Variance of a Discretely Sampled Driftless Geometric Brownian Motion Sum

Summary

The document derives the variance of a sum of discretely sampled asset prices when the underlying follows a driftless geometric Brownian motion. It writes each sampled price as an exponential of Brownian motion with the correction that keeps its expectation equal to the initial price. From this, it obtains the expected sum as the number of observations multiplied by the initial price.

For the second moment, the derivation separates squared individual observations from cross-products at distinct times. It uses independent Brownian increments to evaluate each expectation, then subtracts the square of the mean to obtain the variance. The resulting expression weights each exponential term by the number of later observations that form cross-products with it. The answer assumes the sampling times are indexed in unit steps, volatility is constant, and there is no drift; it does not address irregular sampling or a general time horizon.

Key ideas

• The sampled asset price is represented using an exponential Brownian-motion expression with a drift correction.
• The expected sum of sampled prices equals the number of observations times the initial price.
• The second moment includes both individual squared prices and cross-products across observation times.
• Independent Brownian increments simplify the cross-product expectations.
• The variance follows by subtracting the squared mean from the second moment.

Tags

# Sum of discretely sampled BM


# Sum of discretely sampled BM












If an underlying follows lognormal GM with no drift $dS_t = \sigma S_t dW_t $ and $A_N = \Sigma_{i=1}^{N} S_{t_i}$. How to compute variance of $A_N$?

## Answer by LucaMac (score 2, accepted)

https://quant.stackexchange.com/a/65429

We have $S_t = \sigma S_tdW_t$ and $A_N = \sum_{n=1}^N S_n = S_0\sum_{n=1}^N e^{\sigma W_n-\frac12\sigma^2n}.$

$$\mathbb E[A_N] = S_0\sum_{n=1}^N \mathbb E[e^{\sigma W_n-\frac12\sigma^2n}] = NS_0.$$

$$\mathbb E[A_N^2] = S_0^2\Big(\sum_{n=1}^N\mathbb E[e^{2\sigma W_n-\frac12(2\sigma)^2n+\frac14(2\sigma)^2n}] + 2\sum_{n=1}^{N-1}\sum_{m=n+1}^N\mathbb E[e^{\sigma W_n-\frac12\sigma^2n}e^{\sigma W_m-\frac12\sigma^2m}]\Big) = S_0^2\Big(\sum_{n=1}^Ne^{n\sigma^2}+2\sum_{n=1}^{N-1}\sum_{m=n+1}^Ne^{-\frac12\sigma^2(n+m)}\mathbb E[e^{2\sigma W_n}]\mathbb E[e^{\sigma(W_m-W_n)}]\Big) = S_0^2\Big(\sum_{n=1}^Ne^{n\sigma^2}+2\sum_{n=1}^{N-1}\sum_{m=n+1}^Ne^{-\frac12\sigma^2(n+m)}e^{2\sigma^2n}e^{\frac12\sigma^2(m-n)}\Big) = S_0^2\Big(\sum_{n=1}^Ne^{n\sigma^2}\big(1+2(N-n)\big)\Big).$$ $$Var(A_N) = \mathbb E[A_N^2] - \mathbb E[A_N]^2 = S_0^2\bigg[\Big(\sum_{n=1}^Ne^{n\sigma^2}\big(1+2(N-n)\big)\Big)-N^2\bigg]$$

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