# Tag Info

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### Can optional stopping hold for $\mathcal{L}^1$ bounded martingales?

This is indeed possible. For an example, let $(X_n)$ be a sequence of i.i.d. random variables with $\mathbb{P}(X_n = 2) = \mathbb{P}(X_n = 0) = \frac 12$. Observe that $\mathbb{E}[X_n] = 1$, so the ...
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### Durrett Q4.8.3 on martingales and the Optional Stopping Theorem

You are right that if $\mathbb E[T]=+\infty$ then the inequality is trivial, so we can assume without loss of generatily that $\mathbb E[T]<+\infty$. I am not sure that your justification of ...
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### Uniform Integrability and Conditional Expectations

The sequence $\left(\mathbb E\left[X_n\mid\mathcal F_k\right]\right)_{n\geqslant k}$ is monotonic (non-increasing in the case of a supermartingale, non-decreasing in the case of a submartingale). ...
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### $M_t=\int_0^t e^{-3W_s}dW_s$ properties

Is $M_t$ well defined when $\mathbb EM_t<\infty$? Is this enough or do I need to check any other conditions? Any random variable is well-defined when its expectation is bounded Does finite ...
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### $Cov[X_m, X_n] = \mathbb{E}[(X_m- \mathbb{E}(X_m)) (X_n- \mathbb{E}(X_n))]$ for $S_n := \sum_{i=1}^n X_i$ a martingale

It is true. The fact that $(S_n)$ is a martingale implies that $ES_{n+1}=E[E(S_{n+1}| S_n)]=ES_n$, so $EX_n=E(S_{n+1}-S_n)=0$ for all $n$.
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Accepted

### $u(t,B_t)$ is a martingale if it satisfies a certain condition.

Your self-answer is unsatisfactory. Also: In OP, what do you mean by "a polynomial $u(t,x)$ in $t,x$ such that  \frac{\partial u}{\partial t} + \frac{1}{2}\frac{\partial^2 u}{\partial x^2}"\...
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Here is a full answer to the question in the title: The conditional expectation of a stochastic process is a martingale if the following conditions are satisfied : $X_t$ is $\mathcal{F}_t$-adapted. \$...