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.


1 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.

  • $\begingroup$ thanks! in the last equation, the RHS(exponential) is lognormal, but that that mean $X$ and $Y$ are lognormal as well? They simply have different variance but share the fact that both are lognormal? $\endgroup$
    – Medan
    Mar 24, 2020 at 13:18
  • $\begingroup$ $Y(t)$ is indeed lognormal, because $log(Y(t))=log(r)-\frac{1}{2}\sigma^2t+\sigma W(t)$, but $X(t)$ is not. $X(t)-s$ is lognormal. So it's like a lognormal but shifted by a constant. Note that a normal distribution shifted by a constant is still normal, but a lognormal distribution does not have that property. $\endgroup$ Mar 24, 2020 at 14:02

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