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I would like to find the distributions of the following random variables:

$Z_k= \frac{1}{\pi} \int^{2\pi}_{0} cos(kt) dW_t$ $k=1,2,...$ and $(W_t)_{t\geq 0}$ is a Wiener process.

What is the distribution of $Z_1$, and $(Z_k) $?

I am new to stochastic calculus, I only know how to integrate a Wiener process wrt. an other Wiener process.

Can someone help me, how to do this?

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  • $\begingroup$ Is this a home work? $\endgroup$
    – Aksakal
    Mar 10, 2015 at 16:56
  • $\begingroup$ @Aksakal No, it is not homework. $\endgroup$
    – FelB
    Mar 10, 2015 at 17:25

1 Answer 1

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It's normal distribution. The integral is basically a sum, and a sum of independent normals is normal. Obviously, the mean is zero too.

All you need is to get the variance. Use the fact that $dW_t$ are independent of each other and $(dW_t)^2=dt$ to calculate the variance, which is a square of the integral remembering that the integral is essentially a sum.

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  • $\begingroup$ I tried to calculate the variance, I got this: $VAR(Z_1)=\frac{1}{\pi^2}\sum^{n-1}_{0} cos^2t\cdot(t_{j+1}-t_j)=\frac{1}{\pi^2}\int^{2\pi}_{0}cos^2t \ dt =\frac{1}{\pi}$ Is that correct? $\endgroup$
    – FelB
    Mar 10, 2015 at 18:15
  • $\begingroup$ That's correct. $\endgroup$
    – Aksakal
    Mar 10, 2015 at 18:29

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