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Solution 1: Recall the following two statements. Lemma 1: Let $(Y_t)_{t \geq 0}$ be a supermartingale and $f$ an increasing concave function, then $(f(Y_t))_{t \geq 0}$ is a supermartingale. Lemma 2: Let $(Y_t)_{t \geq 0}$ be a locale martingale such that $Y_t \geq 0$ for all $t \geq 0$. Then $(Y_t)_{t \geq 0}$ is a supermartingale. Lemma 1 is a ...

4

You have that $-\log x$ is a convex function. Then $\mathbb{E}[-\log M_{n+1} | \mathcal{F_n}] \geq -\log(\mathbb{E}(M_{n+1} | \mathcal{F}_n) = -\log(M_n)$ (I used Jensen in the first inequality) Then, $-\log(M_n)$ is a submartingale and then $\log(M_n)$ is a supermartingale!

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$\sigma\{X+Y\}$ is contained in $\sigma\{X,Y\}$. This because measurability of $X$ and $Y$ implies measuribility of $g(X,Y)$ for any Borel-measurable $g:\mathbb R^2\to\mathbb R$. Involved here is the map $\langle x,y\rangle\mapsto x+y$ If e.g. $Y=c-X$ for some constant $c$ then $\sigma\{X+Y\}=\sigma\{c\}=\{\Omega,\varnothing\}$ $\sigma\{X,Y\}$ and the ...

4

In general, there's no connection between $\sigma(X+Y)$ and $\sigma(X)$ and $\sigma(Y)$. If $Y=-X$, for instance, then $X+Y$ is constant so it has a trivial sigma-field, whereas $\sigma(X)$ typically would not. While it is true that $\sigma(X+Y)$ is contained in $\sigma(X,Y)$, the reverse inclusion doesn't hold even if $X$ and $Y$ are independent. For ...

3

We want to show that $X_t = tW_t - \int_0^t W_s \mathrm{d}s$ is a martingale. Let $r\leq t$ then $$\mathbb{E}[X_t\ | \ \mathcal{F}_r]= \mathbb{E}[tW_t\ |\ \mathcal{F}_r] - \mathbb{E}\left[\int_0^t W_s \mathrm{d}s \ | \ \mathcal{F}_r\right].$$ As you noticed $$\mathbb{E}[tW_t\ |\ \mathcal{F}_r] = \mathbb{E}[t(W_t-W_r) + tW_r\ |\ \mathcal{F}_r] = t ... 3 Let a_{m}=\sum_{i=1}^{m}X_i1\{X_i=1\} and b_{m}=-\sum_{i=1}^{m}X_i1\{X_i=-1\}. Then a_m+b_m=m and a_m-b_m=S_m. Solving this system$$ a_m=\frac{m+S_m}{2}\text{ and }b_m=\frac{m-S_m}{2} $$and the probability of getting 1 in step m+1 given S_m is$$ P\{X_{m+1}=1\mid S_m\}=\frac{n-a_m}{2n-m}=\frac{1}{2}-\frac{S_m/2}{2n-m}. $$Hence,$$ ...

2

For any $r\ge 0$ $$\mathbb{E}(M_{\infty}-M_n)^2=\mathbb{E}(M_{\infty}-M_{n+r})^2+\mathbb{E}(M_{n+r}-M_n)^2 \\\overset{(d)}=\mathbb{E}(M_{\infty}-M_{n+r})^2+\sum_{k=n+1}^{n+r}\mathbb{E}(M_k-M_{k-1})^2.$$ Taking $r\to\infty$, the first term converges to $0$ by (f).

2

As already stated in the comment, we usually do not have the formula $\int x dx = x^2/2$, if we replace the ordinary Lebesgue/Riemannian integral by the Itô integral, which is exactly the statement of Itô's lemma, as we have $df(B_t) = f'(B_t) dB_t + \frac{1}{2} f''(B_t) dt$. So in your case, if you are interested in calculating $f(B_s) dB_s$ for a ...

2

Hints: Show (or recall) that any $\mathcal{F}_n$-measurable function $X$ is of the form $$X(\omega) = \sum_{j=0}^{2^n-1} c_j 1_{(j 2^{-n},(j+1)2^{-n}]}(\omega)$$ for suitable constants $c_j \in \mathbb{R}$. Since $X_n$ is $\mathcal{F}_n$-measurable, Step 1 shows that there exist constants such that $$X_n(\omega) = \sum_{j=0}^{2^n-1} c_{j,n} 1_{(j ... 2 Let F_n be the total fortune of gamblers 0,1,2,\ldots,n at time n. Then G_n=F_n-n is a martingale with initial value 0. And F_{\tau_a}=2^4+2. The 2^4 is the fortune of gambler number \tau_a-3, and the 2 is the fortune of gamble number \tau_a. All other gamblers are out of the picture (or have yet to start) at time \tau_a. By Doob's ... 2 In addition to the hints, we need the following identities related to derangements: (1) \displaystyle\sum_{i=0}^{n}\frac{!i}{i!~(n-i)!}=1,~(n\ge0), (2) \displaystyle\sum_{i=0}^{n}\frac{!i~(n-i)}{i!~(n-i)!}=1,~(n\ge1), (3) \displaystyle\sum_{i=0}^{n}\frac{!i~(n-i)~(n-i-1)}{i!~(n-i)!}=1,~(n\ge2), where I have adopted the subfactorial notation. The ... 2 In general no. For example, take (x_n)_{n\geqslant 1} an i.i.d. sequence of integrable random variables and \mathcal F_n:=\sigma(x_1,\dots,x_n). In this case, (x_n,\mathcal F_n) is a martingale differences sequence. However, by independence,$$\mathbb E\left[x_n\mathbf 1\{|x_n|\leqslant a_n\}\mid\mathcal F_{n-1}\right]=\mathbb E\left[x_n\mathbf ...

1

Easy way: we know that $B_t$ is a Gaussian process, i.e. for any $t_1, \dots, t_n$ the random vector $(B_{t_1}, \dots, B_{t_n})$ has a jointly Gaussian distribution. Since any linear transformation of a jointly Gaussian vector is again jointly Gaussian, it follows that $W_t$ is also a Gaussian process, and likewise so is $Y_t$. Then it's easy to compute ...

1

Let $\varphi(\theta)=\mathbb E[e^{\theta X_1}]$ be the moment generating function and $\psi(\theta)=\log\varphi(\theta)$ the cumulant generating function of $X_1$. We compute $$\varphi(\theta) = e^{-\theta}\mathbb P(X_1=-1)+e^{\theta}\mathbb P(X_1=1)=\frac12\left(e^{-\theta}+e^{\theta}\right)$$ and $$\psi(\theta) = \log\left(\frac 12 ... 1 If "exponential martingale" does not ring a bell that you may want to try the following approach. Note that your process S_t := \exp(X_t - \frac{1}{2}Y_t) is a Geometric Brownian motion with a non-constant volatility parameter h(t), that is, is satisfies the following SDE$$\mathrm{d} S_t = S_t h(t)\mathrm{d}W_t $$which reads$$ S_t = 1+ \int_0^t S_s ...

1

Let the filtration be $\mathcal{F}_n=\sigma(X_1,\dots,X_n)$, i.e., the smallest sigma field such that each of $X_1,\dots, X_n$ are measurable with respect to it. Conditioning on $X_1,\dots,X_n$ is equivalent to conditioning on $\mathcal{F}_n$, so according to the second definition we have $E[X_{n+1}|X_1,\dots,X_n]=E[X_{n+1}|\mathcal{F}_n]=X_n$. Note in ...

1

Sketch for the one-period case: Denote $\Pi(X) = \inf_{\mathbb{P}\in \mathcal{P(M)}}\mathbb{E}_{\mathbb{P}}\left(\frac{X}{B_1} \right)$. By FTAP, this is the set of non-arbitrage prices for $X$. Therefore, any $\pi<\Pi(X)$ must be an arbitrage price. Then, denoting $Y = S_1/B_1-S_0$ the increment of discounted share, we have for some $(\xi,\xi_{1})$ that ...

1

This might be helpful: Theorem 29 on page 173 of Protter's "Stochastic integration and differential equations": Let $M$ be a local martingale and let $H\in \mathcal P$ be locally bounded. then the stochastic integral $H\cdot M$ is a local martingale. I'm reading that book, but haven't got that far yet, will be interested to see if someone can provide a ...

1

That's the point; $c(n)$ can't be a random quantity. Therefore you have shown that there is no $c(n)$ such that $M_n$ is a martingale. $$E[S_{n+1}^2|F_n]=E[(S_n+X_{n+1})^2|F_n]$$ $$S_n^2 + 2(2p-1)S_n+ 1 = S_n^2 +(c(n)-c(n+1)) \implies c(n)-c(n+1)=2(2p-1)S_n+1$$ No deterministic $c(n)$ can exist unless $X_i$ were constant, but this contradicts the given ...

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A supermartingale can be thought of as a gambling game which is "super" for the casino, but not for those playing; i.e. $$\mathbb{E}[X_{n+1}|\mathscr{F}_n]\le X_n.$$ Note that for concave functions (like the logarithm), Jensen's inequality reverses, hence we have: $$\mathbb{E}[\log M_{n+1} | \mathscr{F}_n] \le \log(\mathbb{E}[M_{n+1}|\mathscr{F}_n])$$ ...

1

I do not know how one can use the particular Rademacher distribution, but there is a useful reverse martingale trick: it is easy to see that $$E[X_1|S_n, S_{n+1},\dots] = \frac{S_n}{n}.$$ Denoting $\mathcal{F_n} = \sigma\{S_n, S_{n+1},\dots\}$, we thus have a (reverse) Levy martingale $Y_n = E[X_1|\mathcal{F_n}]$. By the reverse martingale convergence, ...

1

Indicating the dependence of $T$ on $c$ explicitly, you have $$\{T_c=\infty\}=\{\sup_n|M_n|\le c\}$$ Therefore, $$\bigcup_{c>0,c\in\Bbb Q}\{T_c=\infty\}=\{\sup_n|M_n|<\infty\}$$

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The proof is correct, but notice that you get something stronger than what you state in the last line of your question, indeed $X_{\infty}$ is finite almost surely, which is the same as saying $-\infty < X_{\infty} < \infty$ (and not only $X_{\infty} < \infty$). To go from the fact that $E(|X_{\infty}|) < \infty$ to the fact that $X_{\infty} \in ... 1 The partial sums of$\sum X_k$are bounded by constant$c$iff$|M_r|\le c$for every$r$. This last occurs iff$T_c=\infty$(where I write$T_c$instead of$T$to make the dependence on$c$explicit). Therefore $$\{\text{partial sums of \sum X_k are bounded}\} = \bigcup_{c=1}^\infty \{T_c=\infty\}.\tag1$$ So if the LHS has positive probability then ... 1 You've shown that there is a$c$such that$P(T=\infty)>0$. So:$$E[A_{T\wedge n}]=E[A_{T\wedge n}|T<\infty]P(T<\infty)+A_n P(T=\infty).$$ Both terms terms on the r.h.s are non-negative and you've also shown that$E[A_{T\wedge n}]$is finite and you can take$n\rightarrow\infty$, so at the very least$A_\infty<\infty$since$P(T=\infty)>0\$

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