I compute $E[X|\sigma\left\{Y\right\}]$ and $E[Y|\sigma\left\{X\right\}]$ correctly.

We have $E[X|\sigma\left\{Y\right\}]=X$ since the sigma algebra generated by

is just all the Borel sets, so

is certainly $\sigma\left\{Y\right\}$ measurable.

Is it right that $E[Y|\sigma\left\{X\right\}](\omega)=\frac{1}{2}\left(e^{-\sqrt{\omega}}+e^{\sqrt{\omega}} \right)$ (since essentially the points a,-a

are equivalent in the sigma algebra generated by

)?

outermeasure · **Posted:** Sat, 10 Dec 2011 10:23:16 UTC

sum wrote:

Thanks for feeding some intuition in this. I still can't get a good explicit picture of it, but I'm understanding little tidbits. Let me make some statements and do an example problem to see if they make sense.

Say we have $E[X|Y]:\Omega\rightarrow\mathbb{R}$ , where X,Y

are random variables. Then, I'm assuming the random variable represents $E[X|Y](\omega)=E[X|Y=Y(\omega)]$ ? If not, then disregard the rest.

Now, let $\Omega=[-1,1]$ with the typical Borel sets, so $P(d\omega)=\frac{1}{2}d\omega$ . Define $X(\omega)=\omega^2$ and $Y(\omega)=e^\omega$ . Then, I want to see if I compute $E[X|\sigma\left\{Y\right\}]$ and $E[Y|\sigma\left\{X\right\}]$ correctly.

We have $E[X|\sigma\left\{Y\right\}]=X$ since the sigma algebra generated by

is just all the Borel sets, so

is certainly $\sigma\left\{Y\right\}$ measurable.

Is it right that $E[Y|\sigma\left\{X\right\}](\omega)=\frac{1}{2}\left(e^{-\sqrt{\omega}}+e^{\sqrt{\omega}} \right)$ (since essentially the points a,-a

are equivalent in the sigma algebra generated by

)?

No. There is a subtle but important difference between $\mathbb{E}[X\mid Y](\omega)$ and $\mathbb{E}[X\mid Y(\omega)]$ .

$\mathbb{E}[X\mid Y](\omega)$ means the L^1

-function $\mathbb{E}[X\mid Y]$ evaluated at $\omega\in\Omega$ (so there is an "almost every $\omega\in\Omega$ " or "almost surely" or ... that you need to add).

On the other hand, $\mathbb{E}[X\mid Y=Y(\omega)]$ is the conditional expectation of

given $Y=Y(\omega)$ occurs, i.e. $E[X\mid A]$ where

is the event $Y^{-1}(\{\omega\})$ . Unfortunately for you, $Y^{-1}(\{\omega\})$ is null, so this is not defined for any $\omega$ .

sum · **Joined:** Tue, 3 May 2011 06:53:24 UTC **Posts:** 8

Okay, I guess I shouldn't have said disregard the rest. Maybe I would understand better with example. As such, can you tell me how to compute $E[X|\sigma\left\{Y\right\}$ and $E[Y|\sigma\left\{X\right\}]$ defined above?

On a sidenote, isn't the event

rather the event $\left\{r\in\Omega: Y(r)=Y(\omega) \right\}$ (We have $\omega\in\Omega$ so $Y^{-1}(\left\{\omega\right\})$ doesn't make sense)? In this case, I don't see how this is null (probability zero, yes) since if we take $e\in\mathbb{R}$ , the event $Y^{-1}(\left\{e\right\})=\left\{1\right\}\subseteq[-1,1]$ .

outermeasure · **Posted:** Sun, 11 Dec 2011 11:42:38 UTC

sum wrote:

Okay, I guess I shouldn't have said disregard the rest. Maybe I would understand better with example. As such, can you tell me how to compute $E[X|\sigma\left\{Y\right\}$ and $E[Y|\sigma\left\{X\right\}]$ defined above?

On a sidenote, isn't the event

rather the event $\left\{r\in\Omega: Y(r)=Y(\omega) \right\}$ (We have $\omega\in\Omega$ so $Y^{-1}(\left\{\omega\right\})$ doesn't make sense)? In this case, I don't see how this is null (probability zero, yes) since if we take $e\in\mathbb{R}$ , the event $Y^{-1}(\left\{e\right\})=\left\{1\right\}\subseteq[-1,1]$ .

Oops, yes, I mean $Y^{-1}Y(\{\omega\})$ .

There is no problem with the $\mathbb{E}[X\mid Y]$ (i.e. your argument

is $\sigma(Y)$ -measurable is fine), so I'll do $\mathbb{E}[Y\mid X]$ from first principle here.

Recall the Dynkin's $\pi\text{-}d$ lemma (also known as the $\pi\text{-}\lambda$ lemma) and Carath\'{e}odory/Hahn-Kolmogorov extension theorem, so to find $\mathbb{E}[Y1_{X^{-1}B}]$ for borel B, it suffices to find
$\mathbb{E}[Y 1_{X^{-1}(-\infty,b]}]=\int_{X^{-1}(-\infty,b]} Y(\omega)\,\mathrm{d}\mathbb{P}(\omega)$
since the Borel sigma algebra is generated by the $\pi$ -system $\{(-\infty,b]\mid b\in\mathbb{R}\}$ (together with $\mathbb{R}$ of course, but that can be done as a limit).

Obviously you only need $0\leq b\leq 1$ by the range of X. Then $X^{-1}(-\infty,b]=[-\sqrt{b},\sqrt{b}]$ and so
$\begin{aligned} \mathbb{E}[Y 1_{X^{-1}(-\infty,b]}]&=\int_{-\sqrt{b}}^{\sqrt{b}} \frac{1}{2}e^{\omega}\,\mathrm{d}\omega\\ &=\frac{1}{2}[e^{\sqrt{b}}-e^{-\sqrt{b}}] \end{aligned}$
Now, what is the unique (in $L^1(\mathbb{P}\vert_{\sigma(X)})$ )

that will give $\mathbb{E}[f 1_{X^{-1}(-\infty,b]}]=\mathbb{E}[Y 1_{X^{-1}(-\infty,b]}]$ ? Obviously we differentiate with respect to

to find out:
$f(b)=\dfrac{\mathrm{d}}{\mathrm{d}b} \mathbb{E}[Y 1_{X^{-1}(-\infty,b]}] =\dfrac{1}{4\sqrt{b}}[e^{\sqrt{b}}+e^{-\sqrt{b}}]\quad \mathbb{P}\vert_{\sigma(X)}\text{-a.s.}$
(abusing notation slightly: f(b)

really mean $f(\omega)$ where $X(\omega)=b$ ). Note the factor $\dfrac{1}{2\sqrt{b}}$ that you gained here! It is the local scaling factor induced by

(or its inverse, depending on which way you think of it, c.f. coarea formula). So finally:
$\mathbb{E}[Y\mid X]=\dfrac{1}{4\sqrt{X}}[e^{\sqrt{X}}+e^{-\sqrt{X}}].$

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Conditional expectation with respect to a sigma algebra

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