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Let $W(t)$ be continuous time white noise, that is, a wide-sense stationary (WSS) zero-mean Gaussian process with autocorrelation function $R_W (\tau) = σ^2\delta(\tau)$. Calculate the auto correlation function of $X(t) = \int_0^t W(r) dr$.

Can anyone help me understand how they get these integration limits in the following step in the calculation?

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By the very definition, we have

$$\delta(\tau) = \begin{cases} 0, & \tau \neq 0, \\ 1, & \tau = 0 \end{cases}.$$

Therefore,

$$\delta(r-q) = \begin{cases} 0, & r \neq q, \\ 1, & r=q \end{cases}.$$

Suppose that $s<t$, i.e. $\min\{s,t\}= s$. For any $q \in (s,t]$ and $r \in [0,s]$ we have

$$\delta(r-q)=0$$

since $r-q<0$. Consequently,

$$\int_0^s \int_s^t \mathbb{E}(W_q W_r) \, dq \, dr = 0.$$

This implies the identity you want to prove.

The argumentation for the case $t \leq s$ works analogously; I leave it to you.

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  • $\begingroup$ Yes, we are left with a 1 (times $\sigma^2$) in the inner integral when q = r = min(s,t). And we start at 0 and go along that line to min(s,t). Maybe whats confusing me is that when you look at the double integral i can imagine what is happening but if you look at the inner integral separately the picture Yves is painting isnt working, right? $\endgroup$ – JKnecht Dec 29 '15 at 15:16
  • $\begingroup$ @JKnecht why ask me about the picture Yves is painting? The identity you want to prove is really straight-forward using the considerations I explained in my answer. $\endgroup$ – saz Dec 29 '15 at 18:29
  • $\begingroup$ I tried to explain that i understand what you are saying but it doesnt become straight forward to me because i think about it in a similar way as Yves does. But maybe i am misunderstanding both of you. $\endgroup$ – JKnecht Dec 29 '15 at 20:15
  • $\begingroup$ Yeah, you are right. I read through your post a couple of more times and it really is straight forward and obvious. I dont know why i didnt see it the first couple of times. Maybe because this is my first math course in ten years... $\endgroup$ – JKnecht Dec 29 '15 at 20:45
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The integration domain on the left is a rectangle $s\times t$ starting from the origin.

Because of the Dirac $\delta$, that on the right is a line segment along the main diagonal, starting from the origin and extending until the first of $s,t$ is met.

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  • $\begingroup$ Thanks for your answer. Im still a little confused tho. Look at my reply to saz. $\endgroup$ – JKnecht Dec 29 '15 at 15:17
  • $\begingroup$ Dont bother...i got it now. $\endgroup$ – JKnecht Dec 29 '15 at 20:45

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