# About the variance of the white noise process

I saw such an example in a book. Assume that the $$w(t)$$ is a white Gaussian noise with zero-mean and power spectrum density $$N_0/2$$. Now, consider the sample function:

$$n(t)=\sqrt{2\over T}\int_{0}^{T}w(t)\cos(2\pi ft)dt, 0

I found that the $$\sqrt {\frac{2}{T}}\cos(2\pi ft),0 is often used as one of the orthogonal basis functions. So, the $$n(t)$$ also can be seen as the coefficient of the basis function. It is said that the $$n(t)$$ is iid zero-mean Gaussian random variables with variance $$\frac{N_0}{2}$$. My question is about the variance. The variance of $$n(t)$$ is given by:

$$var[n(t)]=E[\frac{2}{T} \int_{0}^{T}\int_{0}^{T} w(t_1)\cos(2\pi ft_1)w(t_2)\cos(2\pi ft_2) dt_1dt_2]$$ $$=\frac{2}{T} \int_{0}^{T}\int_{0}^{T} E[w(t_1)w(t_2)]\cos(2\pi ft_1)\cos(2\pi ft_2) dt_1dt_2$$ $$=\frac{2}{T} \int_{0}^{T}\int_{0}^{T} R_w(t_1,t_2)\cos(2\pi ft_1)\cos(2\pi ft_2) dt_1dt_2$$

where $$R_w(t_1,t_2)$$ is the autocorrelation function of the white noise $$w(t)$$. The autocorrelation function is $$\frac{N_0}{2}g(t_1-t_2)$$, where $$g(t)$$ is the dirac delta function.

Then, the variance becomes

$$=\frac{N_0}{2} \frac{2}{T} \int_{0}^{T}\int_{0}^{T} g(t_1-t_2)\cos(2\pi ft_1)\cos(2\pi ft_2) dt_1dt_2$$ $$=\frac{N_0}{2} \frac{2}{T} \int_{0}^{T} \cos^2(2\pi ft_2)dt_2$$ $$=\frac{N_0}{2}$$

Doesn't it mean that the $$n(t)$$ is not independent at different times $$t_1$$ and $$t_2$$? And why it is wrong to calculate the variance by letting $$t_1 = t_2$$? If $$t_1 = t_2$$, the variance of the $$n(t)$$ is infinite, which is the same as the variance of the white Gaussian noise. Thanks.

• To answer your first question: indeed, sampling $n(t)$ will produce samples that are not independent. However, for certain basis functions $n(t_0)$ and $n(t_0+T)$ will be uncorrelated for some value of $T$. – MBaz Feb 27 '19 at 16:54