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I want to use Ito's formula on the following SDE:

$$ dX_t= - \gamma (\log X_t - \theta) X_t d t + \sigma X_t d W_t $$

to obtain an expression for $ \log X_T $ where $T > t $ is some fixed time.

pretty new to this, I tried applying Ito's with $ f(t,X_t) = \log X_t $, but then is $f_t = 0$? My friend is also multiplying by an integrating factor of $e^{\gamma T}$, but I've honestly no idea why he's doing that. Any guidance would be much appreciated.

EDIT: I'm supposed to get:

I'm supposed to get $$\log X_T = e^{-\gamma (T - t)} \log X_t + \left( \theta - \frac{1}{2\gamma} \sigma^2 \right) ( 1 - e^{-\gamma (T - t)} )+ \sigma\int_t^{T} e^{-\gamma (T - s)} d W_s $$

and then make various claims about the mean and the variance as $t \to \infty $, though I think that part is straight forward enough

Christopher Turnbull
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  • I think there is no closed form, but maybe it can help you, divide both sides by $$x_t$$ and you will have $d(\ln x_t)$ in the left-hand side, now it will be like Orenestein-Uhlenbeck equation. – Khosrotash Jul 06 '20 at 05:27
  • yep I don't think it has a closed form either. I've edited it to show the answer one should obtain, still not entirely following – Christopher Turnbull Jul 06 '20 at 08:20

1 Answers1

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Using Ito's lemma for $log(X_t)$

$$dlog(X_t)=- \gamma (\log X_t - \theta) d t + \sigma d W_t-\frac{1}{2}\sigma^2dt=[- \gamma (\log X_t - \theta)-\frac{1}{2}\sigma^2]dt+\sigma d W_t.$$

Set $Z_t=log(X_t)$. You have

$$dZ_t=\gamma(c-Z_t)+\sigma d W_t$$

Here $c=\theta-\frac{1}{2\gamma}\sigma^2$. This is a standard OU-process for $Z_t$. It has the well-known solution (obtained using similar tricks you can find online)

$$Z_T=c+(Z_t-c)e^{-\gamma(T-t)}+\sigma \int_t^{T}e^{-\gamma(T-t)}dW_s$$.

Therefore

$$log(X_T)=\theta-\frac{1}{2\gamma}\sigma^2+(log(X_t)-\theta+\frac{1}{2\gamma}\sigma^2)e^{-\gamma(T-t)}+\sigma \int_t^{T}e^{-\gamma(T-t)}dW_s$$

$$=e^{-\gamma (T - t)} \log X_t + \left( \theta - \frac{1}{2\gamma} \sigma^2 \right) ( 1 - e^{-\gamma (T - t)} )+ \sigma\int_t^{T} e^{-\gamma (T - s)} d W_s ,$$

as was to be shown.

fes
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  • nice one. and just to be clear, what would arguments follow with the expectation. I take it the integral goes to zero? – Christopher Turnbull Jul 06 '20 at 13:03
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    @ChristopherTurnbull if you are interested in the mean and variance of $(_T)$, you can use the formulas for OU process (https://en.wikipedia.org/wiki/Ornstein%E2%80%93Uhlenbeck_process), you should be able to find the derivations. E.g. the expectation tends to (I defined above) as $t \rightarrow \infty$. – fes Jul 06 '20 at 13:44
  • @ChristopherTurnbull And yes when you apply Ito's lemma to $f(X_t)=log(X_t)$, the time derivative is zero $f_t=0$. – fes Jul 06 '20 at 17:37
  • Cool, almost all good except slight confusion over why we deal with $\log(X_T)$ instead of $\log(X_t)$. Since $t \to \infty$, also must $T \to \infty $ since $T > t$? I think $e^{\gamma (T-t)} \to 0$, but what if $T = t + 1$ for example – Christopher Turnbull Jul 06 '20 at 18:15
  • @ChristopherTurnbull The two time points $T>t$ are pretty arbitrary. Supposedly you need to calculate e.g. $\mathbb{E}_t[log(X_T)]$ when $T \rightarrow \infty$, think of $t$ as "today". Alternatively you set today as $t=0$ and calculate $\mathbb{E}_0[log(X_t)]$. I was a bit unclear on this. – fes Jul 06 '20 at 18:48
  • Yeah I just don't really know. Super confused about the $t$ and $T$ notation, as the question said "Let $ t \to \infty $ and find the expectation and the variance of $\log X_T$.". But could you get the same answer by setting $t = 0$ and letting $T \to \infty$? It's so confusing – Christopher Turnbull Jul 06 '20 at 21:40
  • Let us continue this discussion in chat. – Christopher Turnbull Jul 06 '20 at 21:47