I have this process:

$dx_t = -\frac{k}{2}x_tdt + \frac{\beta}{2}dz_t$

and must prove it's normally distributed with first two moments:

$\mu = e^{-\frac{1}{2}kt}x_0$

$\sigma^2 = \frac{\beta^2}{4k}(1-e^{-kt})$

I tried to multiply $x_t$ by $e^{kt}$ and apply Ito's Lemma to this 'product process' in order to eventually recover back $x_t$ by taking exponentials.

The normality is straightfoward; the variance is ok but the mean isn't since I'm left with an integral whose integrand includes $x_t$ and I'm stuck.

I don't know whether I made some mistakes or adopted the wrong approach since the beginning.


1 Answer 1


You have following SDE $$dx_t=-\frac{k}{2}x_t dt+ \frac{\beta}{2}dz_t \tag{1}$$ Consider, $f=e^{\frac{k}{2}t}x_t$. Using Ito: $$df= \frac{k}{2}e^{\frac{k}{2}t}x_t\, dt+ e^{\frac{k}{2}t} \,dx_t \tag{2} $$ So, we have \begin{align} d\left(e^{\frac{k}{2}t}x_t\right)&= \frac{k}{2}e^{\frac{k}{2}t}x_t\, dt + e^{\frac{k}{2}t} \left(-\frac{k}{2}x_t dt+ \frac{\beta}{2}dz_t \right)\\ &= \frac{\beta}{2}e^{\frac{k}{2}t}dz_t \end{align} Note: RHS does not involve $x_t$. Now integrate both side to get your answer.


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