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Given that I have $W_t$ and $W_s$ for some $t>0$ and $s>0$ and $W_n = \sum_{i=1}^n R_i$ I do not understand why $\text{Var}[W_t-W_s]=|t-s|$.

I understand why $\text{Var}[W_n]=n$. This is because $\text{Var}[W_n]=\text{E}[(\sum_{i=1}^n R_i)^2]=\sum_{i=1}^n\text{E}[R_i^2]=n*1=n$. There is a particular way in which $R_i$ is defined in my notes, basically we are looking at a coin toss experiment, so $R_i$ is either a $+1$ or a $-1$. If you apply the above logic to my problem, I get something like:

$$\text{Var}=\text{E}[(\sum_{i=1}^t R_i-\sum_{j=1}^s R_j)^2]=\text{E}[R_1+R_2+...+R_t-(R_1+R_2+..+R_s)]$$

And then I am not sure what to do.

Naz
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  • Note that, if $t>s$, then $\sum_{i=1}^tR_i-\sum_{j=1}^sR_j=\sum_{k=s+1}^tR_k$. – Tom-Tom Dec 23 '15 at 10:49
  • oh... I totally missed that minus sign in there... – Naz Dec 23 '15 at 10:55
  • Ok, I got it. If you type up a brief answer with that I will choose your answer :) – Naz Dec 23 '15 at 10:57
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    There's serious notation issues here, and I think an error in the order of priority of results. The Wiener process is defined as the $0$ mean continuous time independent increment process s.t. $W_t - W_s\sim\mathcal{N}(0,|t-s|)$. It can be shown that $$W_t = \lim_{n\to \infty} \frac{1}{\sqrt{n}} \sum_{k=1}^{\lfloor nt \rfloor} R_k$$ For i.i.d standard gaussians $R_k$. The latter is contingent on the former. In fact, there exist alternate representations of the process. Crux of the ... – stochasticboy321 Dec 23 '15 at 11:06
  • ...matter being $W_t = \sum_{i=1}^t R_i$ is a terrible way to write things, and points to possible errors that you may want to correct with some reading. – stochasticboy321 Dec 23 '15 at 11:06
  • We are following the Wimott's quant finance book and it is not very rigorous at all. And for someone with no formal maths background, it kind of delivers the point well enough I hope. But I hope to do all of this more rigorously later on. – Naz Dec 23 '15 at 11:24

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If $t>s$, then $$\sum_{i=1}^tR_i−\sum_{j=1}^sR_j=\sum_{k=s+1}^tR_k.$$

Tom-Tom
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