# Different definitions of conditional expectation

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?

• 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. Sep 24 '19 at 19:19
• 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. Sep 24 '19 at 22:01
• @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? Sep 24 '19 at 23:44

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 $$h(y) = \mathbb E[X|Y=y]$$