# $\mathrm{E}[e^{u X}|\mathcal{A}] = \mathrm{E}[e^{u X}|\mathcal{B}]$ implies equality of conditional distributions

Let $X$ be a random variable on the probability space $(\Omega, \mathcal{F}, P)$ and $\mathcal{A} \subset \mathcal{B} \subset \mathcal{F}$ be a $\sigma$-subalgebras.

I want to prove that if $$\mathrm{E}[e^{u X}|\mathcal{A}] = \mathrm{E}[e^{u X}|\mathcal{B}]$$ holds for any $u \in \mathbb{C}$ then $P(X\in \Gamma|\mathcal{A}) = P(X\in \Gamma|\mathcal{B})$ for any borel $\Gamma$. Any hints?

I would do something like that: First, your assumptions imply that $\mathbb{E}[e^{uX}] < \infty ~\forall u \in \mathbb{R}$. let $\forall u \in \mathbb{R}:~f_a(u) := \mathbb{E}[e^{u X} | \mathcal{A}]$ and $f_b(u) := \mathbb{E}[e^{u X} | \mathcal{B}]$. Using your assumption, one can see that up to a countable number of modifications the function $f_a$ is $\mathcal{C}^{\infty}(\mathbb{R})$ and that: $$\mathbb{P}(dw) ~a.s.~~f_a^{(n)}(u)= \mathbb{E}(X^n e^{u X} | \mathcal{A})$$

The same apply to $f_b$. From $f_a = f_b$ is follows that $f_a^{(n)}(0) = f_b^{(n)}(0)$. Thus, $\mathbb{P}(dw) \text{ a.s. },~\mathbb{E}[X^n | \mathcal{A}] = \mathbb{E}[X^n | \mathcal{B}]$ where the event of probabilty zero can be choosen independently of $n$.

It thus follows that, if $f$ is a polynomial on $\mathbb{R}$, $\mathbb{P}(dw) \text{ a.s. },~\mathbb{E}[f(X) | \mathcal{A}] = \mathbb{E}[f(X) | \mathcal{B}]$. The result can be extended to any continuous function $f$ with a compact support. It is then not hard to conclude, as the indicator function of any closed set can be arbitrary approximated by a continuous function.

• Sorry can you explain why this process have smooth modification? – qwenty May 7 '18 at 20:49
• You can apply the Kolmogorov continuity criteria (en.wikipedia.org/wiki/Kolmogorov_continuity_theorem). Using Cauchy-Schwartz inequality, you can show, that if $x,y \in [-A, A]$ for some $A > 0$, then $\mathbb{E} |f_a(x) - f_b(y)|^2 \leq C_A |x-y|^2$ so there exists a continuous modification of $f_a$. The same apply to all the derivatives of $f_a$ – Adri May 9 '18 at 11:57
• @Andri i feel stupid but still can't see how to show that $\mathbb{E} |f_a(x) - f_a(y)|^2 \leq C_A |x-y|^2$ – qwenty May 9 '18 at 19:58
• $\mathbb{E} |f_a(x) - f_a(y)|^2 \leq \mathbb{E}|e^{x X} - e^{y X}|^2$ by Cauchy-Schwartz, and you can use for instance: $\forall a, b \geq 0,~ |e^{a} - e^b| \leq e^{max(a,b)} |a-b|$ such that: $\mathbb{E} |f_a(x) - f_a(y)|^2 \leq \mathbb{E} e^{2 A X} X^2 |x - y|^2$, assuming $X$ is a non negative random variable. – Adri May 10 '18 at 18:01

1. A subtlety here is that we are given uncountably many random variables, which is problematic in principle. For instance, it is impossible to discuss the continuity of $f_{\mathcal{A}}(u) = \mathbf{E}[e^{uX}\mid\mathcal{A}]$ because it requires you to look at uncountably many random variables. (This kind of situation is typical when dealing with stochastic processes such as Brownian motion. A typical solution there is to realize the law of a stochastic process on a suitable path space.)

2. Assume that $X$ is real-valued and $e^{u \lvert X \rvert}$ is integrable for all $u>0$. Then there exists a continuous modification of $f_{\mathcal{A}}$, which we again denote by $f_{\mathcal{A}}$. Of course, each value $f_{\mathcal{A}}(u)$ of this modification is a version of $\mathbf{E}[e^{uX}\mid\mathcal{A}]$.

Let $f_{\mathcal{B}}$ be the continuous modification of $u\mapsto\mathbf{E}[e^{uX}\mid\mathcal{B}]$ as well. Then by the continuity, the event $\{ f_{\mathcal{A}}(u) = f_{\mathcal{B}}(u) \text{ for all } u \in \mathbb{C} \}$ is measurable with probability $1$. Using these versions, we can initiate the conditional version of Fourier transform to show the claim.

Let $\mathcal{S}(\mathbb{R})$ the Schwartz space. Since the Fourier transform $\varphi \mapsto \int e^{i\xi u}\varphi(u) \, du$ is an isomorphism on $\mathcal{S}(\mathbb{R})$, we may consider its inverse transform $\varphi \mapsto \check{\varphi}$. Then for each $\varphi \in \mathcal{S}(\mathbb{R})$ and $A \in \mathcal{A}$,

\begin{align*} \mathbf{E}\left[ \left( \int \check{\varphi}(u)f_{\mathcal{A}}(iu) \, du \right) \mathbf{1}_{A}\right] = \int \check{\varphi}(u) \mathbf{E}[e^{iuX}\mathbf{1}_A] \, du = \mathbf{E}\left[ \varphi(X)\mathbf{1}_A \right] \end{align*}

and hence $\int \check{\varphi}(u)f_{\mathcal{A}}(iu) \, du = \mathbf{E}[\varphi(X)\mid\mathcal{A}]$ almost surely. By the indistinguishability of $f_{\mathcal{A}}$ and $f_{\mathcal{B}}$, it follows that $\mathbf{E}[\varphi(X)\mid\mathcal{A}] = \mathbf{E}[\varphi(X)\mid\mathcal{B}]$ almost surely.

Finally, for each Borel $\Gamma \subseteq \mathbb{R}$, pick $\varphi_n \in \mathcal{S}(\mathbb{R})$ such that $\varphi_n(X) \to \mathbf{1}_{\Gamma}(X)$ converges in $L^1(\mathbf{P})$. Then it follows that

• $\mathbf{E}[\varphi_n(X) \mid \mathcal{A}] \to \mathbf{P}[X \in \Gamma \mid \mathcal{A}]$ in $L^1(\mathbf{P})$, and
• $\mathbf{E}[\varphi_n(X) \mid \mathcal{B}] \to \mathbf{P}[X \in \Gamma \mid \mathcal{B}]$ in $L^1(\mathbf{P})$.

Therefore the limit must coincide in $L^1(\mathbf{P})$ and hence $\mathbf{P}$-a.s.

• Sorry can you explain why this process have continuous modification? – qwenty May 7 '18 at 20:49
• @user2715119, Sorry for the late reply. The idea is that you can collect $\mathbf{E}[e^{uX}\mid\mathcal{A}]$ for $u$ in some fixed countable dense subset of $\mathbb{C}$ and then use to interpolate them. The proof will not be so different from that of Kolmogorov's continuity theorem. – Sangchul Lee May 9 '18 at 12:38