離散採樣 BM 的總和
如果底層證券遵循對數正態 GM 且沒有漂移 $ dS_t = \sigma S_t dW_t $ 和 $ A_N = \Sigma_{i=1}^{N} S_{t_i} $ . 如何計算變異數 $ A_N $ ?
我們有 $ S_t = \sigma S_tdW_t $ 和 $ A_N = \sum_{n=1}^N S_n = S_0\sum_{n=1}^N e^{\sigma W_n-\frac12\sigma^2n}. $
$$ \mathbb E[A_N] = S_0\sum_{n=1}^N \mathbb E[e^{\sigma W_n-\frac12\sigma^2n}] = NS_0. $$
$$ \mathbb E[A_N^2] = S_0^2\Big(\sum_{n=1}^N\mathbb E[e^{2\sigma W_n-\frac12(2\sigma)^2n+\frac14(2\sigma)^2n}] + 2\sum_{n=1}^{N-1}\sum_{m=n+1}^N\mathbb E[e^{\sigma W_n-\frac12\sigma^2n}e^{\sigma W_m-\frac12\sigma^2m}]\Big) = S_0^2\Big(\sum_{n=1}^Ne^{n\sigma^2}+2\sum_{n=1}^{N-1}\sum_{m=n+1}^Ne^{-\frac12\sigma^2(n+m)}\mathbb E[e^{2\sigma W_n}]\mathbb E[e^{\sigma(W_m-W_n)}]\Big) = S_0^2\Big(\sum_{n=1}^Ne^{n\sigma^2}+2\sum_{n=1}^{N-1}\sum_{m=n+1}^Ne^{-\frac12\sigma^2(n+m)}e^{2\sigma^2n}e^{\frac12\sigma^2(m-n)}\Big) = S_0^2\Big(\sum_{n=1}^Ne^{n\sigma^2}\big(1+2(N-n)\big)\Big). $$ $$ Var(A_N) = \mathbb E[A_N^2] - \mathbb E[A_N]^2 = S_0^2\bigg[\Big(\sum_{n=1}^Ne^{n\sigma^2}\big(1+2(N-n)\big)\Big)-N^2\bigg] $$