# Continous white noise process: Spectral analysis and Time-series

Priestley has explained why white noise process has auto-covariance function $R(\tau)=C\delta(\tau)$, (where $\delta$ is dirac-delta fucntion,) as follows: Consider $\epsilon(t)$ is a white-noise process with $E[\epsilon(t)]=0$ and $E[\epsilon^2(t)]=\sigma^2_\epsilon$. Now if we form another stochastic process $U=\int_{a}^{b}g(u)\epsilon(u)du$; we will have $$Var(U)=\sigma^2_\epsilon\int_{a}^{b}\int_{a}^{b}g(u)g(v)\delta(u-v)dudv .$$ This is where I don't understand this argument. He writes that since the double integral is zero, it is necessary for $\sigma^2_\epsilon$ to be infinite for $Var(U)$ to be meaningful. I don't see how the double integral is zero, becasue the integral actually becomes $\int_{a}^{b}g^2(u)du >0$, which can be zero only if $g(u)=0$. I am not sure what am I missing here?

If $\varepsilon_t$ is a continuous-time white noise process, then you are absolutely correct when you assert that $E[\varepsilon_t]=0$ and absolutely wrong when you assert that $E[\varepsilon_t^2]=\sigma_\varepsilon^2$. In fact, $$E[\varepsilon_t\varepsilon_{t+\tau}] = R_\varepsilon(\tau) = C\delta(\tau)$$ as you (presumably following Priestley) said in your very first sentence. Bear in mind that Dirac deltas are not ordinary functions, they are distributions and they make sense (for ordinary folks like you and me, not the mathematicians lurking in this forum) only in integrals. Thus, $\delta(3)$ is not the value of the Dirac delta at $3$ and $\delta(0)$ is not the value of the Dirac delta at $0$ and claims that $\delta(0) = \infty$ are spurious. What the Dirac delta means is that $$\int_a^b g(t)\delta(t-c) \,\mathrm dt = g(c)~~ \text{if}~a<c<b~ \text{and}~ g(t)~\text{is continuous at}~c$$ no more and no less (at least for us ordinary folks). The integral above has value $0$ if $c<a$ or $c>b$ and has undefined value if $g(t)$ is discontinuous at $c$.
Turning to your next statement, note that $U$ is not a random process but a random variable whose mean is $$E[U] = E\left[\int_a^b g(u) \varepsilon_u \,\mathrm du\right] = \int_a^b E[g(u) \varepsilon_u]\,\mathrm du = \int_a^b g(u) E[\varepsilon_u]\,\mathrm du = 0.$$ I won't bother justifying the interchange of expectation and integration above but trust me; it is valid. Next, note that \begin{align} \operatorname{var}(U) &= E[(U-E[U])^2]\\ &= E\left[\int_a^b g(u) \varepsilon_u \,\mathrm du\int_a^b g(v) \varepsilon_v \,\mathrm dv\right]\\ &= \int_a^b\int_a^b g(u) g(v) E[\varepsilon_u\varepsilon_v] \mathrm du\,\mathrm dv\\ &= \int_a^b\int_a^b g(u) g(v) C\delta(u-v) \mathrm du\,\mathrm dv \end{align} and not what you claim that Priestly wrote. Now, we get to the crux of the matter: \begin{align} \operatorname{var}(U) &= \int_a^b\int_a^b g(u) g(v) C\delta(u-v) \mathrm du\,\mathrm dv\\ &= \int_a^b g(v)\left[\int_a^b g(u) C\delta(u-v) \mathrm du\right]\,\mathrm dv\\ &= \int_a^b g(v)\cdot g(v) C \,\mathrm dv\\ &= C\int_a^b g^2(v) \,\mathrm dv \end{align} which is pretty close to what you wrote.
• My problem is with the claim that the $Var(U)$ becomes zero if $C$ is finite. This is not answered here. – Abhinav Gupta Jun 27 '18 at 23:05
• @AbhinavGupta If Priestly really wrote that $E[\epsilon_t^2]=\sigma_\epsilon^2$, it is wrong. If Priestly really wrote $$Var(U)=\sigma^2_\epsilon\int_{a}^{b}\int_{a}^{b}g(u)g(v)\delta(u-v)dudv,$$ then that is wrong too. If Priestly really wrote that the above double integral has value $0$ unless $\sigma^2_\epsilon$ is infinite, then that is wrong too. So, double-check to make sure that you are not misinterpreting what your book says, and if you have stated matters correctly, apply Dorothy Parker's dictum: "This is not a book to be put down lightly; it should be thrown with great force! – Dilip Sarwate Jun 28 '18 at 3:33