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5

$\{X_k\}$ cannot be a nonnegative MTG with $X_0=0$. So, I assume it is just a MTG. Using the original Doob's weak inequality, for some $z >-r$ (and noticing that $\mathbb{E}X_n=0$): $$P\{\max_{1\le k\le n}{X_k}\ge r\}\le P\{\max_{1\le k\le n}{(X_k+z)^2}\ge (r+z)^2\}\le\frac{1}{(r+z)^2}\mathbb{E}(X_n+z)^2=\frac{1}{(r+z)^2}(\mathbb{E}X_n^2+z^2)$$ ...


3

(b) It is well-known that the countable supremum of measurable functions is measurable, i.e. if $(Y_n)_{n \in \mathbb{N}}$ is a sequence of $\mathcal{F}$-measurable (real-valued) random variables, then $\sup_{n \geq 1} Y_n$ is also $\mathcal{F}$-measurable. To see this, note that $$\left\{ \sup_{n \geq 1} Y_n > \alpha \right\} = \bigcup_{n \in \mathbb{N}} ...


3

I think both of you and your book are right! The point here is $d(S_t)d(e^{-rt})=0$. The reason is $dt\cdot dW=0$ and $dt\cdot dt=0$.


2

Since $\cosh(x) \ge 1$ for all real $x$ you have $\frac{e^{t/2}}{\cosh X_t} \le e^{t/2}$. So on any finite time interval you have a bounded local martingale, which is a martingale. Alternatively, note that $\frac{\sinh(x)}{(\cosh(x))^2} \le \frac{1}{2}$ for all $x$. So for any $T$, $$\int_0^T E\left[\left|-\frac{e^{t/2} \sinh ...


2

Hints: is correct Suppose that $S_n=0$ for some $n \geq 1$. Since $$S_n = 0 \iff -S_{n-1} = X_n \sqrt{1+S_{n-1}^2},$$ this implies in particular $$S_{n-1}^2 = (X_n)^2 (1+S_{n-1}^2).$$ Show that this in turn implies $1=0$ - which is obviously a contradiction. To prove the second part of (b), note that $$S_n = 1 \iff 1-S_{n-1} = X_n \sqrt{1+S_{n-1}^2}.$$ ...


2

Hints: If you choose a set at random, then with probability at least 1/2 the chromatic number is at least 300 (why?). Azuma's inequality shows that the chromatic number of $G|_S$ is concentrated around its mean. Since the chromatic number is always between 0 and 600, and its median is above 300, its mean can't be too low. Hence the probability that the ...


2

Let's look at $\mathbb{E}[X_{n+1}^2-(n+1)|F_n]$ (take $s=n$). We need show it equals to $X_n^2-n$. Since $X_{n+1}=X_n+ Y_{n+1}$ It should be $$ \mathbb{E}[X_{n}^2|F_n]+2\mathbb{E}[X_{n}Y_{n+1}|F_n]+\mathbb{E}[Y_{n+1}^2|F_n]-\mathbb{E}[n+1|F_n] $$, Since $X_{n}$ is measurable w.r.t. $F_n$, and $Y_{n+1}$ is independent of $F_n$ (for which we can "take out ...


2

For $s\le n$, $$\mathbb{E}(M_{n+1}\mid \mathcal{F}_s)=\mathbb{E}((X_n+Y_{n+1})^2-(n+1)\mid \mathcal{F}_s)\\=X_n^2+2X_n\mathbb{E}(Y_{n+1})+\mathbb{E}(Y_{n+1}^2)-(n+1)\\=X_n^2-n=M_n$$ the steps have used the fact that $Y_n$ is an i.i.d. $\mathcal{N}(0,1)$ sequence. The possible mistake that you are making in your approach is the one rightly pointed out by @saz ...


2

Work on a compact interval. call the function $f$. We then have a constant $L$ for that interval. If $t_0 < t_1 < ... < t_n,$ then $$ \sum |f(t_j)-f(t_{j+1})|^2 \leq L \sum |t_j-t_{j+1}|^2 $$ As the partition size goes to zero, the RHS goes to zero. In particular if the max distance between two of the points is $\delta$ then it is $$ \leq \delta ...


2

Yes, the stopped compensated Poisson process is uniformly integrable: By Doob's maximal inequality, we have $$\mathbb{E} \left( \sup_{t \in [0,K]} |\bar{N}(t \wedge T)|^2 \right) \leq 4 \mathbb{E}(|\bar{N}(K \wedge T)|^2)$$ for any $K>0$. Since $(\bar{N}_t^2-\lambda t)_{t \geq 0}$ is a martingale, we obtain $$\mathbb{E} \left( \sup_{t \in [0,K]} ...


2

In order to identify the limit of $(X_n)$, one can use the strong law of large numbers: we have $$\lim_{n\to \infty}\frac{\log X_n}n=\mathbb E\left[ \log(Y_1)\right].$$ Here we used the not necessarily integrable version of the law of large numbers, since the positive part of $\log (Y_1)$ is integrable. Using Jensen's inequality, note that $\mathbb ...


1

$$\mathbb{E}\left[S_n\right]=\mathbb{E}\left[\mathbb{E}\left[S_n\mid X_{n-1}\right]\right]=\mathbb{E}\left[S_{n-1}\right]=\mathbb{E}\left[\mathbb{E}\left[S_{n-1}\mid X_{n-2}\right]\right]$$ $$=\mathbb{E}\left[S_{n-2}\right]=\ldots=\mathbb{E}\left[\mathbb{E}\left[S_{2}\mid X_{1}\right]\right]=\mathbb{E}\left[S_{1}\right].$$ To show the first equality, we use ...


1

Integrating any locally bounded predictable process (in particular a continuous adapted process) with respect to a local martingale (resp locally square integrable martingale) results in a new process which is a local martingale (resp locally square integrable). A proof of this can be found in Jacod & Shirjaev (Chapter 1; Theorem 4.31). (It basically ...


1

Note that $R_{n+1}=R_n$, if a green ball is chosen at step $n+1$ and $R_{n+1}=R_n+1$ if a red ball is chosen at step $n+1$. So, given $R_0,\cdots,\ R_{n-1},R_n$, we have $R_{n+1}=Z_n R_n+(1-Z_n)(R_n+1)$ where $Z_n\sim Bin(1-S_n)$ which means that $R_{n+1}$ is independent of $R_0,\cdots,\ R_{n-1}$ given $R_n$. So $$\mathbb{E}(R_{n+1}\mid R_0,\cdots,\ ...


1

Define a martingale by $$M_t := \begin{cases} \mathbb{E}(\phi(W_T) \mid \mathcal{F}_t), & t \leq T, \\ \phi(W_T), & t>T. \end{cases}$$ Then, by the martingale representation theorem, there exists a representation of the form $$M_t = c+ \int_0^t \beta_s \, dW_s.$$ For $t=T$, this proves the claim. Remark: One possibility to prove the mentioned ...


1

Are the $X_i$'s suppose to represent win/loss? Anyway, if $X_1$ was the result of a coin toss, we would have $\sigma(X_1) = (\emptyset, \Omega, T, H)$ so $\sigma(X_1) \neq (\emptyset, (-1,1), -1, 1)$. Recall that $\sigma(X_1) = (X^{-1}(B)|B \in \scr{B})$ is the collection of preimages of $X_1$ (I like to think of it as the set of events that determine the ...


1

The first equality is always true. The second is the definition of the conditional expectation, plus the fact that $S_k$ is a deterministic function of $$ X_k \dots X_1 $$


1

We want to prove that $\mathbb E[S_{n+m}\mid X_1,...,X_n]= S_n,\quad n,m\geq 1$ Use induction on $m$. By the definition of the martingale, $\mathbb E[S_{n+1}\mid X_1,...,X_n]= S_n$ so the base case is true. Suppose $\mathbb E[S_{n+m}\mid X_1,...,X_n]= S_n$ for some $m \geq 1$. Then: $$S_n = \mathbb E[S_{n+m}\mid X_1,...,X_n]$$ (by the induction ...


1

Denote by $\mathcal{F}_n := \sigma(X_1,\ldots,X_n)$ the canonical $\sigma$-algebra and fix $i<j$. Then, $$\mathbb{E}(X_j - X_{j-1} \mid \mathcal{F}_i) = \mathbb{E}(X_j \mid \mathcal{F}_i)- \mathbb{E}(X_{j-1} \mid X_i) = X_i-X_i = 0. \tag{1}$$ Using the tower property and pull out, we get $$\begin{align*} \mathbb{E}[(X_i-X_{i-1}) \cdot (X_j-X_{j-1})] ...


1

Hint: (For $\xi > 0$) (I am assuming $X_i$ are independent). 1) $$\{M_n \geq \lambda\} \iff \{e^{\xi M_n} \geq e^{\xi\lambda }\}$$ 2) $$e^{\xi M_n} = e^{\xi \max_i S_i} = \max_i e^{\xi S_i} = \max_i Y_i$$ Can you take it from here? Let me know if you need additional help. Update: For the next park, work with $-X_i$. Then $T_n =-S_n = \sum_{i=1}^n ...


1

I may have a solution to my problem but I think it doesn't need the martingale law of large numbers. As we can note that for all $ n \in \mathbb{N}$ $$ N(n) = \sum_{1 \leq i \leq n} \left(N(i) - N(i-1)\right)$$ is the sum of $n$ iid random variables all with a Poisson$(\lambda)$ distribution and hence are integrable. Then applying the strong law of large ...


1

It's quite common to distinguish between Type A arbitrage and Type B arbitrage. We say that a trading strategy is a type A arbitrage if it has a positive initial cashflow and no risk of future loss. type B arbitrage if it has a nonnegative initial cashflow, no risk of future loss and a positive probability of future profit. I think what you are referring ...


1

Heuritically, $d(S_t)d(e^{-rt})=-re^{-rt}dt(dS_t)\sim O(dt)^{3/2}$. Thus, it is not considered in the SDE for $S_te^{-rt}$.


1

Set $X_t = W_t^2 + at + b\int_0^t W_s^2\, ds$, so that $$dX_t = (1 + a + bW_t^2)\, dt + 2W_t\, dW_t$$ and $$d[X,X]_t = 4W_t^2\, dt.$$ Using Ito's lemma, we get $$d[e^{X_t}] = e^{X_t}\, dX_t + \frac{1}{2} e^{X_t} d[X, X]_t = e^{X_t}\bigl((1 + a + (b + 2)W_t^2)\,dt + 2W_t\, dW_t\bigr).$$ For $e^{X_t}$ to be a martingale, the drift term must vanish, ...


1

In general, we cannot expect $\tau<\infty$ almost surely. Just consider e.g. the trivial martingale $M_t := 0$, $t \geq 0$, then $\tau =\infty$ almost surely. However, since the martingale $(M_{t \wedge \tau})_{t \geq 0}$ is bounded, it follows from the martingale convergence theorem that $M_{t \wedge \infty}$ converges almost surely to some random ...



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