Sum of discretely sampled BM

Question

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 1

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

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

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