# Tag Info

2

Say $X_n = n$ almost surely. Then $\mathbb{E}|X_n| <\infty$, but $\sup_{n}\mathbb{E}|X_n| = \infty$. In other words, we need the expectations of $|X_n|$ to be uniformly bounded.

1

By Fubini's Theorem, (2) is equivalent to $$\int_0^t E[X_s^2] ds < \infty$$ If $$E[X_s^2] = \infty$$ then $$\int_0^t E[X_s^2] ds = \infty \ ↯$$ Hence $$E[X_s^2] < \infty$$ Some specifics with $s$ and $t$ but I think (2) implies (1)

1

(Q1) I think it's helpful to think of it as an integral. For outcome $\omega\in\Omega$ we have that $\mathbb E [X \mid \mathcal G](\omega)=\frac{1}{P(A)}\int_A X(x) P(dx)$, where $A$ is the "smallest" set in $\mathcal G$ that contains $\omega$ (as long as $P(A)\neq0$). If $X$ is constant on $A$, then $\mathbb E [X \mid \mathcal G](\omega)=X(\omega).$ ...

1

The intuition for Q1 I often found useful is that a $\Sigma$-algebra in some sense is a model for known information. The sets in the $\Sigma$-algebra are the ones you have detailed information about, and the sets outside of it are indistinguishable to you, i.e. you don't know what happened or not. When you condition on a random variable, you are really ...

1

Rather, $Z'_\tau(u) = e^{iuX_\tau}/\phi_\tau(u)$. Do notice that the random variable $\omega\mapsto\phi_{\tau(\omega)}(u)$ appearing in the denominator of this fraction is the composition of the function $t\mapsto \phi_t(u)$ with the random variable $\omega\mapsto \tau(\omega)$. This is not the same as the complex number $E[e^{iuX_\tau}]$. $\{Z_t(u)\}_{t\ge ... 1 By the$L^1$-convergence, we have $$L^1-\lim_{s \to \infty} \mathbb{E}(X(t+s) \mid \mathcal{F}_t) = \mathbb{E} \left( X(\infty) \mid \mathcal{F}_t \right). \tag{1}$$ On the other hand, as$(X(t))_{t \geq 0}$is a martingale, we have $$\mathbb{E}(X(t+s) \mid \mathcal{F}_t) = X(t)$$ for all$s \geq 0$. Hence, trivially,$\$L^1-\lim_{s \to \infty} \mathbb{E}...

Only top voted, non community-wiki answers of a minimum length are eligible