Consider a stochastic differential equation
$$Z_ = B(t)+aZ_, $$
where $B_t$ is a Brownian process and $a , > 0$.
Assuming that $Z_0=100, \sigma=0.3, a=0.03$ find the value of $P(Z_1>103.5)$.
My reasoning:
I use Ito's lemma to find the explicit form of the process $Z_t.$ We know that $B_t\sim N(0,t)$, so then we could find the distribution of $Z_1$ and standardize.
But it turned out not so easy. From Ito's lemma when I take $f(t)=lnZ_t$ I got
$$Z(t)=Z(0)e^{(a-\frac{\sigma^2}{2Z_t^2})t+\int\frac{\sigma}{Z_t} dB_t}\\$$
For $f(t)=e^{at}Z_t$ there is not some helpfull result eaither. I think that must be some easier way or a trick here. I would appreciate your help.
The given SDE can be seen in two equivalent ways:
it is a linead SDE and hence an SDE Of the form $$ dZ_t = \big(a(t)Z_t + c(t) \big) \ dt + \big( b(t)Z_t + d(t) \big) \; dW_t $$ with in our case $b=c=0, \; d(t)=\sigma, \; a(t)=a$, whose general solution you can find for example in Wikipedia.
an Ornstein–Uhlenbeck process, i.e. it solves $$ dZ_t = -aZ_tdt + \sigma dW_t $$ with in this case $a \to -a$ whose solution you can find on Wikipedia as well.
In any case the solution is $$ Z_t = e^{at}Z_0 + \sigma \int_{0}^{t}e^{a(t-s)} dW_s $$ which is a normal r.v. with mean $m = e^{at}Z_0$ and variance $v^2 = \frac{\sigma^2}{2a}(e^{2at}-1)$.
Thus the required probability can be found with the usual standardization procedure for a normal r.v. $$ \mathbb{P}(Z_1 > 103.5)= \mathbb{P}\bigg(\frac{Z_1-m}{v} > \frac{103.5-m}{v}\bigg) = 1- \Phi\bigg(\frac{103.5-m}{v}\bigg) $$ with $\Phi(x)$ the CDF of the standard normal.