Distribution of an Itô Integral with a Deterministic Integrand

# SDE Exmaple (no drift)


# SDE Exmaple (no drift)












Assume, $X_t := ∫^t_0e^{μs}dB_s$ ($B_s$ is Brownian motion)

My Reference Said $dX_t = e^{μt}dB_t$.

I tried to Ito's formula to solve this (that is $df = f_tdt+f_{B_t}dB_t + \frac{1}{2}f_{B_tB_t}dt$) But I don't know how to solve it. Is there any reference to solve this example?

## Answer by Matteo Campagnoli (score 3, accepted)

https://quant.stackexchange.com/a/63389

T. Bjork Arbitrage Theory in Continous Time reports the following Lemma at page 57: Let $f(t)$ be a deterministic function of time and define the process $X$ by

$$X(t) = \int_{0}^{t}f(s)dW(s)$$

Then $X(t)$ has a normal distribution with zero mean and variance given by

$$Var[X(t)] = \int_{0}^{t} f^2(s)ds$$

If by "solving" you mean find the distribution, this lemma allows you to do that easily, since $e^{\mu s}$ is clearly a deterministic function.

If you are wondering where does this Lemma come from, here comes a sketchy proof. The mean is obviously zero, because $X(t)$ is an Ito's Integral. Now, let's calculate the expected value of $E[e^{iuX(t)}]$, i.e. the characteristic function. To do so, we first have to find the Ito differential of $Z(t) = e^{iuX(t)}$ for a fixed $t$:

$$dZ(t) = -\frac{1}{2}u^2Z(t)dX^2+iuZ(t)dX = -\frac{1}{2}u^2f^2(t)Z(t)dt+iuf(t)Z(t)dW$$

Integrate both members and take expectations

$$E[Z(t)] = -\frac{1}{2}u^2\int_{0}^{t}f^2(s)E[Z(t)]ds+1$$

We have to add one at the end of it because we've assumed $Z(0)=1$. This looks like an integral equation, by we just need to differentiate both members to end up with a standard differential equation for $m(t) = E[Z(t)]$

$$\dot m(t) = -\frac{u^2}{2}f(t)m(t)$$

$$\Rightarrow m(t) = E[Z(t)] = exp\{-\frac{u^2}{2}\int_{0}^{t}f^2(s)ds\}$$

Once you have the characteristic function, calculate the second momentum, i.e. the variance, and you're done. QED

## Answer by Yoda And Friends (score 3)

https://quant.stackexchange.com/a/63365

I am not sure what do you mean by solving that integral. Stochastic integrals are random variables. In particular, for a nice (in the sense it fulfills certain technical conditions) function $f(t)$ you can find its distribution: $$\int_0^tf(s) \ dB_s \sim N(0, \ \int_0^tf(s)^2 ds)$$