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4

Hint: The (homogeneous) Poisson process has stationary independent increments, so write: $$N_t=N_s+(N_t-N_s)=N_s+N_{t-s}$$ and use that $E[N_s \mid F_s]=N_s$, since $N_s$ is known at time $s$ $E[N_{t-s}\mid F_s]=E[N_{t-s}]$, since $N_{t-s}$ is independent from what happened up to time $s$ $N_{t-s} \sim$ Poisson$(λ(t-s))$, in order to calculate ...


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

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 $0<s<t$ $$P(B_s>0, B_t>0)={\arccos(-\sqrt{s/t})\over 2\pi}.$$


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

In general, if $A,B:(\Omega,\mathcal{F},P)\to\mathbb{R}$ are independent random variables, then $A^2$ and $B^2$ also: Like $f(x)=x^2$ is Borel measurable, then $A^2$ is $\sigma(A)$-measurable. Then $\sigma(A^2)\subseteq\sigma(A)$. By the same way, $\sigma(B^2)\subseteq\sigma(B)$. So, If $C\in\sigma(A^2)$ and $D\in\sigma(B^2)$, then $C\in\sigma(A)$ and ...


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

Try Philip E. Protter's "Stochastic Integration and Differential Equations"


1

Since you mentioned only continuous martingales, note that it is also possible to define stochastic calculus using local martingales and semimartingales. Some textbooks which cover the more general theory are: Limit Theorems for Stochastic Processes by Jacod & Shiryaev Stochastic Integration and Differential Equations by Protter Stochastic Integration ...


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

If you can directly apply the result of Binomial distribution, you may try the following: Let $M_n = \sum_{i=1}^n X_n$ where $X_i$ represent each individual step with support $\Pr\{X_i = 1\} = p, \Pr\{X_i = -1\} = 1 - p$. Note $$ X_i \stackrel {d} {=} 2B_i - 1, i = 1, 2, \ldots, n$$ where $B_i \sim \text{Bernoulli}(p)$. Therefore summing them up $$ M_n ...


1

The minimax argument is the following. For any $μ\ge 0$ $$E\left[{M^*_t-μ\langle M\rangle_L^{r/2}}\right]+μE[\langle M\rangle]_L^{r/2}\le\inf_{μ>0}{\left\{E\left[\sup_{t\ge0}{(M^*_t-μ\langle M\rangle_L^{r/2})}\right]+μE[\langle M\rangle]_L^{r/2}\right\}}$$ The expression on the LHS is bounded by a min-max (inf-sup) expression on the RHS. Specifically, ...


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\\ ...


1

For your first question, you should not mix $df(t, T)|_{T=t}$ with $df(t, t)$. For $df(t, T)|_{T=t}$, there is no differential to the second argument, that is, when taking the differential, the second argument is ignored, while setting $T$ to $t$ after the differential is done. However, for $df(t, t)$, you need to consider the differential for both the ...


1

Yes, it is possible to find a common localizing sequence. By the very definition, there exist localizing stopping times $(\sigma_n)_{n \in \mathbb{N}}$ and $(\tau_n)_{n \in \mathbb{N}}$ for $M$ and $H$, respectively. If we set $$\varrho_n := \min\{\tau_n, \sigma_n\}$$ then $\varrho_n$ is a stopping time satisfying $\varrho_n \uparrow \infty$. Since ...


1

Since $$X_s^2 = (X_{s-}+\Delta X_s)^2 = X_{s-}^2+ 2 X_{s-} \Delta X_s + (\Delta X_s)^2$$ we have $$\Delta (X_s^2) \stackrel{\text{def}}{=} X_s^2-X_{s-}^2 = 2 X_{s-} \Delta X_s + (\Delta X_s)^2.$$ Hence, $$\Delta (X_s^2) - 2 X_{s-} \Delta X_s - (\Delta X_s)^2 = 0$$ for all $s$. This means that the sum $$\sum_{s \leq t} \left( \Delta (X_s^2) - 2 X_{s-} ...



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