9
votes
Calculate value of Integral of Wiener process $\int_{0}^t e^{\lambda u } dZ_u$
This is just a Wiener integral (stochastic integral with respect to a Brownian motion and deterministic integrand), hence a centered Gaussian random variable with variance
$$
\int_0^t{e^{2\lambda u}\...
- 2,570
8
votes
Accepted
Integral of Wiener process w.r.t. time
@Ivan's comment regarding the covariances is the key.
Consider an equally spaced partition $\Pi_n = \left\{ t_0 = 0, t_1 = \Delta_n, \ldots, t_n = t \right\}$ of the interval $[0, t]$, where $t_i = i ...
- 5,959
7
votes
Accepted
conditional expectation of stochastic integral
What a great question! I've had a go at it below, I'd say I'm about 75% sure of the result I've got to but I'd love feedback from others.
I'm going to use the definition of the Ito integral,
\begin{...
- 2,996
7
votes
Accepted
Probability Density Function of a Wiener Process Minimum
Firstly, $m_T=\min\limits_{t\in[0,T]} B_t = -\max\limits_{t\in[0,T]} -B_t \overset{Law}{=} -\max\limits_{t\in[0,T]} B_t = -M_T$. So, you can either consider the running maximum or minimum.
Let $\tau$ ...
- 15.2k
7
votes
Accepted
Proof that $\exp(aW(t)-0.5a^2t)$ is a martingale
Let $(W_t)$ be a standard Brownian motion and $a>0$. We define $X_t=e^{aW_t-\frac{1}{2}a^2t}$. Then, the process $(X_t)$ is adapted and integrable which are the first two conditions of being a ...
- 15.2k
4
votes
Integral of brownian motion wrt. time over [t;T]
The last integral is correct as
$$\int_t^T W_s ds = \int_t^T (T-s) dW_s \sim N\left(0, \int_t^T(T-s)^2ds\right) = N\left(0,\frac{1}{3}(T-t)^3\right).$$
Ref. Arbitrage Theory in Continuos Time (Björk, ...
- 548
4
votes
conditional expectation of stochastic integral
Just wanted to add to @StackG's great answer using a different approach. Please, double-check my solution as well because I'm not 100% sure.
Let $\sigma_t$ be sufficiently regular such that $\dot{\...
- 700
4
votes
Accepted
Can the differential operator be removed to get the mean/variance of an Ito process?
This is wrong! Notice that $dX_t=\mu(t,X_t)dt + \sigma(t,X_t)dW$ is a shorthand for
$$\int_0^tdX_s = \int_0^t \mu(s,X_s)ds + \int_0^t\sigma(s,X_s)dW_s$$
Integrating:
$$X_t-X_0 = \int_0^t \mu(s,X_s)...
- 1,886
4
votes
Integral of the OU (Ornstein Uhlenbeck) process conditioned on hitting a threshold value for the first time
I am curious about the target of your question. It is rarely put as
$$
\mathbb{E}\left[\int_0^{\tau}X_t{\rm d}t\Bigg|X_{\tau}=a\right],
$$
because, as per your statement, $\tau$ is defined as
$$
\tau=\...
- 396
4
votes
Geometric Brownian Motion: Why is the Wiener process multiplied by volatility?
If you examine a standard definition of a Wiener process, $W_t - W_0$ follows the normal distribution with mean zero and variance $t$. If you want the variance to be something else, you have to scale ...
- 6,924
3
votes
Bounded Stochastic discrete process
I believe that the process you postulate has a Beta conditional distribution. If my memory serves me well, I have encountered it in the book by Liptser and Shiryayev "Statistics of Random Processes" ...
- 4,307
3
votes
Are two stochastic processes independent if the Wiener processes inside are uncorrelated
[Edit]
My "answer" below is not a really an answer for I have completely misinterpreted your original question. I thought you asked about the covariance of 2 processes over a given time horizon (i.e. ...
- 14.5k
3
votes
Accepted
Independence of increments of the stochastic process $\frac{1}{t}\int_0^t u dW_u $
Note that, for $t>s>0$,
\begin{align*}
X_t-X_s &= \frac{1}{t}\int_0^t udW_u - \frac{1}{s}\int_0^s udW_u\\
&=\frac{1}{t}\bigg(\int_s^t u dW_u + \int_0^s udW_u \bigg)- \frac{1}{s}\int_0^s ...
- 21k
3
votes
Accepted
Calculation of a process's drift
Your solution is correct. Generally speaking, for any $\alpha,\beta\in\mathbb{R}$, the drift $\mu_{X^{\alpha\beta}}$ of the process:
$$X_t^{\alpha\beta}:=e^{\alpha t+\beta W_t}$$
will be equal to:
$$\...
- 7,989
3
votes
Accepted
What is the difference between standard deviation, volatility and quadratic variation?
Using only words and no equations:
Knowing the Variance (or standard deviation) of a Brownian Motion we can calculate the uncertainty in the future position of a particle. Knowing $\sigma^2$ and ...
- 10.9k
3
votes
Integrated Brownian motion
Your derivation is correct. Even if we fix your obvious typo the formula
$$
\textstyle\int_t^TW_s\,ds=\int_t^T(T-s)\,dW_s
$$
is wrong.
There is no doubt that
\begin{align}
TW_T&=\textstyle\int_0^...
- 2,003
2
votes
Two Wiener process under same martingale measure Q
The following is not a proof but some reasoning:
If you consider the L2-limits then you see something along the lines:
$$
dW^2 = \lim_{n \rightarrow \infty}\sum_{j=1}^n (W_{t_{j+1}} - W_{t_{j}})^2 \...
- 13.6k
2
votes
Accepted
Calculate value of Integral of Wiener process $\int_{0}^t e^{\lambda u } dZ_u$
I'll try to use Ito's Lemma to come up with a solution. Ito's lemma states that:
$$F(Z_t,t)=\int_0^t\left(\frac{\partial F}{\partial u}+\frac{\partial F}{\partial Z}a+0.5\frac{\partial^2 F}{\partial Z^...
- 5,998
2
votes
Accepted
Expectation on a function of Wiener Process
The proof uses the martingale property of the Ito integral. For an adapted stochastic process $X_t$ such that
$$\mathbb{E}\int_0^{t}|X_s|^2ds <\infty$$
we have
$$\mathbb{E}\int_0^{t}X_sdW_s =0$$
...
- 1,692
2
votes
Arbitrage portfolio example
You can choose an arbitrage for the classical 1-dimensional Black-Scholes model, and not use $S_2$ at all.
Such an arbitrage is e.g. presented in Example 3.5 in "Fractional Processes As Models In ...
- 21
1
vote
Accepted
Regression of stochastic integral on Wiener process
By definition,
$$
{\mathbb Cov}(M_t,W_T) = {\mathbb E}[M_t W_T] - {\mathbb E}[M_t] {\mathbb E}[W_T] = {\mathbb E}[M_t W_T]
$$
since ${\mathbb E}[M_t] = {\mathbb E}[W_T] = 0 $. We now consider the ...
- 700
1
vote
Differentiability of solutions of a stochastic differential equation
Rather non rigorously,
$\frac{W(t)}{t} \sim N(0,\frac{1}{t}) $
if $t \to 0$ , we can see the variance goes to infinity. Hence Ito process is not differentiable.
- 341
1
vote
How to Evaluate Expected Value powered 4 of a Wiener Process?
You state $X(t_j) - X(t_{j-1}) \backsim \mathcal{N}(0, \frac{t}{n})$. Thus:
\begin{equation}
X(t_j) - X(t_{j-1}) = \sqrt{\frac{t}{n}} Z ,
\end{equation}
where $Z \backsim \mathcal{N}(0, 1)$. Note that:...
- 739
1
vote
Can anyone explain to how Hull get from stock returns to continuously compounded stock returns?
The key is on the left hand side. Recall that the differential of log of x is:
$d \ln x =\frac{1}{x}dx$
So you get:
$\ln x_t-\ln x_0=at$
Which you will need to exponentiate to get rid of the log:
...
- 6,204
1
vote
Accepted
Interpretation of IV and its use in stock movement prediction
The Bachelier model which assumes that price follows a normal distribution is a correct approximation for the Black-Scholes one for short times t. When time is short it's fine to ignore drift because ...
- 2,187
1
vote
Accepted
Solving a backwards heat equation using stochastic calculus
Based on the form of your equation, we can consider the SDE
\begin{align*}
dX_t = \sigma dW_t,
\end{align*}
where $W$ is a standard Brownian motion. Since, for $0 \leq t \leq T$,
\begin{align*}
X_T =...
- 21k
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