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0

Are you talking about something like this? $$dx(t)=\ldots\ dt+[x(t)]^\gamma\ dW(t)$$ If $\gamma$ is zero then you've got BM, if it's one you get GBM, inbetween you have a 'mix'.


0

Idea Let $B$ be a standard brownian motion starting from $x_0=0$, $m_T = \inf_{u\leq T}B_u$ and $M_T =\sup_{u\leq T}B_u$. Let's define if it exists for $A\in\sigma(B_u,u\leq T)$, $\mathbb{P}(A | B_T=x_T)\stackrel{\rm def}{=}\lim_{\varepsilon\to 0}\mathbb{P}(A|B_T\in(x_T-\varepsilon,x_T+\varepsilon))$ $$\begin{split} \mathbb{P}(\tau_U\leq T \cap \tau_U\leq ...


2

The classical and naïve procedure for generating Poisson Hypersphere samples is by acceptance rejection, which has complexity over $O(N^2)$ and is thus unfeasible for most practical usage with on-the-fly generation. This cost could be improved by space partitioning techniques at low dimensions, but at high ones afaik they become useless again with uniform ...


2

Besides the code's problem, I highly recommend the Brownian Bridge correction method which can compensate the pricing error resulting from discretization of the continuous path.


1

There are many things wrong with your code. I'll leave aside the manner in which it is implemented, but note that it is: (1) not Matlab friendly with all the for loops (you should vectorise), (2) the fact that you have splitted the case j==0 in the main loop is a poor coding practice. for i=1:n I=1; for j = 0:(m-1); Z(j+1)= randn (1 ,1); dW=sqrt (T/...


8

1. This integral is not Ito's Integral. Indeed $Y_t$ is a random time change with time change rate $\frac{W_t}{1+W_t^2}.$ (Oksendal, Sixth edition,page 147) 2. Sometimes this trick is useful.Indeed we assume that we are going to solve Riemann integral !. Let $$f''(x)=\frac{-2x}{(1+x^2)^2}$$ then $$f'(x)=\left(\frac{1}{1+x^2}\right)+c_1$$ and $$f(x)=\...


5

IMHO the problem isn't stated correctly indeed, in the sense that the Radon-Nikodym derivative provided as the "solution" is not the unique way to define a measure $\mathbb{Q}$ equivalent to $\mathbb{P}$ and under which $X_t$ is a martingale. Just take $$\frac {d\mathbb{Q}}{d\mathbb{P}} =\mathcal{E}\left(-\int_0^t \cos(s) dW_s + a\right)$$ for any $a \in \...


6

Let $$Y_t = \int_0^t N_u du$$ where $(N_t)_{t \geq 0}$ figures a Poisson process with intensity $\lambda$. Using the stochastic Fubini theorem we have that: \begin{align} Y_T &= \int_0^T N_t dt \\ &= \int_0^T \int_0^t dN_u dt \\ &\color{lightgray}{= \int_0^T \int_0^T \mathbf{1}\{u \in [0,t]\} dN_u\ dt} \\ &\color{lightgray}{= \int_0^T \...


2

\begin{align*} \int_0^T W(t)\, dt &{}= \int_0^T\!\!\int_0^t dW(u)\,dt \\ &{}= \int_0^T\!\!\int_u^T dt\, dW(u) \\&{}= \int_0^T (T - u)\,dW(u) \\&{}= TW(T) - \int_0^T u\, dW(u) \end{align*} however i am not sure if it is what you are asking for


0

Hint Let $\,H_0(x,t)=1$ , $H_1(x,t)=x$ and for every $n\ge 2$ set $${{H}_{n}}(x,t)=x {{H}_{n -1}}(x,t)-(n-1)\,t\,{{H}_{n-2}}(x,t)$$ then ${{H}_{n }}(W_t ,t)$ is a Martingale. For exapmple $$H_1(W_t,t)=W_t$$ $$\qquad H_2(W_t,t)=W_t^2-t$$ $$\qquad\qquad H_3(W_t,t)=W_t^3-3tW_t$$ $$\vdots $$


1

First, let us formulate the problem mathematically: A symmetric random walk starts at 0 and moves up or down one unit (with equal probability) every 1 second. The are two absorbing barriers located at H and -L, with $H,L>0$. Given infinite time, what is the probability $p_H$ that H will be hit before -L is hit and what is the probability $p_L$ that -L ...



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