Skip to main content
9 votes

Are differentials on their own in stochastic calculus just an abuse of notation?

Differentials in stochastic calculus have very precise interpretations. The stochastic differential equation in differential form $$dX_t = \mu (t, X_t) dt + \sigma (t, X_t) dB_t$$ translates precisely ...
Jose Avilez's user avatar
  • 13.2k
5 votes
Accepted

Stochastic integration by parts of geometric brownian motion

You can't go any further than $\int e^{B_{t}+at}dt$. This is called the integrated Geometric Brownian motion. For references on its law see "The Integral of Geometric Brownian Motion" &...
Thomas Kojar's user avatar
  • 3,716
5 votes
Accepted

$M_t=\int_0^t e^{-3W_s}dW_s$ properties

Is $M_t$ well defined when $\mathbb EM_t<\infty$? Is this enough or do I need to check any other conditions? Any random variable is well-defined when its expectation is bounded Does finite ...
Thomas Kojar's user avatar
  • 3,716
5 votes
Accepted

What is the expectation of the positive part of a stochastic integral?

Being a square-integrable martingale, its expectation is zero as pointed out by other users. Going further, consider the function $f(x) = \frac{1}{2}\max\{0, x\}^2$. Although this is not a $C^2$-...
Sangchul Lee's user avatar
5 votes
Accepted

Is the stochastic integral process $\left(\int_{0}^{t}e^{-\lambda\left(t-s\right)}\mathrm{d}B_s\right)_{t\geq 0}$ a martingale?

The criterion you mentioned is certainly a sufficient condition, but not a necessary condition. So, let's go back to the basic. Let $M_t = \int_{0}^{t} e^{-\lambda (t-s)} \, \mathrm{d}B_s$. Then for $...
Sangchul Lee's user avatar
4 votes

What is the expectation of the positive part of a stochastic integral?

This is to expand a bit on my comment. There are different ways to see that the process $Y_t := \int_0^t W(s)\mathbf 1\{W(s)>0\}dW(s)$ is a martingale, but a very useful one is given by the ...
Stratos supports the strike's user avatar
4 votes
Accepted

Computing the quadratic covariation $\langle B,B^2\rangle_t$ of a Brownian motion and its square

See first this discussion on how to obtain the quadratic variation of two Itô processes. By Itô's formula we obtain that $B_{t}^{2}$ follows the dynamics $$ B_{t}^{2} = \int_{0}^{t}ds + \int_{0}^{t}...
minginator's user avatar
4 votes
Accepted

Is $\int_s^t f(W_u)\mathrm{d}u$ independent to $\mathcal{F}_s$ for all (Lebesgue) measurable $f$?

By integration-by-parts: $$ \textstyle\int_s^tW_u\,du=tW_t-sW_s-\int_s^tu\,dW_u\,. $$ The last integral is independent of ${\cal F}_s$ but the term $tW_t-sW_s$ is not. It is of the form $$ t(W_t-W_s)+(...
Kurt G.'s user avatar
  • 15.3k
3 votes
Accepted

How to compute the expectation of $e^{W_t}\int_0^t e^{-W_s} dW_s$

It is well-known and easy to see by Ito's lemma that $$ M_t=e^{W_t-t/2} $$ is a martingale that satsifes the SDE $$ dM_t=M_t\,dW_t\,. $$ Equivalently, $$ M_t=1+\int_0^tM_s\,dW_s\,. $$ The calculation ...
Kurt G.'s user avatar
  • 15.3k
3 votes

Definition of quadratic variation in Protter

It's not $H_0X_0Y_0$ that you would expect, but $(H\cdot X)_0Y_0$, which is $0$ as $(H\cdot X)_0=0$.
Will's user avatar
  • 7,142
3 votes
Accepted

Stochastic Integral Evaluation

Indeed as mentioned in comments we have $$\int_{z=0}^T \int_{s=0}^z dW(s) dW(z)=\int_{0}^{T}W_{s}dW_{s}$$ and so using Itô-formula we get $$=\frac{1}{2}(W_{T}^{2}-T).$$ Generally, multiple Itô ...
Thomas Kojar's user avatar
  • 3,716
3 votes

Proof that the stochastic exponential is a local martingale

In general, an Ito integral is a local martingale and it is a true martingale provided that the integrand is in $L^2$. This is precisely what the remark is telling you, as saying that the integrand is ...
Jose Avilez's user avatar
  • 13.2k
3 votes
Accepted

How can I find a stochastic process $A_t$ s.t. $\frac{1}{S_t}=\mathcal{E}(A)_t$?

Too long for a comment: You are overthinking it. First let me remark that the solution of the SDE $dS_t=\mu S_t\,dt+\sigma S_t\,dB_t$ is $$ S_t=S_0\,e^{\mu t-\frac{\sigma^2}{2}t+\sigma B_t} $$ which ...
Kurt G.'s user avatar
  • 15.3k
3 votes
Accepted

Missing term in Ito Integral

From scratch. You assume $$ V(y,t)=\frac12g\,y^2 $$ where it seems that $g$ is a function of $t\,.$ Then for $Y_t=\sigma B_t\sim N(0,\sigma^2t)$ we clearly have your boxed equation $$ \mathbb E[V(Y_T,...
Kurt G.'s user avatar
  • 15.3k
3 votes
Accepted

Stochastic integral of continuous integrand with bounded variation

By partial summation/integration it is possible to have the $s_i \in (t_i, t_i+1)$ for all $i$. Indeed we have $$ \sum_{i=0}^{n-1} X_{s_i}(Y_{t_i} - Y_{t_{i-1}}) = \sum_{i=0}^{n-1} X_{s_i} Y_{t_{i+1}} ...
Joris Bierkens's user avatar
3 votes
Accepted

Example of differential form of Ito process

Informally speaking, $\mathrm{d}W_t$ is $O(t^{1/2})$, so $(\mathrm{d}W_t)^2$ is $O(t)$ and should not be neglected when applying Taylor's formula to order $1$ to a stochastic process to express its ...
C. Falcon's user avatar
  • 19.1k
3 votes
Accepted

Canonical lift of a smooth path as a rough path

Surely if $X_t$ is nice and smooth, then $\alpha > \frac{1}{2}$, but then $2\alpha > 1$, and wouldn't this degenerate into saying that $\mathbb{X}$ is constant? From the known result on Holder ...
Thomas Kojar's user avatar
  • 3,716
3 votes
Accepted

Does the Itô integral rely on the martingale structure of the driving Brownian motion?

First, to be clear the Itô integral was specifically designed by Itô so that it preserves the martingale property: the process $$S_{t}:=\int^{t} f(s)dX_{s}\approx \sum f(s_{i})(X_{s_{i+1}}-X_{s_{i}})$$...
Thomas Kojar's user avatar
  • 3,716
3 votes
Accepted

Rigorous extension of Stochastic Integral for processes with $E[\int^{T_n}_0 X_t^2 dt ] < \infty$, $ \forall n \geq 0 $

Let $$ I_n(t,\omega) := \int^t_{0} G^n_sdB_s(\omega), \quad \forall n \geq0, \ t \geq 0, \ \omega \in \Omega, $$ Then we have for each $n$, $I_n$ has continuous sample paths and $\mathcal F$-adapter. ...
LNT's user avatar
  • 720
3 votes
Accepted

Dense subset of $L^2$ in the integral representation of $X \in L^2(\Omega)$

The set $\mathcal U$ is not dense in $L^2$ as it contains only non-negative random variables with unit expectation. However, this shows that $\text{span}(\mathcal U)$ is dense in $L^2$, or ...
user6247850's user avatar
  • 13.8k
3 votes
Accepted

The Explicit Solution to the Stochastic Logistic Equation

Define $$\begin{align} &Z_t =\frac{1}{X_t}\\ &M_t = \exp\left\{(rK-\frac12\alpha^2)t+\alpha B_t\right\} \end{align}$$ By using the Ito's lemma, we have: $$\begin{align} &\frac{dZ_t}{Z_t} =\...
NN2's user avatar
  • 16.3k
3 votes
Accepted

What is the solution to the SDE $X^x_t = B_t + \int_0^t \frac{x − X^x_s}{1-s} ds$

Setting $Z_t$ as what you did in your work, and then writing (2) in Differential form gives us that $$dX^x_t = xdt + dB_t - \frac{B_t}{(1-t)}dt + Z_tdt \tag{4}\label{eq4} $$ And (2) also gives us that ...
LocalNondeterminism's user avatar
3 votes
Accepted

Showing bounds of Stochastic Process

Consider the equation $$ \mathrm{d}Y_t = \biggl( 2 + 8 \frac{e^{Y_t} - 1}{e^{Y_t} + 1} \biggr) \, \mathrm{d}t + 4 \, \mathrm{d}B_t. \tag{1}\label{e:1} $$ The coefficients $\mu(y, t) = 2+\operatorname{...
Sangchul Lee's user avatar
3 votes
Accepted

Quadratic variation of the square of Brownian motion

• Yes, it's correct. • No. • The quadratic variation of $M$ will be the same as that of $B^2$, as these two processes differ by $t$. Write $$ (B_{t_{i}}^2 - B_{t_{i-1}}^2)^2 = (B_{t_{i}} + B_{t_{i-1}})...
John Dawkins's user avatar
  • 26.6k
2 votes
Accepted

Why is Itô integration not a "pathwise solution theory" to SDEs, but rough path theory is?

The solution of an Itö-SDE $X_{t}=y+\int bdt+\int \sigma dB_{t}$ is defined outside a set of measure zero that depends on the whole equation i.e. $\Omega_{y,b,\sigma}$ with $P[\Omega_{y,b,\sigma}]=0$. ...
Thomas Kojar's user avatar
  • 3,716
2 votes

Stochastic integral (integer powers of white noise)

There is a nice way to interpret such stochastic integrals by using scale functions and time changes of Brownian Motion. The idea is as follows: Suppose we have some "usual SDE", i.e. one ...
Small Deviation's user avatar
2 votes

predictable $\Rightarrow$ left continous? progressive $\Rightarrow$ adapted and left continous?

I don't think that a predictable process is always left-continuous. My reasoning depends on the following observation: Let $q_n$ be a sequence of increasing numbers with limit $q$ and such that $q_n&...
P.Jo's user avatar
  • 839
2 votes

Covariance of Stochastic Differential Equation

Here we simply apply Ito isometry. So as explained Quadratic Variations and the Ito Isometry we have $$ EX_{t}X_{r}=E[\int^t b ds\int^r b ds]+E[\int^t b ds \int^r \sigma dB_{s}]+E[\int^r b ds \int^t \...
Thomas Kojar's user avatar
  • 3,716
2 votes

Apply Ito to $exp((\mu-0.5*\sigma^2)t+\sigma*W_t)$ for Brownian Motion $W_t$

The problem here is how you have written Ito's formula, the way you have it written is actually incorrect if we are considering a general stochastic process $X_t$, the $\sigma^2$ term appears from the ...
Emmet's user avatar
  • 495
2 votes

Differentiating Stochastic Integrals (Ito Integrals)

$X_{t}$ is not in general differentiable in the sense of ordinary calculus. The notation $$ dX_{t}=B_{t}^{2}dB_{t} $$ is equivalent to $$ X_{t_{1}}-X_{t_{0}}=\int_{t_{0}}^{t_{1}}B_{t}^{2}dB_{t}\qquad\...
parsiad's user avatar
  • 25.2k

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