3
$\begingroup$

As I'm trying to learn more about conditional expectation, I noticed that there are different definitions for that, depending on the book. For the following, let $(\Omega,\mathcal{F},P)$ be a measure space.

Definition 1:

Let $(X,Y)$ be a two-dimensional random variable on $\Omega$ with a joint distribution function $f_{X,Y}(x,y)$. The $(X=x)$-conditional expectation of $Y$ is given by $$E(Y\mid X=x):=\int_{-\infty}^{\infty} yf_{Y\mid X=x}(y)dy$$

with

$$f_{Y\mid X=x}(y):=\frac{f_{X,Y}(x,y)}{f_X(x)}.$$

($f_{Y\mid X=x}(y)$ is also called conditional density function)

Definition 2:

Let $X$ be a random variable on $\Omega$ and $\mathcal{G}$ a sub $\sigma$-field of $\mathcal{F}$. A $\mathcal{G}$-measurable random variable $U$ with

$$\int_G U \ dP = \int_G X \ dP$$

for all $G \in \mathcal{G}$ is called $\mathcal{G}$-conditional expectation of $X$.

Intuitively, both definitions make sense to me. But how does one show that the definitions are equivalent?

Thank you in advance.

$\endgroup$
  • 5
    $\begingroup$ They aren't. Definition $2$ is a general case, as you can see, there's no need of density ($X$ should be in $L^1$). While definition $1$ is just a special case with densities. If you're interested in proof that when those conditions with densities are meet then def $2$ => def $1$, I can try to write it. $\endgroup$ – Dominik Kutek Sep 24 '19 at 19:19
  • 1
    $\begingroup$ Durrett shows that the general definition in (2) yields the special undergraduate cases of the definition (such as (1)) in his grad probability book if you want a reference. $\endgroup$ – Foobaz John Sep 24 '19 at 22:01
  • $\begingroup$ @DominikKutek Yes, 2 => 1 would be interesting to see, if possible. Can you show me? Is there any way to change either definition to make it equivalent? $\endgroup$ – Yasuduck Sep 24 '19 at 23:44
1
$\begingroup$

Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space. Let $(X,Y)$ be random vector with probability density function $g_{(X,Y)}$. Finally, let $f$ be any borel function, such that $\mathbb E[|f(X,Y)|] < \infty$. Then it holds: $\mathbb E[f(X,Y)|Y] = h(Y)$, where:

$$ h(y) = \frac{\int_{\mathbb R} f(x,y)g_{(X,Y)}(x,y)dx}{\int_{\mathbb R} g_{(X,Y)}(x,y)dx} $$when $\int_{\mathbb R} g_{(X,Y)}(x,y)dx \neq 0$, and $h(y) = 0$ otherwise.

Firstly, we can put $0$ in the second case, because the set $S=\{ \omega \in \Omega : \int_{\mathbb R} g_{(X,Y)}(x,Y(\omega)) dx = 0 \}$ has measure $0$. Clearly $\mathbb P(S) = \mathbb P(Y \in S_Y)$, where $S_Y = \{ y \in \mathbb R: \int_{\mathbb R} g_{(X,Y)}(x,y)dx = 0 \}$. Then $\mathbb P(Y \in S_Y) = \int_{S_Y} g_Y(y) dy $, where $g_Y$ is marginal density (It exists due to Fubini + existence of joint density of rv $(X,Y)$ ). But note $g_Y(y) = \int_{\mathbb R} g_{(X,Y)}(x,y)dx $, so we're just integrating $0$ function (cause we on $S_Y$ where it's $0$), so $\mathbb P(S) = 0$. This + the fact that Conditional Expectation is up to the set of measure $0$ allows us to forget about the case when $g_Y(y) = 0$.

So, we have to prove $2$ things:

1) $h(Y)$ is $\sigma(Y)$ measurable. Clearly both $\int_{\mathbb R} g_{(X,Y)}(x,Y) dx$ and $\int_{\mathbb R} g_{(X,Y)}(x,Y) f(x,Y) dx$ are $\sigma(Y)$ measurable due to Fubinii theorem (integrals of $\sigma(Y) -$ measurable functions are $\sigma(Y)$ measurable (We here used the fact that $g_{(X,Y)}$ is bounded and $\mathbb E[f(X,Y)]$ is finite to be able to apply Fubini's theorem.

2) For any $A \in \sigma(Y)$ we have to show $\int_A f(X,Y) d\mathbb P = \int_A h(Y) d\mathbb P$. Note that $A$ is of the form $Y^{-1}(B)$ where $B \in \mathcal B(\mathbb R)$ (borel set).

Note that $\int_A f(X,Y) d\mathbb P = \mathbb E[ f(X,Y) \cdot \chi_{_{Y \in B}} ]$ and $\int_A h(Y) d\mathbb P = \mathbb E[ h(Y) \cdot \chi_{_{Y \in B}} ]$

We'll use the fact, that if random variable/vector (in $\mathbb R^n$) $V$ has density function $g_V$, then for any borel function $\phi: \mathbb R^n \to \mathbb R^n$, we have $\mathbb E[\phi(V)] = \int_{\mathbb R^n} \phi(v) g_V(v) d\lambda_n(v)$.

Then: $$\mathbb E[ f(X,Y) \cdot \chi_{_{Y \in B}} ] = \int_{\mathbb R^2} f(x,y)\chi_{_{B}} g_{(X,Y)}(x,y) d\lambda_2(x,y) = \int_{B} \int_{\mathbb R} f(x,y)g_{(X,Y)}(x,y)dxdy $$

That last split of integrals due to fubinii (function is integrable due to our assumption with $f$ ).

And now similarly at the beggining:

$$ \mathbb E[ h(Y) \cdot \chi_{_{Y \in B}} ] = \int_{B} h(y) (\int_{\mathbb R} g_{(X,Y)}(x,y)dx)dy$$

Now due to our assumption of $h$ (that is getting rid of that case when denominator is $0$ due to its being $0$-measurable set). We have:

$$ \int_{B} (h(y)) (\int_{\mathbb R} g_{(X,Y)}(x,y)dx) dy = \int_{B} (\frac{\int_{\mathbb R} g_{(X,Y)}(x,y)f(x,y)dx}{\int_{\mathbb R} g_{(X,Y)}(x,y)dx}) (\int_{\mathbb R} g_{(X,Y)}(x,y)dx )dy$$

After simplification we get $\mathbb E[ h(Y) \cdot \chi_{_{Y \in B}} ] = \int_{B} \int_{\mathbb R} f(x,y)g_{(X,Y)}(x,y)dxdy = \mathbb E[ f(X,Y) \cdot \chi_{_{Y \in B}} ]$, what we wanted to prove.

Now your "definition $1$" follows when you take $f(x,y) = x$. Then $ h(y) = \mathbb E[X|Y=y] $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.