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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?

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I agree with your derivation.

$\mathrm{pdf}=\frac{d(\mathrm{CDF})}{dx}=\frac{d\Phi(\frac{ln(x)-m+\sigma^2/2}{\sigma})}{dx}=\frac{1}{x\sigma}\phi(\frac{ln(x)-m+\sigma^2/2}{\sigma})$.

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.

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