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I've been studying stochastic calculus on my own for a while and stumped on some things that I don't fully understand:

Let's start with the geometric brownian motion:

\begin{align} dS_t = \mu S_t dt + \sigma S_t dW_t \end{align}

To solve it for $S_t$ I need to find an equation $f(t,S_t)$ that works with the Itô's Lemma:

\begin{align} df(t,S_t) = \frac{\partial}{\partial t}f(t,S_t)\,dt +\frac{\partial}{\partial S}f(t,S_t)\,dS_t + \frac{1}{2}\frac{\partial^2}{\partial S^2}f(t,S)\,(dS_t)^2 \end{align}

At every place that I look I see that $f(t,S_t) = log(S_t)$, but I don't understand why. Is this just an ansatz?

If I change my SDE to a more generalized short rate model:

\begin{align} dS_t = (\alpha + \beta S_t) dt + \sigma S_t^\gamma dW_t \end{align}

I'll need to find another ansatz? And if I find a function $f(t,S_t)$ that works, I can simply do some algebra and get an expression for $S_t$?

Ѕᴀᴀᴅ
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Seyfzao
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  • So, there are two answers to your question. The first is, yes it's an ansatz, and no, it isn't. Note the relationship between $$\frac{\mathrm{d}S_t}{S_t}$$ and $\log(S_t)$ if you divide both sides by $S_t$, so no it isn't an ansatz. Yet it is. It really exists to avoid the embarrassment of negative numbers when they are not possible. The implications are that if all parameters are known, then your diffusion term will be lognormal. If you change your model, you may have to start over or work with a non-analytic outcome. – Dave Harris Nov 08 '21 at 04:23
  • See a complete answer here-->https://math.stackexchange.com/questions/2832820/geometric-brownian-motion-product-ansatz-rationale?rq=1 – Dave Harris Nov 08 '21 at 04:25

1 Answers1

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For $\gamma=0$, the solution to $$\tag{1} dS_t=(\alpha+\beta S_t)\,ds+\sigma S^\gamma_t\,dW_t $$ is $$ S_t=S_0e^{\beta t}-\frac{\alpha}{\beta}(1-e^{\beta t})+\sigma\int_0^te^{\beta(t-s)}\,dW_s\,. $$ (see Ornstein-Uhlenbeck Process).

For $\gamma=1$, the solution of (1) is $$ S_t= \frac{S_0+\alpha\int_0^t b_s\,ds}{b_t}\, $$ where $$ b_t = e^{-\beta t-\sigma W_t+\frac{1}{2}\sigma^2t} $$ (see this question). For general $\gamma$, I don't know of an explicit solution of (1).

Kurt G.
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