Find the solution $Y_t$ of the following SDE $$d\ln Y_t=(-\beta \ln Y_i+C)dt+\sigma dW_t,Y_0=y$$

I know how to solve this by letting $X_t:=\ln Y_t$ and apply ito's lemma on $e^{-\beta t}\ln Y_t$. This is very similar to Ornstein-Uhlenbeck process.

I just wondering why we cannot directly differentiate $e^{-\beta t}\ln Y_t$. I mean how do we proceed as follows and why the following approach does not work,$$d(e^{-\beta t}\ln Y_t)=-\beta e^{-\beta t}\ln Y_tdt+\frac{e^{-\beta t}}{Y_t}dY_t-\frac12\frac{e^{-\beta t}}{Y_t^2}dY_tdY_t $$ Another question is the following always true? $$dY_tdY_t=d[Y,Y]_t$$ Is this just two different notation for quadratic variation? Further on that in this case is the following true?$$ d\ln Y_td\ln Y_t=((-\beta \ln Y_i+C)dt+\sigma dW_t)^2=\sigma^2dt$$ My understanding for Stochastic calculus is a bit shaky. Thanks for any clarification.

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    $\begingroup$ We do not know the dynamic $Y_t$ but only the dynamic of $ln(Y_t)$. Bare in mind we need to find the process $Y$. $\endgroup$ – Sesame Nov 27 '19 at 15:51
  • $\begingroup$ I think you have a typo because first you say $d \ln Y_t=(\beta \ln Y_t +C)dt +\sigma dW_t$ but at the end you say $dY_tdY_t=((\beta \ln Y_t +C)dt +\sigma dW_t)^2$ which would imply $dY_t=(\beta \ln Y_t +C)dt +\sigma dW_t$. So which one is the correct one? $\endgroup$ – UBM Nov 27 '19 at 21:01


$$d(e^{\beta t}\ln Y_t) = e^{\beta t}(Cdt+\sigma dW_t)$$

Integrate to get,

$$e^{\beta t} \ln Y_t - \ln Y_0 = C\int_0^t e^{\beta \tau}d\tau + \sigma \int_0^t e^{\beta \tau}dW\tau$$


$$ \ln Y_t = e^{-\beta t}\ln y + \frac C\beta(1-e^{-\beta t}) +\sigma \int_0^t e^{-\beta (t-\tau)}dW\tau$$

Thus, the solution of the SDE is

$$ Y_t = \exp\left(e^{-\beta t}\ln y + \frac C\beta(1-e^{-\beta t}) +\sigma \int_0^t e^{-\beta (t-\tau)}dW\tau\right)$$

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  • $\begingroup$ i know how to solve it by this method. Could you give me some help for my second and third question? $\endgroup$ – Kenneth Nye Nov 29 '19 at 4:44
  • $\begingroup$ Thanks for your effort, i should mention it clearly that i knew how to solve it in the first place. Sorry for that. $\endgroup$ – Kenneth Nye Nov 29 '19 at 4:49
  • $\begingroup$ @KennethNye - Not sure why you want to differentiate $d(e^{-\beta t}\ln Y_t)=e^{-\beta t}(-2\beta\ln Y_tdt+Cdt +\sigma dW_t) $, which can not integrated afterwards. $\endgroup$ – Quanto Nov 29 '19 at 4:55
  • $\begingroup$ i understand already for that, at that time i think i was just being stupid since if we differentiate $d \ln Y_t$ we could not do anything about $dY_t$. For my second question, generally to say is the notation $d[X,X]_t$ the same as $dX_t dX_t$? or are they fundamentally different? $\endgroup$ – Kenneth Nye Nov 29 '19 at 5:01
  • $\begingroup$ @KennethNye - The notation $dX_tdX_t$ is standard and well-defined. The other one is not and should be avoided. $\endgroup$ – Quanto Nov 29 '19 at 5:06

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