I am reading through Oksendals SDEs. I think there may be a mistake in question 5.18b and I can not find an errata so I was looking for some confirmation. The problem concerns the following SDE
$\mathrm{d}X_t = \kappa (\alpha - \log X_t)X_t\mathrm{d}t + \sigma X_t \mathrm{d}B_t,\;\;\;X_0 = x>0 $
There is a hint to change variables to $Y_t = \log X_t$. This gives the following SDE for $Y_t$
$\mathrm{d}Y_t = \kappa(\alpha - Y_t)\mathrm{d}t + \sigma \mathrm{d}B_t - \frac{1}{2} \sigma^2 \mathrm{d}t$
This can be solved using an integrating factor
$Y_t = Y_0 e^{-\kappa t}+(\alpha-\frac{1}{2\kappa}\sigma^2)(1-e^{-\kappa t})+\sigma e^{-\kappa t}\int_0^{t}e^{\kappa s}\mathrm{d}B_s$
Okay now part b wants us compute $E[X_t]$. For a fixed time all the terms in $Y_t$ but the last one are constants hence
$E[X_t] = E[\exp(Y_t)]\\ = \exp(e^{-\kappa t }\ln x + (\alpha-\frac{1}{2\kappa}\sigma^2)(1-e^{-\kappa t}))E[\exp(\sigma e^{-\kappa t}\int_0^{t}e^{\kappa s}\mathrm{d}B_s)]$
Okay now it seems to me that the random variable $\sigma e^{-\kappa t}\int_0^{t}e^{\kappa s}\mathrm{d}B_s$ is normally distributed, since it is the limit of a sum of normally distributed random variables, I haven't rigorously proven this to myself but I do not think there is any subtlety that would mess things up. Clearly it must have mean $0$, and the variance can be computed using Ito isometry. If I do that I get that the variance is
$E[(\sigma e^{-\kappa t}\int_0^{t}e^{\kappa s}\mathrm{d}B_s)^2]=\frac{\sigma^2(1-e^{-2\kappa t})}{2\kappa}$
Hence now the expectation term above becomes a moment generating function of a Gaussian with a known mean and variance, evaluated at $1$. It seems to me that this will give us
$E[X_t] = \exp(e^{-\kappa t }\ln x + (\alpha-\frac{1}{2\kappa}\sigma^2)(1-e^{-\kappa t}))\exp(\frac{1}{2}\frac{\sigma^2(1-e^{-2\kappa t})}{2\kappa})\\ = \exp(e^{-\kappa t }\ln x + (\alpha-\frac{1}{2\kappa}\sigma^2)(1-e^{-\kappa t})+ \frac{\sigma^2(1-e^{-2\kappa t})}{4\kappa})$
The problem is that Oksendal wants us to prove that $E[X_t] = \exp(e^{-\kappa t }\ln x + (\alpha-\frac{1}{2\kappa}\sigma^2)(1-e^{-\kappa t})+ \frac{\sigma^2(1-e^{-2\kappa t})}{2\kappa})$
Is there a mistake in his answer or did I make a mistake somewhere?
I did some simulations and they seemed to be in pretty strong agreement with my result.
Thanks!
I think you are right. The stochastic Itô integral is a Gaussian random variable with mean zero, since the integrand is an $L_a^2([0,T]\times \Omega)$ process (in fact, deterministic)
The mean is 0, (it is a martingale) and the variance (second moment), by Itô's isometry, is
$$\sigma^2 e^{-2\kappa t} E\left[\int_0^t e^{-2\kappa s}ds\right] = \frac{\sigma^2}{2\kappa} \left( 1- e^{-2\kappa t}\right).$$
Then the exponential is a $log$-normal and so on.
I do not have the book with me right now but if you copied everything correctly it should be a typo. It is not a major error though.