Expanding Squared Lognormal Brownian Terms and Their Moments

Summary

The question asks how to expand a sum of squared terms built from an exponential Brownian-motion model and relates the calculation to the martingale exponential. Squaring each term doubles its drift and Brownian exponent; expanding the resulting sum as a product with itself introduces a double sum over pairs of observation times. The covariance structure of Brownian motion, in which the covariance at two times is their minimum, is relevant when calculating expectations of those paired exponentials.

The displayed double sum in the question is not itself a direct algebraic expansion of the original sum: it concerns a paired expression and may represent an expected value under suitable assumptions. A derivation must distinguish the random sum from its expectation and apply the Gaussian exponential moment formula, retaining the time-dependent drift terms. The prompt supplies no answer or verification, so the exact claimed expression should be checked against the intended quantity and assumptions.

Key ideas

• Squaring an exponential term doubles both its drift exponent and its Brownian exponent.
• Multiplying a sum by itself yields terms indexed by pairs of observation times.
• Brownian-motion covariance at two times is determined by the earlier time.
• A random expression and its expected value are different quantities and should not be conflated.
• The proposed expression requires a Gaussian exponential-moment calculation and consistent drift terms.

Tags

# How to expand lognormal approximation of Brownian motion


# 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 $m_0=1$.

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