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3

Hint: By the strong law of large numbers, $$\frac{1}{n} \sum_{i=1}^n X_i \to \mathbb{E}X_1 = 0$$ almost surely. Write $$\exp \left( \sum_{i=}^n X_i - \frac{n \sigma^2}{2} \right) = \exp \left( n \left[ \frac{1}{n} \sum_{i=1}^n X_i - \frac{\sigma^2}{2} \right] \right)$$ in order to deduce that $M=0$ a.s.


2

The existence of the limit $S$ follows from the martingale convergence theorem (because the expectation of $|S_n|$ is uniformly bounded), see for instance Durrett's book Probability: Theory and Examples page 235. It remains to show that the limit is almost surely constant. To this aim, notice that $\mathbb E(S_n)\to \mathbb E(S)$ and apply Fatou's lemma to ...


2

Hints: Use Jensen's inequality in order to prove that $$\mathbb{E}(|X_n|^p) \leq \mathbb{E}(|Y|^p).$$ Deduce from $$\sup_{n \geq 1} \mathbb{E}(|X_n|^p)< \infty$$ and Doob's maximal inequality that $$\sup_{n \geq 1} |X_n| \in L^p.$$ Apply the dominated convergence theorem.


2

We have to show that if $ X\in\mathbb L^p$ and $X_n:=\mathbb E[X\mid \mathcal F_n]$, then $\mathbb E|X_n-\mathbb E[X\mid \mathcal F_\infty|^p\to 0$. We can use a truncation argument: for a fixed $R$ we have $$\mathbb E|X_n-Y|^p\leqslant 2^{p-1}\mathbb E\left|\mathbb E[X\chi\{|X|\leqslant R\}\mid\mathcal F_n]-\mathbb E[X\chi\{|X|\leqslant R\}\mid\mathcal ...


2

This is how I did it: $S_{n+1}=X_1+...+\frac{1}{2(1-p)}X_{n}+ \frac{2(1-p)-1}{2(1-p)}X_n+\frac{1}{2(1-p)}X_{n+1}=$ $=S_n+\frac{2(1-p)-1}{2(1-p)}X_n+\frac{1}{2(1-p)}X_{n+1}$ Then the conditional expectation: $E[S_{n+1}|\mathcal{F}_{n}]= E[S_n+\frac{2(1-p)-1}{2(1-p)}X_n+\frac{1}{2(1-p)}X_{n+1}|\mathcal{F}_n]=$ ...


2

For simplicity, lets say that we have two constants, $a,b: a\geq 1, b\leq 1$ and that $P(Y=a)=p, P(Y=b)=(1-p)=q$ with the constraint that $ap+bq=1=(a-b)p+b \implies p=\frac{1-b}{a-b}$ Now, $\lim_{n\rightarrow \infty}X_n=0\; a.s. \iff P(\lim_{n\rightarrow \infty}X_n>0)=0$. As you pointed out, that is equivalent to requiring $\lim_{n\rightarrow ...


2

Since the random walk has $+-1$ valued steps, converging to an interior point means that the walker has to remain at that point for every step after a certain time, which is impossible.


2

Notice that $M_n$ is $\mathcal F_n$ measurable hence so is $S_n$, and $M_n$ is integrable since it is bounded. It remains to check that $\mathbb E(S_{n+1} \mid \mathcal F_n)=S_n$. To this aim, we have to compute $\mathbb E(e^{bM_{n+1}} \mid \mathcal F_n)$. As $$e^{bM_{n+1}}=\underbrace{e^{bM_{n+1}-bM_n}}_{\mbox{independent of }\mathcal F_n}\cdot ...


2

Let $X_1:=0$, $\ X_n:=X_{n-1}-(n-1)+Y_{n}$, where $Y_n$ is independent, $\mathcal{F}_{n}$ measurable and $P(Y_n=0)=1/n$, $P(Y_n=n)=1-1/n$. Then $E[Y_n]=n-1$, so $E[X_n|\mathcal{F}_{n-1}]=X_{n-1}$. Since $X_n\rightarrow\infty$ is a 0-1 event, we just need to show that it occurs with positive probability to conclude $X_n\rightarrow\infty$ almost surely. If ...


1

It follows from Itô's formula that the solution to the SDE $$dM_t = M_t \sigma_t dW_t$$ equals $$M_t = M_0 \exp \left( \int_0^t \sigma_s \, dW_s - \frac{1}{2} \int_0^t \sigma_s^2 \, ds \right).$$ By assumption, $M_0 \geq 0$. Hence, $$\begin{align*} \sqrt{M_T} &= \sqrt{M_0} \exp \left( \frac{1}{2} \int_0^T \sigma_s \, dW_s - \frac{1}{4} \int_0^T ...


1

Hint: Fix $s \leq t$. Show that $$\langle M,N \rangle_t^s := \langle M,N \rangle_t - \langle M,N \rangle_s $$ defines a symmetric positiv definite bilinear form. Alternatively: Apply the first inequality (the one you already proved) to the shifted processes $$\tilde{M}_t := M_{t+s}-M_s \qquad \quad \tilde{N}_t := N_{t+s}-M_s, \qquad t \geq 0.$$


1

Fix $s \le t$. Let $\{\tau_k\}$ be a localizing sequence for $M_t$; i.e. $\tau_k$ are stopping times with $\tau_k \uparrow \infty$ almost surely and $M_{t \wedge \tau_k}$ is a martingale for each $k$. By uniform integrability, we have $X_{t \wedge \tau_k} \to X_t$ in $L^1$, and hence $E[X_{t \wedge \tau_k} \mid \mathcal{F}_s] \to E[X_t \mid ...


1

Take the expectation of both sides of the definition. They should simplify...


1

Suppose that $(X_t)_{t \geq 0}$ is a local martingale. Since $$X_0 + \int_0^t \alpha_s dW_s$$ is also a martingale, this means that $$M_t := X_t - \left( X_0 + \int_0^t \alpha_s \, dW_s \right) = \int_0^t \beta_s \, ds$$ is a local martingale. Moreover, $(M_t)_{t \geq 0}$ is of bounded variation and has continuous sample paths. It is widely known that ...


1

Hint: Solve the stochastic differential equation $$dM_t = M_t \, dW_t$$ in order to find an explicit formula for $M_t$ in terms of $t$ and $W_t$.


1

We can see that $S_{n+1} = S_n + X_{n+1} - (n+1)$ therefore the expectation is given by $$E[S_{n+1}|S_n] = E[S_n|S_n] + E[X_{n+1}] - n-1 $$ $$=S_n + (n+1)-(n+1)$$ $$=S_n\qquad\qquad\qquad\qquad\,\,\,$$ Where we have used that $E[Pois(n)] = n$ to get the expecation of $X_{n+1}$. Note also, that $E[S_n] = 0$ but since we already know $S_n$, $E[S_n|S_n] = ...


1

The purpose of these notations is to define $X_n$ rigorously. The suggestion after Q1, while trying to achieve the same goal based on the same idea, is, as you note, more confused. The rephrasing in Q2 is correct. Re Q3, the Bernoulli case might be enlightening. Assume that $\mu$ and $\nu$ are products of Bernoulli measures on $\{0,1\}$, ...


1

Looks to me as you "pulled out" a bit too much: Pull out gives $$\mathbb{E}\big[ (Z_i-Z_{i-1}) \cdot (Z_j-Z_{j-1}) \mid \mathcal{F}_j \big] = (Z_j-Z_{j-1}) \cdot \mathbb{E}(Z_i-Z_{i-1} \mid \mathcal{F}_j).$$ Note that there is still the conditional expectation on the right-hand side. Now the martingale property gives $$\mathbb{E}(Z_i-Z_{i-1} \mid ...



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