# Showing that $X\sim E[X\mid\mathcal{G}]$ implies $X=E[X\mid\mathcal{G}]$ almost surely

Suppose $(\Omega,\mathcal F,P)$ is a probability space and that $\mathcal G$ is a sub-sigma-algebra of $\mathcal F$. If $X$ is an integrable, non-negative random variable with the same distribution as $E[X\mid \mathcal G]$, how does one show that $X=E[X\mid \mathcal G]$ a.s.?

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Let $u:[0,\infty)\to[0,\infty)$ denote the function defined by $u(x)=x/\sqrt{1+x^2}$. Then $u$ is smooth, strictly concave and bounded, its derivative $u'$ is bounded, and $u(x)\leqslant u(y)+u'(y)(x-y)$ for every nonnegative $(x,y)$.

Let $Y=E[X\mid\mathcal G]$ and $Z=u(Y)-u(X)+u'(Y)(X-Y)$, then $Z\geqslant0$ almost surely. The distributions of $X$ and $Y$ coincide hence $E[u(X)]=E[u(Y)]$. By definition of conditional expectation and because $u'$ is bounded, $E[u'(Y)X]$ and $E[u'(Y)Y]$ exist and coincide. This implies that $E[Z]=0$ hence $Z=0$ almost surely.

The inequality $u(x)\leqslant u(y)+u'(y)(x-y)$ is strict as soon as $x\ne y$ hence $Z\gt0$ on the event $[X\ne Y]$. Thus, $X=Y$ almost surely.

Plan B (using the function exponential):

Let $Y=E[X\mid\mathcal G]$ and $T=\mathrm e^{-X}-\mathrm e^{-Y}+\mathrm e^{-Y}(X-Y)$, then:

• Since $\mathrm e^{-x}-\mathrm e^{-y}+\mathrm e^{-y}(x-y)\geqslant0$ for every $(x,y)$, $T\geqslant0$ almost surely.
• Each term in $T$ is integrable because $X$ and $Y$ are integrable and $\mathrm e^{-X}$ and $\mathrm e^{-Y}$ are bounded thanks to the hypothesis that $X\geqslant0$ almost surely, which implies also that $Y\geqslant0$ almost surely.
• The distributions of $X$ and $Y$ coincide hence $E[\mathrm e^{-X}]=E[\mathrm e^{-Y}]$.
• By definition of conditional expectation, $E[\mathrm e^{-Y}X]=E[\mathrm e^{-Y}Y]$.

All this implies that $E[T]=0$ hence $T=0$ almost surely.

The inequality $\mathrm e^{-x}-\mathrm e^{-y}+\mathrm e^{-y}(x-y)\geqslant0$ is strict as soon as $x\ne y$ hence $T\gt0$ on the event $[X\ne Y]$. Thus, if $P(X\ne Y)\gt0$ then $E[T]\gt0$. By contraposition, $X=Y$ almost surely.

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