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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} $?

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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
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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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