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4

Hint. Observe that if you set $$ f(x)=\int_{-\infty}^{-x} e^{- \frac{1}{2}y^{2}} dy\quad \text{then} \quad f'(x)=- e^{- \frac{1}{2}x^{2}} $$ and the initial integral takes the form $$ \begin{align} \int_{0}^{+\infty} \int_{-\infty}^{-x} \frac{1}{2 \pi} e^{- \frac{1}{2}(x^{2}+y^{2})} dx dy&=-\frac{1}{2 \pi}\int_{0}^{+\infty}f'(x)\cdot f(x)\:dx\\\\ ...


2

The distribution of $(X,Y)$ is rotationally invariant, so the chance that it lies in the region $\{(x,y): x>0, x+y<0\}$ (shaded area below) is $1/8$.


2

Let $S(t)$ be governed by the SDE $$dS(t)=\mu S(t)dt+\sigma S(t)dW_t$$ Let $f(S)=\log(S)$. Heuristically, we can write $$\begin{align} d\log(S)&=\frac{\partial f(S)}{\partial t}\,(dt)+\frac{\partial f(S)}{\partial S}\,(dS)+\frac12\frac{\partial ^2f(S)}{\partial S^2}(dS)^2\\\\ &=\frac{\partial \log(S)}{\partial t}\,(dt)+\frac{\partial ...


2

Since $\phi$ is continuous with compact support, it suffices to show that for each $R$, $$\mathbb E\left[\int_{[0,R]}|B_t |\mathrm d\lambda(t)\right]$$ is finite. By Schwarz inequality, it suffices to check finiteness of $$I:=\mathbb E\left[\int_{[0,R]}B_t^2\mathrm d\lambda(t)\right].$$ Using Fubini-Tonnelli theorem (that is, switching integral when the ...


2

Hint: Use the summation by parts formula (also known as Abel's summation formula): $$\sum_{k=m}^n f_k (g_{k+1}-g_k) = (f_{n+1} g_{n+1}-f_m g_m) - \sum_{k=m}^n g_{k+1} (f_{k+1}-f_k).$$


2

The trouble is that there are two types of brackets, $\langle X \rangle_t$ (which I call angle bracket) and $[X]_t$ (sharp/square bracket). In this particular case they do not coincide because $(X_t)_{t \geq 0}$ is not a martingale (see also this answer). $[X]$ is the unique process such that $X^2-[X]$ is a martingale. $d(X_t^2) = 2X_t \, dX_t + ...


2

If $(X,Y)$ are jointly Gaussian with mean zero, then $P(X>0,Y>0)={\arccos(-\rho)\over 2\pi}$ where $\rho$ is the correlation of $X$ and $Y$. Therefore, for Brownian motion and $0<s<t$ $$P(B_s>0, B_t>0)={\arccos(-\sqrt{s/t})\over 2\pi}.$$


2

If we set $$\phi := \sum_{j=1}^n \alpha_j \phi_j \in \mathcal{D},$$ then by the linearity of the integral $$\alpha_1 \langle W,\phi_1 \rangle + \ldots + \alpha_n \langle W, \phi_n \rangle = \langle W,\phi \rangle.$$ Consequently, it suffices to show that $\langle W,\phi \rangle$ is Gaussian for all $\phi \in \mathcal{D}$. Fix $\phi \in \mathcal{D}$. ...


1

Here's another approach, based on an old result of M. Kac (If $X$ and $Y$ are independent random variables such that $X+Y$ is independent of $X-Y$, then $X$ and $Y$ are normally distributed with the same variance.). Let $B'$ be a second Brownian motion independent of $B$ (make a product space construction if necessary), and define $\langle ...


1

Let $S(t)$ be governed by the SDE $$dS(t)=\mu S(t)dt+\sigma S(t)dW_t$$ Let $f(S)=S^n$. Heuristically, we can write $$\begin{align} d(S^n)&=\frac{\partial f(S)}{\partial t}\,(dt)+\frac{\partial f(S)}{\partial S}\,(dS)+\frac12\frac{\partial ^2f(S)}{\partial S^2}(dS)^2\\\\ &=\frac{\partial S^n}{\partial t}\,(dt)+\frac{\partial S^n}{\partial ...


1

The first part is right. If you reduced it correctly, then recall that $W(t)\sim N(0,t)$. Then you could argue/recall that odd moments of a normal distribution are equal to zero, or you could integrate \begin{align*} E[\{W(t)\}^3]&= \int_{-\infty}^\infty x^3\cdot\frac{1}{\sqrt{2\pi t}}\exp\left\{-\frac{1}{2}\cdot\frac{x^2}{t}\right\}\,dx. \end{align*}


1

Yes, if $c$ is such that $\mathbb{E}[c^{W_{t+1}} \mid F_t] = c^{W_t}$ then it will also be true that $E[c^{W_{t+1}-W_t} \mid F_t] = 1$. This is simply the well-known property of conditional expectation that if $Y$ is measurable with respect to $\mathcal{G}$, then $Y E[X \mid \mathcal{G}] = E[XY \mid \mathcal{G}]$. If I am not mistaken, you will find that ...


1

Using summation by parts, as suggested, it actually becomes quite clear. $$\sum_{n=1}^{N}t_n(W(t_{n+1})-W(t_n))=t_N W(t_N)-t_1W(t_1)-\sum_{n=1}^N W(t_n)(t_{n+1}-t_n)$$ With $t_N=T$ and $t_1=0$. This gives: $$\int_0^TtdW(t)= TW(T)-\sum_{n=1}^N W(t_n)(t_{n+1}-t_n)$$ Taking $n \rightarrow \infty$ gives the Riemann integral. Thus the results ends up being: ...


1

This is a settled question but I like to add another solution :-) We have $W_t \mid \mathcal{F}_s \sim \mathcal{N}(W_s,t-s)$ hence using this we have $$\mathbb{E}(W_t^3 \mid \mathcal{F}_s)=W_s^3+3W_s(t-s).$$ Therefore \begin{align} \mathbb{E}(W_t^3 - 3t W_t \mid \mathcal{F}_s)&=\mathbb{E}(W_t^3 \mid \mathcal{F}_s)-3t\mathbb{E}( W_t \mid ...


1

Consider the function $f(x,t) = \frac {e^{r(T-t)}}{x}$. Then $X = f(t, S)$ and you can use Ito lemma to say that $$dX_t = \frac{\partial}{\partial t}f(t,x)\mid_{ t,S} dt + \frac{\partial}{\partial x}f(t,x)\mid_{ t,S} dS_t+\frac 12\frac{\partial^2}{\partial x^2}f(t,x)\mid_{ t,S} d \langle S \rangle _t$$ Which results in $$dX_t = \frac ...


1

The formula you use in your matlab program is different from the one you propose in the statments. What you want to calculate can be derived theoretically. Let the running maximum be $M_t = \sup_{s\in [0\ t]} W_t$ We know that pdf of $M_t$ is twice of that of $W_t$, but only defined on $[0 \ \infty)$. From this, you can calculate that $E(M_t) = ...


1

For the second part, let $ X_t^* = \max_{s\in [0\ t]}X_s$ note that: $P(\hat T \le t) = P(X_t^*\ge 1)$ and also, by reflection principle, you can show that the pdf of $X_t^*$ is twice the pdf of $X_t$, but defined on $[0\ \infty)$ only. Put these together, you get the cdf of $\hat T$ $P(\hat T \le t) = 2(1- \phi(1/{\sqrt{t}}))$ where $\phi$ is the ...


1

As the title of the question says, it's just a straightforward application of Ito's lemma: Since $S$ satisfies the given SDE, $(\log)'(x)=x^{-1}$, and $(\log)''(x)=-x^{-2}$, we have $$ \begin{split} d(\log (S_t))&= \frac{1}{S(t)}dS_t-\frac{1}{2}\frac{1}{S_t^2}\sigma^2S_t^2dt\\ &= \mu dt+\sigma dW_t-\frac{\sigma^2}{2}dt\\ ...



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