Where is the mistake in my idea? (Conditional Expectation)

Question

Let $X_t$ be a stochastic process. $\mathcal{F}_t:=\sigma(X_s:0\leq s\leq t)$.

Suppose $\mathrm{E}[e^{ik(X_t-X_s)}|X_s]=e^{-\frac{1}{2}k^2(t-s)}$.

My idea

1. By the tower property of conditional expectation: if $\mathcal{H}$ is a sub $\sigma$-algebra of $\mathcal{G}$, $\mathrm{E}[\mathrm{E}[Y|\mathcal{G}]|\mathcal{H}]=\mathrm{E}[Y|\mathcal{H}]$, I get \begin{equation} \mathrm{E}[\mathrm{E}[e^{ik(X_t-X_s)}|\mathcal{F}_s]|X_s]=e^{-\frac{1}{2}k^2(t-s)}. \end{equation}

2. That is, $\mathrm{E}[\mathrm{E}[e^{ik(X_t-X_s)}|\mathcal{F}_s]-e^{-\frac{1}{2}k^2(t-s)}|X_s]=0.$ Consequently, by uniqueness of conditional expectation, I get $\mathrm{E}[e^{ik(X_t-X_s)}|\mathcal{F}_s]=e^{-\frac{1}{2}k^2(t-s)}$

But, clearly this is absurd because assumption say only about $\sigma(X_s)$ which included in $\mathcal{F}_s$ . I don't know what is wrong. Thank you for your cooperation.

Answer 1

There is no such 'uniquness of conditional expectation' result. You are saying that if $E(Y|X_s)=0$ then $Y=0$. This is false. For example $Y$ could be any random variable with mean $0$ independent of $X_s$.

From $E(Y|X)=0$ we can only conclude that $EY=0$.