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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?

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    $\begingroup$ Do you agree that, if you start with the second equation, and put an integral sign $\int$ on both sides of the $=$ you get the first equation. $\endgroup$
    – nbbo2
    Apr 13, 2021 at 12:14
  • $\begingroup$ @noob2 Yeah I intuitively agree your reply. But when I tried to 'prove' it, it gets me in trouble. My examination (1) Let's assume $X_t = f$ then using formula $df = dX_t = \frac{∂X_t}{∂t}dt + \frac{∂X_t}{∂B_t}dB_t + \frac{1}{2}\frac{∂^2X_t}{∂B_t^2}dt$ when organizing it, $dX_t = \frac{∂X_t}{∂t}dt + e^{μt}dB_t$... My alternative examination (2) Just intuitively differentiate $X_t$ but when I differentiate it, $X_t$ is ito integral form. So again it revert to my examination (1).. What is my problem and is there any reference of proving it?.. $\endgroup$
    – user13232877
    Apr 14, 2021 at 0:25
  • $\begingroup$ @noob2 And do you know any elementary SDE reference dealing with basis which have useful example and solution? Because it would be bother you if I ask question anytime. Also I'm not native in English. So I don't know which reference has some good reputation. The Problem is I don't know totally how to approach deterministic calculus, stochastic calculus differently..! In the elementary calculus (I mean deterministic.. normally applied to polynomial term) when we differentiate integral term it give back integrands.(That is $\frac{dX_t}{dB_t} = e^{μt}$, because integrands $e^{μt}$ $\endgroup$
    – user13232877
    Apr 14, 2021 at 0:33
  • $\begingroup$ are integrated with $B_s$ So it is $∫^t_0e^{μs}dB_s$ and when we differentiate it with $B_t$ again then it put back the $e^{μt}$. But when I approach it stochastic (because it has $dB_t$ term), as I said earlier in my reply, I cannot find solution it... $\endgroup$
    – user13232877
    Apr 14, 2021 at 0:35

2 Answers 2

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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

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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: $$\int_0^tf(s) \ dB_s \sim N(0, \ \int_0^tf(s)^2 ds)$$

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  • $\begingroup$ When I consider definition of integral, $∫^t_0 f(s)dB_s$ can be divided limit and discrete sum. So considering just $∑^t_{s=1}f(s)(B_{s+dt}-B_s)$. Using brownian motion notion, $f(s)(B_{s+dt}-B_s)$ follows $N(0,f(s)^2dt)$. Also Sum of Normal variable it follows also Normal. Is it right solution for your reply? $\endgroup$
    – user13232877
    Apr 14, 2021 at 0:41
  • $\begingroup$ I suggest you reading this Wikipedia page. It should clarify most of what you need: en.wikipedia.org/wiki/Itô_calculus $\endgroup$
    – Yoda And Friends
    Apr 19, 2021 at 17:14

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