$\mathbb{E}[X | \mathcal{F}]^{2}<\infty$ if $\mathbb{E}[X^{2}]<\infty$

Question

In Durrett's Probability Theory and Examples theorem 5.1.8 says:

Suppose that $\mathbb{E} X^2 <\infty$. Then $\mathbb{E}(X| \mathcal{F})$ is the variable $Y \in \mathcal{F}$ that minimizes the "mean square error" $\mathbb{E}(X-Y)^2$.

To explain notation, it is assumed we are working with the probability space $(\Omega, \mathcal{F}_{0}, P)$, and $\mathcal{F}\subset \mathcal{F}_{0}$ is a sub $\sigma-$field of the $\sigma-$field $\mathcal{F}_{0}$.

Durrett then goes on to explain that $\mathbb{E}(X| \mathcal{F})$ is the projection of $X$ onto the Hilbert space $L^{2}(\mathcal{F}) = \{ Y \in \mathcal{F} : \mathbb{E} Y^2 <\infty \}$.

As far as I can see, for this to be true it must be that $\mathbb{E}[X^{2}]<\infty$ implies $\mathbb{E}[X | \mathcal{F}]^{2}<\infty$. However, I have not been able to show this using the measure-theoretic definition of conditional expectation.

Can anyone provide a proof of this using the measure-theoretic definition of conditional expectation? Any help is appreciated.

Answer 1

See Durrett's Theorem 5.1.4 which, taking $p=2$, says that $E[E[X \mid \mathcal{F}]^2] \le E[X^2]$. So $E[X \mid \mathcal{F}]$ is in $L^2$. In particular, since the random variable $E[X \mid \mathcal{F}]$ has finite expectation, it must be finite almost surely.

Answer 2

I'm making a bit different way.

$X^2 \in L^1(\Omega, \mathcal{F}_0, \mathbb{P})$ therefore by Jensen's inequality $\mathbb{E}(X^2) \geq \mathbb{E}(X)^2$, i.e. $\mathbb{E}(X) < \infty$, i.e. $X \in L^1(\mathcal{F}_0)$, then $\mathbb{E}(X | \mathcal{F})$ exists and $L^1(\mathcal{F})$ for $\mathcal{F} \subseteq \mathcal{F}_0$ - $\sigma$-algebra.

By Jensen's inequality we have $\mathbb{E}(X | \mathcal{F})^2 \leq \mathbb{E}(X^2 | \mathcal{F})$ a.s. $\mathbb{E}(X | \mathcal{F})^2$ is positive and measurable, so we can integrate, hence $\mathbb{E}(\mathbb{E}(X | \mathcal{F})^2) \leq \mathbb{E}(\mathbb{E}(X^2 | \mathcal{F})) < \infty$, i.e. $\mathbb{E}(X|\mathcal{F})^2 < \infty$ a.s.