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The two basic models of finance are the following:

$\textbf{The Samuelson SDE (aka Black - Scholes - Merton model):}$ Suppose that $Z=\left(Z_t, t\in\mathbb{R}^{+}\right)$ is a Wiener process (aka standard Brownian motion). Following Samuelson $(1965)$ and Black - Scholes and Merton $(1973)$, we assume that the stock price process $s$ evolves according to the following SDE

$$ds_t=\mu s_tdt+\sigma s_tdZ_t\tag{1}$$

Solving $(1)$ we can find that the distribution $s_t$ is

$$s_t\sim N\left(s_oe^{\left(\mu-\frac{\sigma^2}{2}\right)t}, e^{\sigma^2t} \right)\tag{*}$$

Therefore $s_t$ is normally distributed with mean and variance that are functions of time.

$\textbf{The Ornstein-Uhlenbeck SDE (OU model):}$ Suppose that the price of a stock s satisfies the following Stochastic Differential Equation

$$ds_t=k(\theta-s_t)dt+\sigma dZ_t\tag{2}$$

Solving $(2)$ we can find that the distribution $s_t$ is

$$s_t\sim N\left(s_oe^{-kt}+\theta(1-e^{-kt}),\frac{\sigma^2}{k}(1-e^{-2kt})\right)\tag{**}$$

and hence, $s_t$ is normally distributed with mean and variance that are functions of time.

$\textbf{Question:}$ How do we prove $(*)$ and $(**)$ from $(1)$ and $(2)$ respectively and what does it mean that $s$ follows a normal distribution where the mean and the variance are functions of time?

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    $\begingroup$ Hint: For $(*)$, use Itô's lemma with the transformation $f(S_t)=\ln S_t$. This should give an explicit solution for $S_t$ and from there it is relatively straightforward to derive the distribution. $\endgroup$
    – UNOwen
    Nov 9, 2022 at 15:44
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    $\begingroup$ A stochastic process that is normally distributed with mean $\mu_t$ and variance $\sigma_t^2$ as functions of time simply means that the distribution changes/evolves with time. As an example, one would intuitively expect the variance of standard brownian motion to increase w.r.t. time (since we are adding more random perturbations from some initial point over time and there is no tendency to revert back to our initial point). $\endgroup$
    – UNOwen
    Nov 9, 2022 at 15:52
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    $\begingroup$ For $(**)$, use the transformation $f(S_t)=S_t e^{k t}$. Note that this derivation will be slightly more intricate than the one for geometric brownian motion. Referring to this may help. As a reminder, when using Itô's lemma, it is useful to know that $\mathop{dW_t}^2=\mathop{dt}$, $\mathop{dW_t}\cdot \mathop{dt}=0$, and $\mathop{dt}^2=0$ such that $W_t$ is a Wiener process (which is the same as your $Z_t$). $\endgroup$
    – UNOwen
    Nov 9, 2022 at 16:03
  • $\begingroup$ @UNOwen I will find time and I will do the calculations. After that I will post the solution here to be checked. Thanks for the hints! $\endgroup$
    – Oliver Queen
    Nov 9, 2022 at 20:54

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