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I have the following in my notes and I'm not sure if it's true or not. Any help would be highly appreciated.

If $\{W_t\}_{t\geq0}$ is a standard Brownian motion stochastic process, $\Delta>0$ and $n\in\mathbb{N}$ then $W_{n\Delta}-W_{(n-1)\Delta}$ is mean zero and variance $\Delta$ normal random variable.

I am using the following definition of Brownian motion, as taken from Harrison's Stochastic flow systems:

A stochastic process $\{W_t\}_{t\geq0}$ has independent increments if the random variables $X_{t_1}-X_{t_0},\ldots,X_{t_n}-X_{t_{n-1}}$ are independent for all $n\geq 1$ and all $0\leq t_0\leq\ldots\leq t_n<\infty$. It has stationary independent increments if the distribution of $X_t-X_s$ depends only on $t-s$. A Standard Brownian motion is a stochastic process that has continuous sample paths, stationary independent increments and $X_t\sim N(0,t)$.

That the difference is mean zero, I can see. What I'm not convinced about is that variance wil be $\Delta$. Any thoughts?

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  • $\begingroup$ you mean: normal random variable, not just random variable. $\endgroup$
    – Sergio Parreiras
    Apr 6, 2014 at 22:32
  • $\begingroup$ yes, thanks for pointing out $\endgroup$
    – mathemagician
    Apr 6, 2014 at 22:34
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    $\begingroup$ What is your definition of Brownian motion? To me, this is true by definition. $\endgroup$
    – Potato
    Apr 6, 2014 at 22:35
  • $\begingroup$ I'm starting to understand the source of my confusion. On Wikipedia, a Wiener process is defined as satisfying $W_t-W_s\sim N(0,t-s)$ for $0\leq s<t$. However, Harrison which is the textbook I'm studying from defines the Wiener process (page 1) as $W_t\sim N(0,t)$ Under this definition, we would however have that $W_t-W_s\sim N(0,t+s)$, hence my confusion. Am I missing something here? $\endgroup$
    – mathemagician
    Apr 6, 2014 at 22:44
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    $\begingroup$ Your last assertion is not correct, because $W_t$ and $W_s$ are not independent. Can you please type the definition you are using into the body of your question? $\endgroup$
    – Potato
    Apr 6, 2014 at 22:49

2 Answers 2

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Since $X_t$ is $N(0,t)$ for every $t$, one knows that $X_0=0$. Since the distribution of $X_t-X_s$ depends only on $t-s$, one knows that, for every $n$, $X_{n\Delta}-X_{(n-1)\Delta}$ is distributed like $X_{\Delta}-X_0=X_\Delta$, which is $N(0,\Delta)$. In particular, $X_{n\Delta}-X_{(n-1)\Delta}$ is centered with variance $\Delta$, QED.

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  • $\begingroup$ Even simpler. Didn't notice that he was given that $X_t - X_s$ depends only on $t-s$. Is that a neccessary part of the definition then? Since I didn't use it and I think my calculation is also correct. $\endgroup$
    – Kai Sikorski
    Apr 6, 2014 at 23:56
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$$E[ (W_{n\Delta} - W_{(n-1)\Delta})^2] = E[W_{n\Delta}^2] - 2 E[W_{n\Delta}W_{(n-1)\Delta}] + E[W_{(n-1)\Delta}^2]$$ Use what we know about the second moment of $W_t$, and rewrite the second term. $$ =n \Delta - 2 E\left[\left(W_{(n-1)\Delta} + (W_{n\Delta}-W_{(n-1)\Delta})\right)W_{(n-1)\Delta}\right] + (n-1)\Delta$$ Expand second term and use independence of non overlapping intervals $$ = (2n-1) \Delta - 2 E\left[W_{(n-1)\Delta}^2\right] - 2E[(W_{n\Delta}-W_{(n-1)\Delta})]E[W_{(n-1)\Delta}]$$ The third term above has to be $0$ by properties we prescribed on the first moment of $W_t$. Now just do algebra $$ = (2n -1) \Delta -2(n-1)\Delta = \Delta$$

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