Let $X_t$ be the (strong) solution of the SDE \begin{align} {dX_{t}=a\,dt+\sigma X_{t}\,dW_{t}},\ \ X_0=x_0.\tag{$\star$} \end{align} where $a,\sigma,x_0\in \mathbb{R}$, and $W$ is the 1-dimensional standard Brownian motion.

Question: For fixed $t$, is the random variable $X_t-x_0e^{at}$ lognormal? Or is it not?

Attempt: We have the mild solution of $(\star)$: $$X_t=x_0e^{ta}+\int_0^t e^{a(t-s)} \sigma X_s dW_s,$$ and thus suffices to show $\int_0^t e^{a(t-s)} \sigma X_s dW_s$ is lognormal.

Now, consider the SDE $dS_{t}=\sigma S_{t}\,dW_{t},\ \ S_0=x_0.$ Then, we have $\sigma S_{t}\,dW_{t}=0S_{t}dt+\sigma S_{t}\,dW_{t}$, i.e., the solution is given by the geometric BM:$$S_{t}=S_{0}\exp \left(\left(-{\frac {\sigma ^{2}}{2}}\right)t+\sigma W_{t}\right).$$ But we also have the solution as the solution of the integral equation $S_{t}=\int_0^t e^{a(t-s)} \sigma S_s dW_s$. Thus $\int_0^t e^{a(t-s)} \sigma S_s dW_s$ is lognormal.

But $X$ satisfies a "shifted" integral equation and not sure consequences of that. Also, for the geometric BM we could divide the both sides by $S_t$ which leads to the nice form to which It\^{o}'s formula can be applied, but not sure how to do that for $(\star)$.

  • $\begingroup$ You usually apply a change in measure (drift) to yield the second SDE. This is because you can integrate easier, but also it represents a risk neutral process (drift less) or martingale. $\endgroup$ – Chinny84 Feb 16 '17 at 15:00
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    $\begingroup$ Note that the solution to the SDE $$dX_t = a \, dt + \sigma X_t \, dW_t, \qquad X_0 = x_0$$ can be calculated explicitly. The (unique) solution is given by $$X_t = \left( x_0 + a \int_0^t \exp \left( \frac{1}{2} \sigma^2 s - \sigma W_s \right) \, ds \right) \exp \left( - \frac{1}{2} \sigma^2 t + \sigma W_t \right) .$$ $\endgroup$ – saz Feb 16 '17 at 16:02
  • $\begingroup$ @Chinny84 Ah, thanks for the pointer, I will look into it. $\endgroup$ – shall.i.am Feb 17 '17 at 22:51
  • $\begingroup$ @saz I see, $x_0$ is annoying to be lognormal, huh. Could I ask what I should read to derive the solution? Change of measure and Itô's formula? $\endgroup$ – shall.i.am Feb 17 '17 at 22:54
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    $\begingroup$ @shall.i.am Itô's formula is enough; see e.g. this question for more details: math.stackexchange.com/q/813755 $\endgroup$ – saz Feb 18 '17 at 7:05

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