5 votes
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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 ...
user6247850's user avatar
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5 votes

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 ...
Will's user avatar
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5 votes

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). ...
Davide Giraudo's user avatar
5 votes
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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 ...
Thomas Kojar's user avatar
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5 votes
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Show that for $0<b<1<a$: $\mathbb P(\tau_a < \tau_b)=\frac{1-b^4}{a^4-b^4}$

Let $\rho=\min(\tau_a, \tau_b)$. Observe that $\rho$ is a stopping time, since $$ \rho=\inf \{ t>0: X_t \in \{ a,b \} \}. $$ By the optional stopping theorem, $$ \mathbb E(X_{\rho}^4)=\mathbb E(X_0^...
pre-kidney's user avatar
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5 votes
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Is the stochastic integral process $\left(\int_{0}^{t}e^{-\lambda\left(t-s\right)}\mathrm{d}B_s\right)_{t\geq 0}$ a martingale?

The criterion you mentioned is certainly a sufficient condition, but not a necessary condition. So, let's go back to the basic. Let $M_t = \int_{0}^{t} e^{-\lambda (t-s)} \, \mathrm{d}B_s$. Then for $...
Sangchul Lee's user avatar
4 votes

Finding a Doob decomposition

I thank Snoop for a comment and John Dawkins for his other answer which led me to elaborate a bit to show what can go wrong (and in fact was going wrong in my earlier versions): Since the $X_k$ are i....
Kurt G.'s user avatar
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4 votes
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Finite stopping time

Let $p$ be the probability that $X_1>0.5$. Note that $p$ is strictly positive. Notice that, if there ever exist $\lceil 4K\rceil $ consecutive indices, $i$, for which $X_i>0.5$, then $Z_K$ will ...
Mike Earnest's user avatar
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4 votes
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Computing Expectation of Summation Involving IID Uniform Random Variables with Conditional Stopping

Let $W_k$ be the indicator variable of the event that $X_1 \ge X_2 \ge \cdots \ge X_{k-1} \ge X_k$. Then the sum can be written as $$ S= \sum_{k=1}^\infty \frac{X_k W_k}{2^k}$$ Now $E[X_k W_k] = P(W_k=...
leonbloy's user avatar
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4 votes
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Equivalent characterization of the martingale property

Recall the monotone class theorem. Let $\mathscr A$ be a collection of subsets that contains $\Omega$ and is closed under intersection. Let $\mathscr H$ be a vector space of real-valued functions on $\...
Andrew's user avatar
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4 votes
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Game involving envelopes

Here is a sketch of an approach, assuming you are trying to maximise the expected amount taken: If you have only one envelope, you have to take the amount you see. Find the expected amount to be ...
Henry's user avatar
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4 votes
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Conditional Expectation of 3 random variables

Let $X_1,X_0$ be iid Bernoulli fair variables: $P(X_i=0)=P(X_i=1)=1/2$ Let $X_2 = 2 X_1 X_0$. We get $E[X_2| X_1]= 2 X_1 E[X_0] = 2 X_1 \frac12 = X_1$ and $E[X_2]= E[E[X_2| X_1]]= 1/2$. Also $$E[X_1 | ...
leonbloy's user avatar
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Proof of Doob's Maximal Inequality for Positive Supermartingales

To show that $\mathbb{E}[X_T] \le \mathbb{E}[X_0]$, note that we have $\mathbb{E}[X_{T \wedge n}] \le \mathbb{E}[X_0]$ for all $n$ because $T \wedge n$ is a bounded stopping time. Then, since $X_{T \...
user6247850's user avatar
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4 votes

Uniform Integrability and Proving martingale of Poisson product process

$E(M_{n})=n!E(\prod_{k=1}^{n}X_{k})=n!\prod_{k=1}^{n}E(X_{k})=n!\prod_{k=1}^{n}\frac{1}{k}=\frac{n!}{n!}=1$. The expectation of a martingale cannot vary with $n$, it has to be constant. $X_{k}$'s are ...
Mr.Gandalf Sauron's user avatar
3 votes
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Show $Y_t = e^{\frac{t}{2}} \sin(B_t)$ is a martingale using conditional expectations.

Recall that a BM $(W_t)_{t\geq 0}$ is Markov and $$E[e^{i\xi W_t}|W_s]=E[e^{i\xi (W_t-W_s)}]e^{i\xi W_s}=e^{i\xi W_s-\xi^2(t-s)/2}$$ So $$\begin{aligned}E[Y_t|\mathscr{F}_s]&=e^{t/2}E[\sin (W_t)|\...
Snoop's user avatar
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3 votes
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How can I compute this expectation using the optional stopping theorem?

You have guessed all steps correctly but you are having trouble to finish the proof. Well, contrary to my comment, I am here again posting an answer to your question :) . To answer your first question,...
Mr.Gandalf Sauron's user avatar
3 votes

Quadratic variation of the compensated Poisson process $N_t - \lambda t$

Since the expectation and the variance of $N_t-N_s$ are both equal to $\lambda(t-s)$ and since the increments of $N$ are independent we have $$ \mathbb{E}[(N_t-N_s)^2|{\cal F}_s]=\mathbb{E}[(N_t-N_s)^...
Kurt G.'s user avatar
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3 votes

$M_n$ converges almost surely to 0

Note that $M_n \to 0 \iff \ln(M_n) \to -\infty$. As mentioned in the comment, the main point is to apply the Strong Law of Large Numbers to $\ln(X_n)$. Further, you need to show that $\mathbb{E}\ln(...
Paul Gullesh's user avatar
3 votes
Accepted

Choquet-Deny theorem

A set $A\in\mathcal F_n$ can be written as $(X_1,\dots,X_n)\in B$ for some $B\in\mathcal B(\mathbb R)$. Therefore, one has $$ \mathbb E\left[f\left(Y_{n+1}\right)\mathbf{1}_A\right]= \mathbb E\left[f\...
Davide Giraudo's user avatar
3 votes

Is the product/sum of a martingale with a constant still a martingale?

Yes. Technically, the process $(M_i)$ is a martingale with respect to some filtration $(\mathcal{F}_i)$, i.e. martingale property is that $\mathbb{E}[M_{i+1} \mid \mathcal{F}_i] = M_i$. Usually the ...
JKL's user avatar
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3 votes
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Expected number of cards to draw before first ace using stopping times?

Consider a player who bets each time a card is drawn 1\$: he keeps playing as long as an ace is drawn, and wins $\frac{52-t+1}{4}$ if an ace is drawn at time $t$. Let's check that his total profit $S_{...
Léo Aparisi de Lannoy's user avatar
3 votes
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Proving $X_t = 1 + \int_0^t X_s \, dN_s$ is a supermartingale

First, note that these are all about the case where $X$ is an exponential martingale, so $N$ is a local martingale. This also implies $X$ is a local martingale. For 1), let $(\tau_n) \rightarrow \...
user6247850's user avatar
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3 votes
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Polya's urn, should I use martingales or LLN

Let $r,b$ the number of red and blue balls which we have initially. Define $R_n=r+S_n$ where $S_n=X_1+...+X_n$ where $X_k$ is $1$ if red ball is drawn at turn $k$, and $0$ otherwise; set $S_0=0$. If ...
Snoop's user avatar
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3 votes
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Showing that a specific stochastic process is a martingale.

It's much simpler than that. Let $m>n$ and note that \begin{align} E(X_m | \mathcal{F}_n ) &= E(X_n + (X_m-X_n)| \mathcal{F}_n) \\ &= X_n + E( X_m-X_n| \mathcal{F}_n) \\ &= X_n + E( X_m-...
Jose Avilez's user avatar
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3 votes

Proof that the stochastic exponential is a local martingale

In general, an Ito integral is a local martingale and it is a true martingale provided that the integrand is in $L^2$. This is precisely what the remark is telling you, as saying that the integrand is ...
Jose Avilez's user avatar
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3 votes
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Expected number of coin flips to get THTHTT

Below i will use $0,1$ instead of $H,T$. Consider the space $\Omega$ of all $0,1$-words, and the $\sigma$-algebra generated by all events $A(w)$, where $A(w)$ is the set of words starting with $w$. We ...
dan_fulea's user avatar
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3 votes
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Square Integrable Martingales Applied to Power Expression

Note $$\begin{aligned}E[M^2_n]&=E[2^{2S_n}]\\ &=E[2^{2(S_n-S_{n-1})}2^{2S_{n-1}}]\\ &\stackrel{\textrm{Tower prop.}}=E[E[2^{2(S_n-S_{n-1})}|\mathscr{F}_{n-1}]2^{2S_{n-1}}]\\ &\stackrel{...
Snoop's user avatar
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3 votes
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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 $n$.
geetha290krm's user avatar
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2 votes
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$u(t,B_t)$ 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}"\...
Kurt G.'s user avatar
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2 votes
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Is Conditional Expectation of stochastic process a martingale?

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. $...
Gabriel dLN's user avatar

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