# How to expand lognormal approximation of Brownian motion

How can we expand this sum? $$\sum_{i=1}^n (e^{rt_i-\frac{1}{2}\sigma^2t_i+\sigma w_{t_i}})^2$$ where: $$w_{t_i}$$ is a standard Brownian motion.

If we let $$m_t=e^{-\frac{1}{2}\sigma^2t_i+\sigma w_{t_i}}$$ is a martingale where $$m_0$$ = 1, why does it reduce to $$\sum_{i,j=1}^n e^{r(t_i+t_j)+\sigma^2\;min(t_i, t_j)}$$.

I know that $$\sigma w_{t_i} \sim N(0, \sigma^2 t)$$ and the $$\text{cov}(w_{t_i},w_{t_j}) = \min\;(t_i,t_j)$$ so im essentially asking how the expansion of the sum works and how to put it together since $$m_t$$ is a martingale and $$m_0=1$$.

• Your first sum appears wrong. Please double check. – Gordon May 15 '20 at 14:27
• yeap its correct its the solution the stock price $x_t$ under the risk-neutral measure. i just left the $x_0$ out since it is a constant and im just interested in expanding the sum – Ryantstrong May 16 '20 at 5:18