11

Besides @StackG's splendid answer, I would like to offer an answer that is based on the notion that the multivariate Brownian motion is of course multivariate normally distributed, and on its moment generating function. We know that $$ \mathbb{E}\left(W_{i,t}W_{j,t}\right)=\rho_{i,j}t $$ i.e. an $N$-dimensional vector $X$ of correlated Brownian motions has ...


11

You need to rotate them so we can find some orthogonal axes. A simple way to think about this is by remembering that we can decompose the second of two brownian motions into a sum of the first brownian and an independent component, using the expression \begin{align} W_{t,2} = \rho_{12} W_{t,1} + \sqrt{1-\rho_{12}^2} \tilde{W}_{t,2} \end{align} where $\tilde{...


11

By construction, the Itô integral, $I_t=\int_0^t X_s\text{d}W_s$, is a martingale if $\int_0^t \mathbb{E}[X_s^2]\text{d}s<\infty$. The martingale property, $\mathbb{E}_s[I_t]=I_s$ implies $\mathbb{E}[I_t]=I_0=0$. Because $W_s\overset{d}{=}\sqrt{s}Z$, where $Z\sim N(0,1)$, we indeed have \begin{align*} \int_0^t\mathbb{E}\left[\frac{1}{(1+W_s^2)^2}\right]\...


6

It is covered very nicely in Iain Clark's Foreign Exchange Option Pricing, A Practitioner’s Guide (pages 98-104). The book also contains references to the relevant literature including Feller's original paper.


6

$$ d Y \left(t\right) := d \left[\int_0^t{a \left(s\right)\mathrm{d}W_s}\right] = a \left(t\right) dW_t $$ Note that since $Y$ is a driftless process, it is a local martingale, and because $a$ is bounded, a true martingale. Its quadratic variation is given by $$ \langle Y \left(\cdot\right)\rangle_t = \int_0^t{a^2 \left(s\right)\mathrm{d}s} $$ by definition ...


5

For part 1 of your question, the short answer is no, calculating conditional density is a looong way of doing it. Possible but not the easiest. Here is the sketch for a shorter version. We note that $(X_{T/2},X_{T})$ is a jointly Gaussian vector with mean $\mu = (X_0 + aT/2,X_0 + aT)$, and the variance-covariance matrix $$ \begin{pmatrix} b^2 T/2 & b^2 T/...


5

An alternative way is using the Stratonovich integral. By definition, we have $$\int_0^t X_s \, \circ dW_s = \lim_{n\rightarrow \infty} \sum_{i=1}^n \frac{X_{t_i} +X_{t_{i-1}}}{2}\left( W_{t_i} -W_{t_{i-1}}\right) \; \; (1)$$ One can then show that for a deterministic smooth functions $f$ and $g$ we have: $$ \int_0^t g'(W_s)\, \circ dW_s = g(W_t)- g(W_0)\; \;...


5

Integrating $W_t$ Consider the partition $t_i=it/n$ with $t_0=0$ and $t_n=t$. Then, by definition, \begin{align*} \int_0^t W_s\text{d}W_s &= \lim_{n\to\infty} \sum_{i=0}^{n-1} W_{t_i}\left(W_{t_{i+1}}-W_{t_i}\right). \end{align*} You can do the limit by using the identity $$ W_{t_i}\left(W_{t_{i+1}}-W_{t_i}\right)=\frac{1}{2}\left(W_{t_{i+1}}^2-W_{t_i}^2-...


5

As shown in Credit Risk Modeling Notes (Bielecki, Jeanblanc, Rutkowski), Corollary 1.3.1, for $t < s$, we have: $$ P(\tau \leq s | {\cal F}_t) = N\left( -Y_t \sigma^{-1}(s-t)^{-1/2}- \nu(s-t)^{1/2}\right ) + {\rm e}^{-2\nu \sigma^{-2}Y_t} N\left( -Y_t \sigma^{-1}(s-t)^{-1/2}+ \nu(s-t)^{1/2}\right ),$$ where $$ Y_t = y_0+ \nu t +\sigma W_t, \: \sigma >0,...


5

As stated here, for $f = f(t, x) ∈ C^{1,2}(\mathbb{R}^2)$ a deterministic function and Ito process $$X_t = W_t^2,$$ the stochastic process $$Y_t = f(t,X_t)$$ is an Ito process and we have $$df (t,X_t) = \partial_tf(t,X_t)\,dt + \partial_xf(t,X_t)\,dX_t + \frac{1}{2} \partial_{xx}^2f(t,X_t)(dX_t)^2. $$ Since $$ dX_t = 2W_t dW_t + dt $$ and $$ (dX_t)^2 = ...


4

Alternatively, we can use Ito isometry ($X$'s integrability and adaptability are assured by $a$'s boundness and adaptability, respectively): $$E[X_t|{\cal F}_s] = E[X_s\big|{\cal F}_s] + E\left[\left(\int_s^t a_udW_u\right)^2 - \int_s^t a_u^2du \big|{\cal F}_s \right] $$ $$ = X_s + E\left[\left(\int_s^t a_udW_u\right)^2\big|{\cal F}_s\right] - E \left[ \...


4

With your SDE for $F$, I get: $$ dXdF = -a^2XFdt $$ $$FdX = rFdt + aXFdW $$ $$XdF = a^2XF dt -aXF dW$$ So, adding up: $$ d(XF) = rF dt, $$ giving $$ X_t = F^{-1}_t X_0 + rF^{-1}_t \int_0^t F_u du $$


4

In all honesty, Quadratic Variation for Stochastic Processes is an advanced topic, and computing it rigorously from first principles is a graduate-level probability question. Part 1: Quadratic Variation: Informal "proof" First, how is Quadratic Variation Defined? For a stochastic process $X_t$, the quadratic variation, denoted $<X_t>$, is ...


4

Since $\text{Cov}(X, Y) = E(XY) - EX EY$, we have \begin{align} \text{Cov}(tB_{3t} - B_{2t} + 5, B_s - 1) &= E[tB_{3t}B_s - tB_{3t} - B_{2t}B_s + B_{2t} + 5B_s - 5] - (5)(-1) \\ &= tE[B_{3t}B_s] - E[B_{2t}B_s] \\ &= ts - s \end{align} where the first equaltiy is just mutliplying out the product, the second equality comes from discarding zero ...


4

Answering the title question: Let $f(t,W_t)=W_t^2-t$, then it is easier to derive the dynamics using the "general formula" for Itô's lemma (reference, see eq. 10): $$df(t,W_t)=\frac{\partial f}{\partial t} dt + \frac{\partial f}{\partial W_t} dW_t + \frac{1}{2}\frac{\partial^2f}{\partial W_t^2} dW_t^2$$ where, $$\frac{\partial f}{\partial t} = -1, \...


3

That looks correct, but a bit complicated. We know that under Black-Scholes with no dividends, $E^Q(S_t) = Forward = S_0 e^{rt}$ $e^{-rT}E^Q(\int_0^TS_tdt) = e^{-rT}\int_0^TE^Q(S_t)dt \\ = e^{-rT}\int_0^T S_0 e^{rt} dt = S_0 e^{-rT}\int_0^T e^{rt} dt \\ = S_0 e^{-rT} \frac{1}{r}(e^{rT} - 1) = S_0\frac{1-e^{-rT}}{r}$ It is straightforward to generalize to the ...


3

Write $X_t = W^2_t$, then you are trying to find $\int_0^t dX_t = X_t - X_0 $. Now use Ito's Lemma to find the dynamics of $W^2_t$ and try to solve the SDE.


3

Echoing some of the comments to the OP above, the only real difference between random walks and Brownian motions is a question of time frequency. IE a Brownian motion is just an aggregation of a (binary) random walk with higher frequency. Given both will always be at best an approximation of reality, asking for which is "better" becomes a bit of a ...


3

I believe your SDE has an unintended error. It should be: $$ dr_t = a \cdot (b - r_t) \cdot dt + \sigma \cdot \sqrt{r_t} \cdot dB_t. $$ On the other hand, the Feller condition is discussed and explained in Section 10.2.1.2 (pg. 432) of the Andersen and Piterbarg book: Interest Rate Modeling. Hope it helps!


3

Our market has a tradeable asset $S$ and a risk-less money market account $B$, that is, the numéraire of the risk-neutral measure. We assume the following standard conditions, which are widely applicable to most common models: We work in an Itô diffusion setting, and neglect jump modelling: $$\begin{align} & dS_t=\mu(t,S_t)dt+\sigma(t,S_t)dW^S_t \\ &...


3

That the expectation is zero is often called the martingale property of Ito integral (see e.g. Oksendal Theorem 3.2.1). The formal proof consists of showing this for "simple" integrand functions and then generalising this by taking limits. This requires that the integrand process is adapted (i.e. not forward looking) and square integrable. Square ...


3

Providing only a sketch here, using Ito integral definition (and commuting limit, summations and expectation), the result boils down to studying the expectation term: $$ E\left[ f_{t_{i-1}} (W_{t_i}-W_{t_{i-1}}) \cdot g_{t_{j-1}} (W_{t_j} - W_{t_{j-1}}) \right]. $$ If the intervals don't overlap, $i\not= j$, and say $t_i \leq t_{j-1}$, then $f_{t_{i-1}} g_{...


3

(As said in the comments, you need to put down some of your thoughts regarding the question too, like specifying the tools/theorems you would use or actual attempts to apply them, even if you can only cover early steps, not just the question itself.) Hints: We are given $$X_t^\theta:=\exp \left(\theta W_t-\frac{1}{2} \theta^{2} t\right)=\sum_{n=0}^{\infty} \...


3

Since the process $e^{-\mu t}$ is continuously differntiable, then it has finite variation. Thus, Ito's lemma essentially implies the 'normal' product rule: \begin{align} dU_t &= d(S_te^{-\mu t}) \\ &= e^{-\mu t}dS_t + S_td(e^{-\mu t}) \\ &= e^{-\mu t}\mu S_t dt + e^{-\mu t}\sigma S_t dw_t - \mu e^{-\mu t} S_t dt \\ &= e^{-\mu t}\sigma S_t ...


3

We need to show that $$\lim_{t\to1^-} (1-t)\int_0^t\frac1{1-s}dW_s \stackrel{\text{a.s.}}= 0.$$ $(M_t)_{t<1}=\Big(\int_0^t\frac1{1-s}dW_s\Big)_{t<1}$ is a martingale, so we can use Dambis-Dubins-Schwarz and say that $M_t = B_{\langle M\rangle_t}$ for a Brownian motion $B$ (with a different filtration of course). However, $\langle M\rangle_t = \int_0^t \...


3

Hints: Use Ito theorem for $\ln Z_t$ to get: $$dZ_t = -\theta_t Z_t dW_t $$ Then use the product rule to compute $d(U_tZ_t)$. I introduced $U_t:=V_t B_t^{-1}$ to keep things cleaner. $$dU_t = h'_t B_t^{-1}\sigma_t dW_t + h'_t B_t^{-1}\sigma_t \theta_t dt $$ $$ dU_t\cdot dZ_t = -h'_t B_t^{-1}\sigma_t \theta_t Z_t dt $$ $$ d(U_tZ_t) = U_tdZ_t + Z_tdU_t +dU_t\...


3

$$ X_t = B_t 1_{t<0.5} + (x+ B_t) 1_{t\geq 0.5} = B_t + x1_{t\geq 0.5}$$ $$ [X, X]_t = [B, B]_t + x^2 1_{t\geq 0.5} = t+ x^2 1_{t\geq 0.5}$$ (the author probably intended to use $0.5$ as jump size too)


2

An Ito integral is a martingale, and thus its expectation at anytime is it's value at t=0 - which is trivially 0; because the lower and upper limit of the integral would be 0. For proof of martingality, you can refer to Shreve. It uses the definition of the ito integral by looking at it as the sum of many random variables generated from slicing the time axis....


2

A couple of things are required to make this work, the two key points are: The Ito Integral is a Martingale only when the integrand is not forward-looking ie. when we DEFINE the summation to be this: \begin{align} \int^t_0 W_t dW_t = \sum^N_{i=1} W_{i-1}\bigl( W_i - W_{i-1}\bigr) \end{align} As pointed out in the comments, this wouldn't matter in the ...


2

That is because of sigma_tilde_squared == 0, You could add 0.01 at the add to avoid it == 0


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