CDF&density of stock price modeled by standard brownian motion

Assume that the price of the stock follows the model $$S(t) = S(0) exp ( mt − ((σ^2)/2 ) t + σW(t) )$$ , (1) where W(t) is a standard Brownian motion; σ > 0, S(0) > 0, m are some constants.

Derive the CDF and PDF for $$S(t)/S(t-1)$$.

CDF:

$$S(t)/S(t-1) = \frac{S(0)exp(mt-\frac{\sigma^2}{2}t+\sigma W(t))}{S(0)exp(m(t-1) -\frac{\sigma^2}{2}(t-1)+\sigma W(t-1))}$$ using standard algebra and rewriting I get

$$S(t)/S(t-1)=exp(m-\frac{\sigma^2}{2}+\sigma(W(t)-W(t-1))$$. Using that $$W(t)-W(t-1)$$ is $$N(0,1)$$ I get that $$S(t)/S(t-1) = exp(m-\frac{\sigma^2}{2}+\sigma Z)$$, where Z is $$N(0,1)$$

CDF is thus $$F(x)=P(exp(m-\frac{\sigma^2}{2}+\sigma Z)\le x)$$ = $$F(x)=P(Z\le \frac{ln(x)-m+\frac{\sigma^2}{2}}{\sigma})$$ which is $$\Phi(\frac{ln(x)-m+\frac{\sigma^2}{2}}{\sigma})$$.

Is this correct?

If the CDF is correct, can I use it to derive the PDF? Or how would I go about calculating the PDF?

is it possible to calculate the correlation between $$S(t)/S(t − 1)$$ and $$S(t − 1)/S(t − 2)$$ from this? Or how can it be done?

$$\mathrm{pdf}=\frac{d(\mathrm{CDF})}{dx}=\frac{d\Phi(\frac{ln(x)-m+\sigma^2/2}{\sigma})}{dx}=\frac{1}{x\sigma}\phi(\frac{ln(x)-m+\sigma^2/2}{\sigma})$$.
As for your question on the correlation between $$S(t)/S(t-1)$$ and $$S(t-1)/S(t-2)$$, there is none because the Z~(0,1) are not serially correlated.