Why isn't $E(\vert X\vert |\mathcal{G})=\vert E(X|\mathcal{G})\vert$?

Question

Let $(\Omega,\mathcal{F},P)$ be a probability space, $\mathcal{G}$ sub-$\sigma$-algebra of $\mathcal{F}$, and $X\in L^1(P)$.

We have $$\begin{aligned} Y & := E(X|\mathcal{G})\\ & = E(X^+ - X^-|\mathcal{G})\\ & = E(X^+|\mathcal{G}) - E(X^-|\mathcal{G})\\ \end{aligned} $$

As $E(X^+|\mathcal{G})$ is non-negative and $-E(X^-|\mathcal{G})$ is non-positive we have $Y^+=(E(X^+|\mathcal{G}) - E(X^-|\mathcal{G}))\vee 0 = E(X^+|\mathcal{G})$ and similarily $Y^-=E(X^-|\mathcal{G})$. Thus, $$ \vert Y\vert = Y^+ +Y^- = E(X^++X^-|\mathcal{G}) = E(\vert X\Vert\mathcal{G})$ $$

In most texts, I just see $\vert Y \vert \leq E(\vert X\Vert |\mathcal{G})$, so whats wrong about the above?

Thank you.

Answer 1

$Y^+=(\mathbb E[X^+\vert\mathcal G]-\mathbb E[X^-\vert\mathcal G])\vee0$ is wrong. If $a,b>0$ it is not true that $(a-b)^+=a$. Take for instance $a=b=1$.

If you want a counterexample for this triangle inequality, just take $\mathcal G$ the trivial $\sigma$-algebra on $\mathbb R$, that is $\mathcal G=\{\emptyset,\mathbb R\}$ and $\mathbb P(X=1)=\mathbb P(X=-1)=1/2$. Then $\vert\mathbb E[X\vert\mathcal G]\vert=\vert\mathbb E[X]\vert=0$ but $\mathbb E[\vert X\vert\vert\mathcal G]=\mathbb E[\vert X\vert]=1$.

Answer 2

It is not true that $E(X^{+}|\mathcal G)$ is non-positive. It is non-negative.

You get a counter-example to the identity by taking $X$ to be a fair coin tossing random variable and $\mathcal G$ to be the trivial sigma field consisting of just the empty set and the whole space.

Answer for the edited version: you are saying that if $x \geq 0$ and $ y\geq 0 $ then $(x-y)^{+}=\max \{x-y, 0\}=x$. This is clearly false. Take $x=1$ and $y=2$.

Answer 3

Of course you know Jensen's inequality, but I'm guessing the point is to explain in this particular case.

We have

$$E[|X| | \mathcal G] = E[X^+ + X^- | \mathcal G] = E[X^+ | \mathcal G] + E[X^- | \mathcal G]$$

By triangle inequality,

$$|E[X | \mathcal G]| = |E[X^+ - X^- | \mathcal G]| = |E[X^+ | \mathcal G] - E[X^- | \mathcal G]| \le |E[X^+ | \mathcal G]| + |E[X^- | \mathcal G]| = E[X^+ | \mathcal G] + E[X^- | \mathcal G] = E[|X| | \mathcal G]$$

It looks like you're aware that $E[X | \mathcal G] = (E[X | \mathcal G])^+ - (E[X | \mathcal G])^-$ and $|E[X | \mathcal G]| = (E[X | \mathcal G])^+ + (E[X | \mathcal G])^-$, but it looks like you've made the error that $(E[X | \mathcal G])^+ = E[X^+ | \mathcal G]$. This is kind of like saying $5=3$ or $4=3$ because $8=4+4=5+3$.

Consider flipping a single coin. Let $\Omega=\{H,T\}$, $\mathcal G = \mathcal F = 2^{\Omega}$ and $X=1_H-1_T$ with $X^+=1_H$ and $X^-=1_T$. Then $E[X^+] = P(H)$ while $(E[X])^+=\min\{P(H)-P(T),0\}$