scaling log-normal process

Question

I would like to define the process in 2 different ways: The first one is for any given positive const $s$ and $r$ $$X(t) = (s+r)e^{-0.5\sigma^2t+\sigma W(t)}-s,$$ While the second $$Y(t) = re^{-0.5\sigma^2t+\sigma W(t)}$$ with a standard Brownian motion. What can I say about these 2 processes $X$ and $Y$? the mean is the same $r$? The variance of $X$ is larger(what is it exactly?)? Do they have the same log normal distribution but perhaps different mean and variance? Trying to get an understanding as to how they relate to each other and their properties.

Answer 1

Note that $$X(t)=(s+r)e^{-\frac{1}{2}\sigma^2 t}e^{\sigma W(t)}-s$$ $$Y(t)=re^{-\frac{1}{2}\sigma^2t}e^{\sigma W(t)}$$ We know that since $W(t)\sim N(0,t)$, $$\mathbb{E}[e^{\sigma W(t)}]=e^{\frac{1}{2}\sigma^2t}$$ $$Var(e^{\sigma W(t)})=\mathbb{E}[(e^{\sigma W(t)})^2]-\mathbb{E}[e^{\sigma W(t)}]^2$$ $$=e^{2\sigma^2t}-e^{\sigma^2t}$$ Everything else in the two expressions are constant. So you can see that $X(t)$ has higher variance. Another relationship between them is simply that $$\frac{X(t)+s}{s+r}=\frac{Y(t)}{r}=e^{-\frac{1}{2}\sigma^2 t+\sigma W(t)}$$ and this quantity has log normal distribution.