In this post we will define the following results:
| Stochastic integral | Result | Variance |
|---|---|---|
| $\int_0^t \mathrm{d}B_s$ | $B_t$ | $t$ |
| $\int_0^t s \mathrm{d}B_s$ | $tB_t - \mathrm{ibm}(t, \omega)$ | $\tfrac{1}{3}t^3$ |
| $\int_0^t B_s \mathrm{d}B_s$ | $\tfrac{1}{2}B_t^2 - \tfrac{1}{2}t$ | $\tfrac{1}{2}t^2$ |
| $\int_0^t B_s^2 \mathrm{d}B_s$ | $\tfrac{1}{3}B_t^3 - \mathrm{ibm}_t$ | $t^3$ |
| $\int_0^t B_s^k\mathrm{d}B_s$ | $\tfrac{1}{k+1}B_t^{k+1} - k \int_0^t B_s^{k-1} \mathrm{d}s$ | |
| $\int_0^t \exp\left(B_s - \tfrac{s}{2}\right)\mathrm{d}B_s$ | $\exp(B_t - \tfrac{t}{2}) - 1$ | $e^t - 1$ |
Introduction
Recall that the integrated Brownian motion (IBM), which is defined as
$$\begin{aligned} \mathrm{ibm}(t, \omega) = \int_0^t B_s \mathrm{d}s, \end{aligned}$$
is a random variable which follows the normal distribution with zero mean and variance $t^3/3$. We will use the notation $\mathrm{ibm}_t$ hereafter.
This MSE thread is worth reading.
Hereafter, we will give a few examples of stochastic integrals of the Itô type which are given by
$$I(t, \omega) {}={} \int_0^t f(t, \omega) \mathrm{d}B_t,$$
where $f$ is an Itô-integrable function. We shall also use the shorthand notation $I_t(\omega) = I(t, \omega)$.
Let us recall some important results from stochastic calculus:
Theorem 1 (Itô isometry). Let $f:(t_1, t_2)\times \Omega \to {\rm I\!R}$ be an Itô-itegral function on $(t_1, t_2)$. Then
$${\rm I\!E}\left[\left(\int_{t_1}^{t_2}f(t, \omega){\rm d}B_t\right)^2\right] {}={} {\rm I\!E}\left[\int_{t_1}^{t_2}f^2(t, \omega){\rm d}t\right].$$
Theorem 2 (Itô's lemma). Let $X_t$ be an Itô process with
$${\rm d}X_t = u{\rm d}t + v{\rm d}B_t.$$
Let $g:[0, \infty) \times {\rm I\!R} \to {\rm I\!R}$ be twice continuously differentiable on $[0, \infty) \times {\rm I\!R} $. Then, the process $Y_t = g(t, X_t)$ is an Itô process with
$${\rm d}Y_t = \frac{\partial g}{\partial t}(t, X_t){\rm d}t + \frac{\partial g}{\partial x}(t, X_t){\rm d}X_t + \tfrac{1}{2} \frac{\partial^2 g}{\partial x^2}(t, X_t)({\rm d}X_t)^2,$$
where $({\rm d}X_t)^2 = ({\rm d}X_t) \cdot ({\rm d}X_t)$ is computed according to the rules
$${\rm d}t \cdot {\rm d}t = {\rm d}t\cdot {\rm d}B_t = {\rm d}B_t \cdot {\rm d}t = 0, \ {\rm d}B_t \cdot {\rm d}B_t = {\rm d}t.$$
Stochastic Integrals
Stochastic integral No 1
Consider the stochastic integral
$$\begin{aligned}I_1(t, \omega) = \int_0^t s \mathrm{d}B_s = t B_t - \mathrm{ibm}_t, \end{aligned}$$
where $\mathrm{ibm}_t$ denotes the integrated Brownian motion given above.
The expected value of $I_1$ is ${\rm I\!E}[I_1] = {\rm I\!E}[\int_0^t s \mathrm{d}\mathrm{B}_s] = 0$ and its variance is given by
$$\begin{aligned} \mathrm{Var}[I_1] = {\rm I\!E}[I_1^2] {}={}& {\rm I\!E}\left[\left(\int_0^t s \mathrm{d}B_s\right)^2\right]\\{}={}& {\rm I\!E}\left[\int_0^t s^2 \mathrm{d}s\right] {}={}\frac{t^3}{3}.\end{aligned}$$
We have used the Itô isometry.
Note that the above computation is easier than trying to do
$${\rm I\!E}[I_1^2] {}={} {\rm I\!E}[(tB_t - \mathrm{ibm}_t)^2].$$
Stochastic integral No 2
Define
$$\begin{aligned}I_2 = \int_0^t B_s \mathrm{d}B_s\end{aligned}$$
Using Itô's lemma, we obtain
$$\begin{aligned}I_2 = \tfrac{1}{2}B_t^2 - \tfrac{1}{2}t\end{aligned}$$
and ${\rm I\!E}[I_2] = 0$, while the variance is
$$\begin{aligned} \mathrm{Var}[I_2] = {\rm I\!E}[I_2^2] {}={}& {\rm I\!E}[(\tfrac{1}{2}B_t^2 - \tfrac{1}{2}t)^2] \\ {}={}& \tfrac{1}{4}{\rm I\!E}[(B_t^2 - t)^2] \\ {}={}& \tfrac{1}{4}{\rm I\!E}[B_t^4 + t^2 - 2tB_t^2]\end{aligned}$$
and using the fact that ${\rm I\!E}[B_t^2] = t$ and ${\rm I\!E}[B_t^4] = 3t^2$, it is ${\rm I\!E}[I_2^2] = \tfrac{1}{2}t^2$.
Stochastic integral No 3
Define the following stochastic integral which involves the squared Brownian motion
$$\begin{aligned}I_3 = \int_0^t B_s^2 \mathrm{d}B_s\end{aligned}$$
This can be computed by applying Itô's formula to $f(t, x) = \tfrac{1}{3}x^3$ and gives
$$\begin{aligned}I_3 = \tfrac{1}{3}B_t^3 -\mathrm{ibm}_t,\end{aligned}$$
where $\mathrm{ibm}_t$ is the integrated Brownian motion.
The expectation of $I_3$ is clearly ${\rm I\!E}[\int_0^t B_s^2 \mathrm{d}B_s] = 0$ and its variance is
$$\begin{aligned}\mathrm{Var}[I_3] = {\rm I\!E}[I_3^2] {}={}& {\rm I\!E}[(\tfrac{1}{3}B_t^3 -I_t)^2]\\ {}={}& \tfrac{1}{9}{\rm I\!E}[B_t^6] + {\rm I\!E}[I_t^2] - \tfrac{2}{3} {\rm I\!E}[B_t^3 I_t],\end{aligned}$$
and ${\rm I\!E}[B_t^6] = 15 t^3$, ${\rm I\!E}[I_t^2] = t^3/3$ and
$$\begin{aligned}{\rm I\!E}[B_t^3 I_t] = {\rm I\!E} \left[ B_t^3 \int_0^t B_s \mathrm{d}s\right] = {\rm I\!E} \left[ \int_0^t B_t^3 B_s \mathrm{d}s\right] = \int_0^t {\rm I\!E}[B_t^3 B_s]\mathrm{d}s,\end{aligned}$$
Now using Isserlis's Theorem and the fact that $0 \leq s \leq t$ and ${\rm I\!E}[B_t B_s] = s$ and ${\rm I\!E}[B_t^2] = t$ we have ${\rm I\!E}[B_t^3 B_s] = 3ts$, therefore
$${\rm I\!E}[B_t^3 I_t] = \int_0^t {\rm I\!E}[B_t^3 B_s]\mathrm{d}s = \int_0^t 3ts \mathrm{d}s = \tfrac{3}{2}t^3.$$
Overall,
$$\begin{aligned} \mathrm{Var}[I_3] = t^3.\end{aligned}$$
We can generalise the above result using the formula
$$\begin{aligned}\int_0^t B_s^k \mathrm{d}B_s = \tfrac{1}{k+1}B_t^{k+1} - k \int_0^t B_s^{k-1}\mathrm{d}s\end{aligned}$$
and proceeding as above.
Stochastic integral No 4
Consider the stochastic integral
$$\begin{aligned}I_4 = \int_0^t \exp (B_s - \tfrac{1}{2}s) \mathrm{d}B_s\end{aligned}$$
Using Itô’s formula, we may show that this is equal to
$$\begin{aligned}I_4 = \exp (B_t - \tfrac{1}{2}t) - 1 = e^{B_t}e^{-\tfrac{1}{2}t} - 1.\end{aligned}$$
Since $B_t$ follows the normal distribution with zero mean and variance $t$, the random variable $e^{B_t}$ follows the log-normal distribution.
The expectation of $I_4$ is ${\rm I\!E}[\int_0^t \exp (B_s - \tfrac{1}{2}s) \mathrm{d}B_s] = 0$ and its variance is
$$\begin{aligned}\mathrm{Var}[I_4] {}={} {\rm I\!E}[I_4^2] {}={}& {\rm I\!E}[(\exp (B_t - \tfrac{1}{2}t) - 1)^2]\\ {}={}& {\rm I\!E}[\exp (2B_t - t) - 2\exp (B_t - \tfrac{1}{2}t) + 1]\end{aligned}$$
where ${\rm I\!E}[\exp (2B_t - t)] = e^{-t}{\rm I\!E}[\exp (2B_t)]$. Since $2B_t \sim \mathcal{N}(0, 4t)$, ${\rm I\!E}[e^{2B_t}] = e^{\frac{4t}{2}} = e^{2t}$, so ${\rm I\!E}[\exp (2B_t - t)] = e^t$.
Similarly, ${\rm I\!E}[2\exp (B_t - \tfrac{1}{2}t)] = 2e^{-\frac{t}{2}}{\rm I\!E}[e^{B_t}] = 2$.
Overall,
$$\begin{aligned}\mathrm{Var}[I_4] {}={} e^t - 1.\end{aligned}$$