10

Itô's Lemma The standard version of Itô's Lemma applies to a single Itô process $\text{d}X_t=\mu(t,X_t)\mathrm{d}t+\sigma(t,X_t)\mathrm dW_t$. Then, $$\mathrm{d}f(t,X_t) = \left(f_t+\mu(t,X_t)f_x + \frac{1}{2}\sigma(t,X_t)^2f_{xx}\right)\mathrm{d}t+\sigma(t,X_t)f_x\mathrm dW_t.$$ Let $\text{d}Y_t=m(t,Y_t)\mathrm{d}t+s(t,Y_t)\mathrm dW_t^{(2)}$ be a second ...


7

What a great question! I've had a go at it below, I'd say I'm about 75% sure of the result I've got to but I'd love feedback from others. I'm going to use the definition of the Ito integral, \begin{align} \int^t_0 \sigma_s dW_s = \lim_{n \to \infty} \sum_{i=1}^n \sigma_{t_{i-1}} \bigl( W_{t_i} - W_{t_{i-1}} \bigr) \end{align} where $t_n = t$. Then, using the ...


7

This is impossible unless you are very intelligent with good memory-retention skills and already mathemathically proficient in the field of analysis and statistics (and no, a single course in basic probability theory does not suffice). And even if you are, two weeks is an extremely short amount of time. However, assuming these criteria are satisfied, I would ...


7

Hints: First show (using Ito Lemma) that $$ \exp(iuX_t) = 1 + iu \int_0^t\exp(iuX_s) h(s) d W_s -2^{-1}u^2 \int_0^t\exp(iuX_s) h(s)^2ds$$ Then show (by taking expectations): $$ E[\exp(iuX_t)] = 1 - 2^{-1}u^2 \int_0^tE[\exp(iuX_s)] h(s)^2ds $$ Finally note that this is an ODE in unknown variable $x(t):=E[\exp(iuX_t)]$, $x(0)=1$: $$ x'(t) =- 2^{-1}u^2h(t)^2 ...


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 ...


6

No. Itō’s formula helps you derive the dynamics of $f (S_\cdot )$ given the SDE followed by $S$. Here this is not the case. You simply have: $$ \mathrm{d} \left[\int{g(S_t)\mathrm{d}S_t}\right] = g(S_t) dS_t $$


5

If $M$ and $N$ are independent (your references appear to make this assumption), then $M+N$ is also a Poisson process. So, using the polarization identity: $$ dMdN = 2^{-1}\left[(d(M+N))^2 - (dM)^2 - (dN)^2\right] $$ $$ = 2^{-1}\left[d(M+N) - dM - dN \right] = 0 $$ (A proof of $(dX)^2 = dX$ for a Poisson process $X$ is available here.)


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 = ...


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-...


4

The integral of the LHS is just $$\int_0^T d\left(\left(T - t\right)W_t\right) = \left[\left(T - t\right)W_t\right]_0^T = \left(T - T\right)W_T - \left(T - 0\right)W_0 = 0$$. You need $W_0 = 0$ to obtain the last equality.


4

Just wanted to add to @StackG's great answer using a different approach. Please, double-check my solution as well because I'm not 100% sure. Let $\sigma_t$ be sufficiently regular such that $\dot{\sigma}_t \stackrel{def}{=}\frac{d \sigma}{dt}$ is well defined. Then, Ito's lemma: $$ d(\sigma_t W_t) = \dot{\sigma}_t W_t dt + \sigma_t dW_t $$ which in integral ...


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, \...


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

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[ \...


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

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

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

Ito's lemma is for twice differentiable functions of the form $f(t,W(t))$. You speak of $W(t)dW(t)$ - this is informal notation and doesn't have a mathematical meaning. Although once you put the integral sign, it becomes mathematically precise. So there's nothing known as $W(t)dW(t)$, but $I(t)=∫_0^{t}W(u)dW(u)$ is well defined via the definition of an Ito ...


3

I'll give it a try, but am not yet 100% sure that it's the way to go. Ansatz: Let's find the distribution of the integral of a Brownian motion with respect to time (call it $x$) and find the expectation of the product of two such integrals $x$ and $y$. Then, calculate the covariance as $Cov(x,y)=E(xy)-E(x)E(y)$. 1. Distribution of $I(t,T)\equiv\int\limits_{s=...


2

Hints: Show first that $$E[((W^1_t + W^2_t)-(W^1_s + W^2_s))^2] = (2+2\rho)(t-s) $$ Then conclude that $$ [(2+2\rho)^{-1/2} (W^1 + W^2)]_t =t $$ On the other hand, show (using bilinearity of quadratic covariation) that $$ [(2+2\rho)^{-1/2} (W^1 + W^2)]_t = [(2+2\rho)^{-1/2} (W^1 + W^2), (2+2\rho)^{-1/2} (W^1 + W^2) ]_t $$ $$ = (1+\rho)^{-1} (t+ [W^1, W^2]_t)...


2

Note that SDE (4) does have a "closed-form" representation. Let $X$ be $$X := S^p, $$ so (4) is a geometric Brownian motion SDE $$dX = (p\alpha + 2^{-1}p(p-1) \sigma^2) X dt + p \sigma X dW, $$ which, again due to Ito Lemma, is equivalent to $$ d \ln X = (p\alpha + 2^{-1}p(p-1) \sigma^2 - 2^{-1}p^2 \sigma^2) dt + p \sigma dW $$ or $$ d \ln X = ...


2

There are two questions that you are asking: How to prove $\text{Var}[I] = \frac{1}{3}$? and What is $\text{Cov}[I, W_1]$? For Question 1, you absolutely can use Ito isometry. First, note that we can use integration by parts to obtain the formula: \begin{align} \int_0^1 W_t dt &= W_1 - \int_0^1tdW_t \\ &= \int_0^1(1-t)dW_t \end{align} So we can ...


2

For the first question, equality $$\mathbb{E}\left[\int_{[0,1]\times[0,1]} W_sW_tdsdt\right] = \int_{[0,1]\times[0,1]}\mathbb{E}[W_sW_t]dsdt \left(= \int_{[0,1]\times[0,1]}\min(s,t)dsdt\right) $$ is due to commuting expectation and integral (not to Ito isometry), which in turn is allowed by Fubini's theorem condition being met: $$ \left(\int_{[0,1]\times[0,1]...


2

With integral definitions of $r_t$ and $R_t$, we do have: $$ r_t := \int_0^t d(\ln S_u) = \ln S_t - \ln S_0 \color{green}= \ln \left( \frac{S_t}{S_0}\right),$$ but: $$ R_t := \int_0^t \frac{dS_u}{S_u} \color{red}{\not=} \frac{S_t - S_0}{S_0} \color{green}= {\rm e}^{r_t} -1. $$ In symbolic differentials language: $$ \frac{dS_u}{S_u} \color{red}{\not=} \frac{...


1

The basic Ito formula for a Poisson process is $$ dY_t = \mu_t dt + g_t dN_t $$ $$ df(Y_t) = \mu_t f'(Y_t) dt + (f(Y_{t-}+g_t) - f(Y_{t-}))dN_t $$ (dropped $f$'s direct dependence on the time variable to avoid the partial derivative clutter). Case $\mu_t = -\lambda g_t$ (this is your original case): $$ df(Y_t) = -\lambda g_t f'(Y_t) dt+ (f(Y_{t-}+g_t) - f(...


1

That no drift is a martingale: That ito integrals are martingales requires a simple but algebraically cubersome proof. You can refer to Shreve (continuous time) for the proof. You can also intuitively observe it as Brownian increments that are multiplied with their respective integrands are allocated independently of the integrand value. Thus, when all terms ...


1

I don't see where in the book Shreve justifies (3) directly from (1). But he would have almost surely referred to the discretized versions of the "differentials". That is: $$ X(t_{i+1})-X(t_{i}) \approx \sigma(t_i)(W(t_{i+1})-W(t_{i}))+\alpha(t_{i})(t_{i+1}-t_{i}) $$ implies $$ S(t_{i})(X(t_{i+1})-X(t_{i})) \approx S(t_{i})\sigma(t_i)(W(t_{i+1})-W(...


1

I personally find it hard to work with short-hand notation in Stochastic calculus, so for my own record, and for anyone who has similar issues with the short-hand notation, I add the following answer. On the wikipedia link provided, they start the argument you are referring to by stating: ...note that: $$d\left[(T-t)W_t\right]=(T-t)dW_t-W_tdt$$ How I like ...


1

Heuristically but highly non-rigorously speaking, in real-world we have continuous processes that is random (or unpredictable). We, therefore, want to develop a process to develop a way to describe it. The bread and butter of continuous random variable is a normal distribution. It is only natural we try to frame a random process based on normal distribution. ...


Only top voted, non community-wiki answers of a minimum length are eligible