I want to calculate the expected value of a Wiener process with $1 \ge t \ge 0$ but i can't find a formula for it. I don't know how I am supposed to calculate the value with the definition of Wiener process.

For example why is $W_{1/2} = 1/2(W_0+W_1)+1/2Z_2 $ where $Z_2$ is a standard normal random variable?

And how do you get $W_1$ without formula then?

Or how do you calculate $(W_{1/2})^6$ or $e^{W_ {1/2}}$?

Is it $(W_{1/2})^6 = (1/2(W_0+W_1)+1/2Z_2)^6 $ and $e^{W_ {1/2}}= e^{1/2(W_0+W_1)+1/2Z_2} $?


closed as unclear what you're asking by Did, астон вілла олоф мэллбэрг, Shailesh, user91500, JonMark Perry Feb 9 '17 at 7:37

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  • $\begingroup$ You say you want expected values, but you are writing down formulas for the value, which is random, not the expected value, which is a number. Expected values are much easier. Which do you want? You ask how to calculate $W_{1/2}^6.$ It's a random variable, so 'calculate' is a complicated notion. But if you want its expected value that's a question with an uncomplicated answer. $\endgroup$ – spaceisdarkgreen Jan 21 '17 at 7:43
  • $\begingroup$ Are you allowed/supposed to assume $W_0 = 0$? $\endgroup$ – spaceisdarkgreen Jan 21 '17 at 8:17
  • $\begingroup$ Sorry you are right, I mean expected values. I mixed something up... you can ignore the questions with the formulas. I still have no idea how to calculate $E(W_{1_2})$ for example because I don't know how $W_{1_2}$ looks though (if I ignore the formula above). $\endgroup$ – Septime44 Jan 21 '17 at 8:17
  • $\begingroup$ Yes, we can assume that $(W)_t$ is a Wiener process. $\endgroup$ – Septime44 Jan 21 '17 at 8:18
  • $\begingroup$ Note that, by definition, each $W_t$ is Gaussian with mean $0$ and variance $t$... hence $\mathbb{E}(W_t)=0$ for all $t$. $\endgroup$ – saz Jan 21 '17 at 9:21

The expression for $W_{1/2}$ you wrote down is part of one approach to constructing Brownian motion on $[0,1]$. You asked about $W_1.$ $W_1$ is constructed before $W_{1/2}$ so its value can be used to construct $W_{1/2}.$ The definition of $W_1$ is likewise in terms of $W_0$ (which is usually taken to be zero). The definition is $$W_1 = W_0 + Z_1$$ where $Z_1$ is a standard normal.

After $W_1$ is defined, we define $W_{1/2}$ in terms of $W_0$ and $W_1$ as $$ W_{1/2} = \frac{W_0 +W_1}{2} + \frac{1}{2}Z_2$$ where $Z_2$ is a standard normal independent from $Z_1.$ From this definition we can see that the three values we have constructed are consistent the axioms of Brownian motion. Namely, $W_1-W_0=Z_1$ is $N(0,1)$ and $W_{1/2}-W_0$ and $W_1-W_{1/2}$ are independent $N(0,1/2)$ 's. (They come out to $(Z_1+Z_2)/2$ and $(Z_2-Z_1)/2$, which fit the bill.) Then you could go on to define $W_{1/4}$ and $W_{3/4}$ in terms of the other values and new independent normals $Z_3$ and $Z_4.$ And then the eighths and sixteenths, etc.

That's all well and good, but it looks like you have a simpler problem of calculating some expected values. For this, you just need one of the basic properties of Brownian motion: $W_t$ is distributed as $N(0,t).$ So in your case $W_{1/2}$ is $N(0,1/2)$ so has PDF $$ f(x) = \frac{1}{\sqrt{\pi}}e^{-x^2}$$

So $$ E(W_{1/2}^6) = \int_{-\infty}^\infty x^6\frac{1}{\sqrt{\pi}}e^{-x^2}dx = \frac{15}{8}$$ and $$ E(e^{W_{1/2}}) = \int_{-\infty}^\infty e^x\frac{1}{\sqrt{\pi}}e^{-x^2}dx = e^{1/4}.$$


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