# Tag Info

5

It's a basic substitution $s=1/t$. $$t \rightarrow 0\implies s\rightarrow\infty$$

4

Using Ito's Lemma we have $$d\log X_t=\left(v-\frac12\sigma^2\right)dt+\sigma dW_t \tag 1$$ Integrating $(1)$ between $t_1$ and $t_2$ yields \begin{align} \log(X_{t_2}/X_{t_1})&=\left(v-\frac12\sigma^2\right)\left(t_2-t_1\right)+\sigma\int_{t_1}^{t_2}dW_t\\\\ &=\left(v-\frac12\sigma^2\right)\left(t_2-t_1\right)+\sigma\left(W_{t_2}-W_{t_1}\right) ... 3 E^x is the expectation under the path measure of W(\cdot) started at W(0) = x. Example: E^x[W(t)] = E^x[W(t)-W(0)+W(0)] = E^x[W(t)-W(0)] + x = x since W(t)-W(0) \sim \mathcal{N}(0,t). In plain English, E^x is just ordinary expectation with the understanding that the starting point of BM is a variable, x. 3 Let me give you first a non-rigorous "proof" of the statement; afterwards I'll explain how to justify it. If we definef_n(x) := n x 1_{[0,1/n]}(x) + 1_{[1/n,\infty)}(x),$$then f_n \uparrow 1_{(0,\infty)}. Obviously, f_n is not differentiable at x=0 at x=1/n, but for the moment we forget about this (that's the non-rigorous part) and apply ... 3$$E\left[H_{t_{i-1}}^2\left(B_{t_i}-B_{t_{i-1}}\right)^2\right] = E\left[E\left[H_{t_{i-1}}^2\left(B_{t_i}-B_{t_{i-1}}\right)^2 \mid \mathbb{F}_{t_{i-1}}\right] \right] = E\left[H_{t_{i-1}}^2E\left[\left(B_{t_i}-B_{t_{i-1}}\right)^2 \mid \mathbb{F}_{t_{i-1}}\right] \right] = E\left[H_{t_{i-1}}^2E\left[\left(B_{t_i}-B_{t_{i-1}}\right)^2 \right] \right] = ...

3

The means add regardless of any other assumptions. Assuming (as usual in random walks) that the increments are independent, the variances also add. Hence the variance at time $t$ is $\sigma^2t$, and the mean is $\mu t$. Now $$\text{Var}(x_t)=E[x_t^2]-(E[x_t])^2=E[x_t^2]-\mu^2t^2=\sigma^2t$$ hence $$E[x_t^2]=\sigma^2t + \mu^2t^2.$$

3

Without loss of generality, we may assume $x>0$. By the very definition of the stopping time, we have \begin{align*} \mathbb{P}_x(\tau_0>t) &= \mathbb{P}_x \left( \left\{w \in C[0,1]; \inf_{s \in [0,t]} (w(s)-x) +x >0 \right\} \right) \end{align*} Since $\mathbb{P}_x$ is defined by translational invariance (it is the image measure of the ...

3

It is true that the second property can be deduced from the first one. Indeed, the first property implies that $(B_t)$ has stationary, independent increments. Hence, the following statement would give you the implication in the forward direction: Proposition: Any real-valued stochastic process with stationary, independent increments has the elementary ...

3

What you've written is the result that you get when you perform the simplest approximation of Ito's formula. In this case Ito's formula reads $$e^{B_t} = 1 + \int_0^t e^{B_s} dB_s + \int_0^t \frac{1}{2} e^{B_s} ds.$$ For small $t$, $\int_0^t e^{B_s} dB_s \approx B_t$ (replacing the integrand by its value at $0$, which is $1$) and $\int_0^t \frac{1}{2} ... 2 If$X_n\to X$almost surely and$X_n$is independent of$\mathcal G$, then$X$is independent of$\mathcal G$. We use actually this. Call$X$the left hand side of (2). We can show that for each continuous and bounded function$f\colon\mathbf R^n\to\mathbf R$and$E\in\mathcal F_t$, we have$\mathbb E\left[f(X)\mathbf 1(E)\right]=\mathbb E[f(X)]\mu(E)$... 2 Although it is correct to do it this way, it's straight overkill; from my point of view. Here is a much easier solution: Since$W_t \sim N(0,t)$, we have $$\mathbb{E}e^{\lambda W_t} = \exp \left(\frac{1}{2} \lambda^2 t \right).$$ Differentiating both sides with respect to$\lambda$yields $$\mathbb{E}(W_t e^{\lambda W_t}) = \lambda t \exp \left( ... 2 To help hone your intuition, it might help to think about the following simple example, which shows that the Blumenthal law is using much more than just the continuity of Brownian motion. Let Z be a random variable with P(Z=1) = P(Z=-1) = 1/2, i.e. a coin flip. Set X_t = tZ, so that the process X_t just moves with constant speed \pm 1 depending ... 2 The increments are stationary. Since B_t - B_s is the increment over the interval [s, t], it is the same in distribution as the incremeent over the interval [s-s, t-s] = [0,t-s]. Hence,$$B_t-B_s \sim B_{t-s}-B_0.$$But B_0 = 0 almost surely, so that:$$B_t-B_s \sim B_{t-s}.$$Finally, B_{t-s} \sim \mathcal{N} (0,t-s). We didn't use the ... 1 No, we cannot apply the dominated convergence theorem in this way (see all the comments above). For fixed R>0 denote by$$\tau := \inf\{t>0; |B_t| \geq R\}$$the exit time from (-R,R). Moreover, we denote by$$H \bullet B(T) := (H \bullet B)(T) := \int_0^T H(s) \, dB_s$$the stochastic integral of H. By Markov's inequality and Itô's ... 1 Here is a somewhat silly example. Consider the trivial probability measure on the trivial \sigma-algebra \mathcal{X} = \{ \emptyset, \mathbb{R} \} on the reals. Then any (non-random) function f: [0,\infty) \to \mathbb{R} gives rise to a stochastic process A_t = f(t) with trivial finite-dimensional distributions. So there are continuous and ... 1 As \mathcal{F}_{t_n} \supseteq \mathcal{F}_{t+}, we know that X_n is independent of \mathcal{F}_{t+}. Therefore,$$(B_{t+s_1}-B_t, \ldots,B_{t+s_m}-B_{t}) = \lim_{n \to \infty} (B_{t_n+s}-B_{t_n},\ldots,B_{t_n+s_m}-B_{t_n}) = \lim_{n \to \infty} X_n$$is also independent of \mathcal{F}_{t+}. Lemma: Let \mathcal{F} be a \sigma-algebra and ... 1 If you compare your expression with$$ u(x,t) = E^Q\left[ \int_t^T e^{- \int_t^r V(X_\tau,\tau)\, d\tau}f(X_r,r)dr + e^{-\int_t^T V(X_\tau,\tau)\, d\tau}\psi(X_T) \Bigg| X_t=x \right] $$where dX = \mu(X,t)\,dt + \sigma(X,t)\,dW^Q (this is pasted from the wikipedia article ), we see that in your case the process X_t is simply B_t so \mu =0 and ... 1 In some sense, there is an x on the right-hand side because the equality$$\mathbb{E}_x(Y \circ \theta_s \mid \mathcal{F}_s^+)(\omega) = \mathbb{E}_{B_s(\omega)}(Y)$$holds only \mathbb{P}_x-almost surely. For a function f: \mathbb{R} \to \mathbb{R} the Markov property reads$$\mathbb{E}_x (f(B_{t+s}) \mid \mathcal{F}_s^+) = \mathbb{E}_{B_s} f(B_{t}), ... 1 Here we see that the new process is continuous and 0 at origin. We have$E[Y_t]=tE[W_{1/t}]= 0$and$ Y_t$is a Gaussian process also we have the covariance function$ E[Y_sY_t]= st(\dfrac{1}{s} \wedge \dfrac{1}{t}) = s \wedge t.$1 Lemma: Let$X_0,\ldots,X_n$be arbitrary random variables. Then $$\sigma(X_0,X_1-X_0,\ldots,X_n-X_{n-1}) = \sigma(X_0,X_1,\ldots,X_n).$$ Proof: The mapping$X_j-X_{j-1}$is clearly measurable with respect to$\sigma(X_0,\ldots,X_n)$(as difference of two measurable random variables) for all$j$. Consequently,$\sigma(X_0,X_1-X_0,\ldots,X_n-X_{n-1} ...

1

You have $$(1)=\mathbb E[1_{\{\tau_a\leq t\}}1_{\{X_{t-\tau_a}+W_{\tau_a}\leq a\}}] = \mathbb E[\mathbb E[1_{\{\tau_a\leq t\}}1_{\{X_{t-\tau_a}+W_{\tau_a}\leq a\}}|F_{\tau_a}]] = \mathbb E[1_{\{\tau_a\leq t\}}\Pr(X_{t-\tau_a}\leq 0]|F_{\tau_a})],$$ where you use that $W_{\tau_a}=a$ and measurability of $\tau_a$ wrt. $F_{\tau_a}$. Moreover, ...

1

To show independence you just need to show that $\Bbb{P} (B_t- B_s \in A_0, B_{r_1} \in A_1,B_{r_2} \in A_2\ldots B_{r_k} \in A_k) = \Bbb{P} (B_t- B_s \in A_0)\Bbb{P} ( B_{r_1} \in A_1,B_{r_2} \in A_2\ldots B_{r_k} \in A_k)$ for every $r_1, \ldots r_k \leq s$ and $A_1, \ldots, A_k \in \mathcal{B}$. This is because $\sigma(\{B_r, r \leq s\})$ is generated ...

1

We know the distribution of the vector $V:=(X(7)-X(5), X(5)-X(1), X(1))$, since the increments are independent. Defining the function $f\colon (x,y,z)\mapsto (x+y+z)(y+z)z$, the wanted expectation is $\mathbb E\left[f(V)\right]$.

1

Doob says: $B_{\tau \wedge t} = E[B_{t \wedge t}\mid \mathcal{F}_\tau]$. Jensen says: $$E[\lvert B_{\tau \wedge t} \rvert] = E[\lvert E[B_{t \wedge t}\mid \mathcal{F}_\tau]\rvert] \leq E[ E[\lvert B_{t \wedge t}\rvert\mid \mathcal{F}_\tau]] = E[\lvert B_t \rvert] < \infty$$

1

so we want to compute the Stratonovich $S$ integral, which is defined $$S:=\int_0^TW_t\circ dW_t:=\lim_{n\to\infty}\sum_{i=0}^{n-1}\frac{1}{2}(W_{t_{i+1}}+W_{t_{i}})(W_{t_{i+1}}-W_{t_{i}})$$ while we have a partition $P$ of $[0,T]$ with stepsize $h=T/n$, so $$t_0=0,t_1=h,\dots,t_k=kh,\dots,t_n=T$$ so as $n\to\infty$ we have $|P|\to 0$. Now let's ...

1

Fubini's theorem on $[0,t]\times\Omega$ gives $$\mathbb{E}\left[\int^{t}_{0} W^2_s\, ds\right] =\int_\Omega \int^t_0 W_s^2(\omega)\,ds\,\mathbb{P}(d\omega) =\int^t_0\int_\Omega W_s^2(\omega)\,\mathbb{P}(d\omega)\,ds = \int^{t}_{0} \mathbb{E}[W^2_s]\, ds.$$

1

To keep notation simple, we write $S_t$ instead of $S_t^i$, $\sigma_t^j$ instead of $\sigma_{t}^{ij}$ and so on. That's okay, because $i$ is a fixed number throughout this calculation. Suppose $(X_t)_{t \geq 0}$ is an Itô process of the form $$dX_t = b(t) \, dt + \eta(t) \, dW_t$$ where $\eta = (\eta_1,\ldots,\eta_m)$ and $(W_t)_{t \geq 0}$ is an ...

1

There is not one answer to this question. Your question is really one about modeling, rather than being strictly about mathematics, so the best answer depends on what you're trying to model. Two answers that come to mind for me are as follows. One would be to instead consider applying small iid normally distributed perturbations every $\Delta t$ and ...

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