Why this is Normal Distribution?

Question

$$ \log W_t = \int_0^t \left(r + \pi(\mu-r) - \frac{1}{2} \pi^2\sigma^2\right) du + \int_0^t \pi \sigma \,d Z_u $$

And that $$\log W_t \sim \operatorname{Normal} (\left(r + \pi(\mu-r) - \frac{1}{2}\pi^2\sigma^2\right), \pi^2\sigma^2)$$

I don't see how come this can be converted onto Normal distribution (i.e., I don't see how can first line be converted onto second line), can you please help?

Answer 1

Since $\int_0^tdZ_u\sim N(0,\,t)$, a deterministic $\mu$ satisfies$$\mu+\int_0^t\pi\sigma dZ_u=\mu+\pi\sigma\int_0^tdZ_u=\mu+\pi\sigma N(0,\,t)=N(\mu,\,\pi^2\sigma^2t).$$In this case, $\mu=\int_0^tcdu=ct$ for $u$-independent integrand $c=r+\pi(\mu-r)-\frac12\pi^2\sigma^2$.