# Tag Info

0

Introduce imaginary extra random variables. We define a martingale difference sequence $V_1,V_2,\ldots$ as follows. For each instance of the problem let $\tau$ be the number of flips and $x_1,x_2,\ldots, x_\tau$ the results. Note $n$ is different in each instance. Define $V_i = x_i -1/2 \$for $i=1,2,\ldots, \tau$. For each $i > \tau$ define $V_i = 0$. ...

1

Formally you are right. What Williams means is that the process is bounded almost surely. This is enough to prove the theorem anyway. (he proves that a limit exists almost surely, so it's not a problem to ignore a set of probability zero here)

0

The stopping time $\tau(\omega)$ of one particular path $\omega$ is the time this path is further than $K$ away from the origin for the first time. The line you are interested in says the collection of all paths with stopping time smaller than $t$ is the union over all $s$ smaller than $t$ of all path that at time $s$ are further than $K$ from the origin. ...

0

Equivalent are the statements: $\omega\in\bigcup_{s\in\left(0,t\right)}\left\{ |W_{s}|\geq K\right\}$ For some $s\in\left(0,t\right)$ we have $|W_{s}\left(\omega\right)|\geq K$ $\inf\left\{ s>0\mid |W_{s}\left(\omega\right)|\geq K\right\} <t$ $\tau\left(\omega\right)<t$

1

What you've described is the negative binomial distribution. We say $X$ has a negative binomial distribution $X \sim NB(r, p)$ if $X$ is the number of independent bernoulli trials of parameter $p$ until we get $r$ 'successes'. So in your example, if $X$ is the number of trials until we get 10 heads, $X$ will have $NB(10, 1/2)$ distribution. To calculate its ...

1

This is the probability that we get at most $9$ heads in $19$ tosses. That is, $$\sum_{j=0}^9\binom{19}j\frac1{2^{19}}$$

1

The probability of getting at least $m$ heads after $n$ flips is just $P(X \geq 10)$ where $X ~ Bin(n,\frac{1}{2})$. Does this help?

1

Your proof is correct. The boundness of the stopping times is needed in optional stopping theorem because the martingale $M=(M_n)$ may not converge as $n\to\infty$. In other words, if the martingale $M_n$ does not converge as $n\to\infty$ and also the stopping time $\tau$ is unbounded, then the random variable $M_{\tau(\omega)}(\omega)$ cannot be well-...

0

Unfortunately the link is broken ... Sigma-martingales are covered in Philip Protters' book: Stochastic Integral and Differential Equations. In the second edition (Vers 2.1) Chapter IV.9 treats this topic. Corollary 2 of this chapter establishes that a local martingale is a sigma martingale. Theorem 91 establishes that a sigma martingale which is a ...

0

Consider the polarization identity for covariance given by $\langle X,Y \rangle = \frac{1}{4} (\langle X+Y \rangle -\langle X-Y \rangle )$. You can show that both X+Y and X-Y also have finite variation (use the definition of variation and then triangle inequality) so then both terms in the polarisation identity vanish as finite variation implies zero ...

0

Hint Using Itô formula, if $f(x,t)=e^{\sigma x+\mu t}$ $$X_T=1+\int_0^T \left(\mu+\frac{1}{2}\sigma ^2\right)e^{\sigma B_t+\mu t}\,\mathrm d t+\int_0^T\sigma e^{\sigma B_t+\mu t}\,\mathrm d W_t.$$ So, $X_t$ is a martingale if $$\int_0^T \left(\mu+\frac{1}{2}\sigma ^2\right)e^{\sigma B_t+\mu t}\,\mathrm d t=0\quad \text{and}\quad \sigma e^{\sigma B_{\cdot }... 1 This is not an answer, but is too long for comment. I will elaborate on my comment. We have for t\le T$$X_t = x + \int_0^t b(X_s)\,d s + \int_0^t \sigma(X_s) \, d W_s in \mathbb{R}. In the following C changes from line to line. For each t \le T then, \begin{align} E (X_t^2) &\le C \left[x^2 + E \left\{\left( \int_0^t b(X_s)\,d s\right)^2\... 1 As Brownian motion has independent increments, we have for s < t and \sigma > 0 that\mathbb{E}[e^{\sigma B_t}|\mathcal{F_s}] = \mathbb{E}[e^{\sigma (B_t - B_s)}\cdot e^{\sigma B_s} | \mathcal{F_s}] = e^{\sigma B_s} \mathbb{E}[e^{\sigma (B_t - B_s)}].$$This expectation can be computed with help of the moment generating function of the normal ... 0 Here is a case where it is more convenient to take the conditional expectation with respect to a filtration which may be an other one than the natural one: truncation arguments. Suppose that we are studying a martingale differences sequence \left(X_k\right)_{k\geqslant 1} with respect to a filtration \left(\mathcal F_k\right)_{k\geqslant 0}. For a ... 1 For  0 <s<t, E(e^{\int_0^{t} W_u du}e^{\int_0^{s} W_u du}|\mathcal F_s)=e^{2\int_0^{s} W_u du}E(e^{\int_s^{t} (W_u-W_s) du}|\mathcal F_s) e^{(t-s)W_s}. The conditioning in the last expression is unnecessary by independence. Also, \int_s^{t} (W_u-W_s) du has the same distribution as \int_0^{t-s} W_u du (because (W_{r+s}-W_s)_{r \geq 0} is also ... 0 You may use the fact that:  \mathbb{E}[e^{\int_0^tW_udu}e^{\int_0^sW_udu}] = \mathbb{E}[e^{2\int_0^sW_udu + \int_s^tW_udu}]  and that \{W_u\}_{0 \leq u \leq s} and \{W_u\}_{s \leq u \leq t} are independent. 0 To expand on my comment for Question 1: Z_t is not measurable with respect to \mathcal{F}_t as you have currently written it, as it depends on values of W_s for all s\leq t, not just W_t. It is, however, measurable with respect to \sigma(\cup_{s\leq t} W_s), that is, the sigma-algebra generated by all previous values of W_s. To understand why, ... 2 Question 1: Yes. According to Ito product rule d(tW)=W dt+tdW, then we could get$$ \int_{0}^t W(s)ds=tW(t)-\int_{0}^{t}s dW(s) $$For the right-hand side of first term, it is \mathcal{F}_t-measurable;for the second term, it is also \mathcal{F}_t-measurable. For how to determine whether a random variable is \mathcal{F}_t-measurable, there is an ... 0 The mistake in the first approach occurred in the second equality. It is not true (when s<t) that \mathbb E[e^{W_t}\mid \mathcal F_s]=e^{W_t}. In fact, the simplest way I see to compute this conditional expectation (for s<t) is to use the fact that W_t is a Markov process, and therefore so is e^{W_t}. Thus,$$ \mathbb E[e^{W_t}\mid \mathcal ...

1

There is Theorem 9 p.142 in Liptser and Shiryaev's "Theory of Martingales" of which a particular case is the following : Let $M = (M_t)_{t \geq 0}$ be a local martingale and define the process $B = (B_t)_{t \geq 0}$ by : $$B_t = \sum_{0 < s \leq t}\frac{(\Delta M_s / (1+s))^2}{1 + |\Delta M_s / (1 + s)|},$$ where $\Delta M_t$ is the jump of $M$ at time $... 3 Note that$\mathcal{F}_s^{XY} \subseteq \mathcal{F}_s^X \vee \mathcal{F}_s^Y$for all$s \geq 0$; this follows from the fact that the product$X_s Y_s$is measurable with respect to$\mathcal{F}_s^X \vee \mathcal{F}_s^Y$. Hence, by the tower property of conditional expectation, $$\mathbb{E}(X_t Y_t \mid \mathcal{F}_s^{XY}) = \mathbb{E} \bigg[ \underbrace{\... 1 I would rather use the \pi - \lambda theorem (and not monotone class theorem). Sets of the type A\cap B with A \in \mathcal F^{X}_s and B \in \mathcal F^{Y}_s form a \pi system and the class defined in that reference is a \lambda system containing this \pi system. Hence it contains the sigma algebra generated, which is \sigma(\mathcal F^{X}_s \... 1 If \ \mathcal{F}, \mathcal{G}\ are \sigma-algebras, then \mathcal{F}\vee\mathcal{G} is the \sigma-algebra generated by \ \mathcal{F}\cup\mathcal{G}\ , known as the join of \ \mathcal{F}\ and \ \mathcal{G}\ . 1 For two \sigma-algebras \mathcal{A} and \mathcal{B} on the same set, one usually denotes$$ \mathcal{A} \vee \mathcal{B} := \sigma(\mathcal{A}\cup\mathcal{B}).$$This is a pretty common notation when talking about filtrations. 1 If \mathbb{P}(S<\infty)=1 or if M_\infty is defined almost everywhere then obviously M_{S\wedge n}\to M_S almost surely. But if at some point M_n does not converge and S=\infty then at this point M_{S\wedge n}=M_n for all n and this sequence does not converge. So at such points M_S is not really defined. So the problem is when the set of ... 3 A martingale is uniformly integrable if and only if it converges in L^1. A uniformly integrable martingale converges almost surely. For convergence in L^2 we need to know a bit more. (for example the martingale being bounded in L^2 is sufficient) 0 First, it is important to notice that (not unusually) Roger and Williams have, at the point you are looking at, restricted their integrator M to be a continuous L^2-bounded martingale. This means that at this stage, Brownian motion is not an allowed integrator, since it is not L^2-bounded. The punchline will be that by localisation, once we have the ... 0 Yes, this is an immediate consequence of what is commonly referred to as Doob's Optional Sampling Theorem. 0 E|X_t| \leq E|W_t^{3}| +3\int_0^{t} E|W_s| ds. The first term is finite because normal distributions have finite moments of all orders. Also Y=\frac 1 {\sqrt s} W_s has standard normal distribution so E|W_s|=\sqrt s E|Y|. Hence \int_0^{t} E|W_s| ds=E|Y| \int_0^{t} \sqrt s ds <\infty. 1 You are right that there might be overshooting, but we have some bounds on how much. First, since there might be overshooting now, we need to replace a and b with E(M_\tau\mid \tau = \tau_a) and E(M_\tau\mid \tau=\tau_b), and then we get$$ P(M_\tau\ge b) = P(\tau=\tau_b) = \frac{1-E(M_\tau\mid \tau=\tau_a)}{E(M_\tau\mid \tau=\tau_b)-E(M_\tau\mid \... 2 Re pointwise convergence: Kolmogorov three series theorem is somewhat overkill here. Just note that $$|S_n(\omega)-S_{n-1}(\omega)| = 1, \qquad n \geq 1,$$ for almost all$\omega \in \Omega$. In particular, for any$\epsilon \in (0,1)$there does not exist$N \in \mathbb{N}$such that$|S_n(\omega)-S_{n-1}(\omega)| \leq \epsilon$for all$n \geq N$; this ... 1 Since you mention that you do not fully understand what "independent increments" means, I believe it is more appropriate to start with a simpler example that illustrates the idea. Instead of a Brownian motion, consider its discrete analogue: the random walk. I will start from a sequence$X_1,X_2,\ldots$consisting of independent random variables. This means ... 1 If$X$is independent of$\mathcal G$then$X^{2}$is also independent of$\mathcal G$so the answer is YES. 2 If$X$is independent of$\mathcal{G}$then$f(X)$is also independent of$\mathcal{G}$for every Borel function$f$. This is easy to verify from the fact that$\sigma(X)$is independent of$\mathcal{G}$, and that$(f(X))^{-1}(B) = X^{-1}(f^{-1}(B))$. In particular we have$E[f(X) \mid \mathcal{G}] = E[f(X)]$whenever$E[f(X)]$exists. Apply this with$f(...

1

Independent increments means that for $t>s$, the random variable $B_t-B_s$ is independent of the $\sigma$-algebra at time $s$, i.e. independent of $\mathcal{F}_s=\sigma(\cup_{v\leq s} B_v)$. In fact, for Brownian motion, $B_t-B_s\sim \mathcal{N}(0,t-s)$ and is independent of $\mathcal{F}_s$. Of course, this implies $(B_t-B_s)^2$ is also independent of $\... 1 Your answer is correct in the first two cases. If$t <r<s$then$W_s-W_r$is independent of$F_t$. [This can be proved from the fact that BM has independent increments]. Hence the answer in this case is$E(W_s-W_r)^{2}=s-r. 1 First of all, we notice that \begin{align} E[(X_t - X_s)(Y_t - Y_s)| \mathcal F_s] &= E[X_tY_t - X_tY_s -X_sY_t +\underbrace{X_sY_s}_{\mathcal{F}_s-\text{mesurable}}|\mathcal F_s]\\ &= E[X_tY_t|\mathcal F_s] - Y_sE[X_t|\mathcal F_s] - X_sE[Y_t|\mathcal F_s] + X_sY_s \\ &= E[X_tY_t|\mathcal F_s] - Y_sX_s - X_sY_s + X_sY_s \\ &=E[X_tY_t-... 0 It turns out that this can be done using the strong decoupling theorem (7.3.1) in "Decoupling: From Dependence to Independence" by Hitczenko. We introduce independent copies of theX_i$named$Y_i$, such that$E[Y_i \mid X_{i-1}, \dots, X_1] = E[X_i \mid X_{i-1}, \dots, X_1]$. The$Y_i$are conditionally independent given$X_{n}, \dots, X_1$.$E[Y_i \mid ...

Top 50 recent answers are included