I have a statement like this;
A zero mean Gaussian random process $X(t)$ is wide sense stationary with the auto-correlation function $R_x(\tau) = 4e^{-2|\tau|}$
And I want to find the probability density function of this $X(t)$…
So I started with the definition of auto-correlation function;
$$R_x(\tau) = E\left[X(t_1)X(t_2)\right]$$
$$R_x(\tau) = \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}{X(t_1)X(t_2).f_{{X_1}{X_2}}(x_1, x_2;t_1, t_2)dx_1dx_2}$$
Since $X(t)$ is a WSS process, $X(t_1) = X(t_2)$
$\therefore$
$$4e^{-2|\tau|} = \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}{X^2.f_{{X_1}{X_2}}(x_1, x_2;t_1, t_2)dxdx}$$
$$\frac{\dfrac{d^2}{dx^2}\left(4e^{-2|\tau|}\right)}{X^2} = f_{{X_1}{X_2}}(x_1, x_2;t_1, t_2)$$
But since the differentiation always produces a zero, the PDF is calculated as zero. I'm pretty sure there's a problem with my steps. Can anyone point me the right direction? 🙂 Thanks in advance!!
EDIT::: I thought about trying multivariate Gaussian distribution;
Since this process is WSS, we can say the auto-correlation function is the same as covariance matrix for the Gaussian. (Am I right?)
And then we have, $$C_X = \begin{bmatrix}4 & 4e^{-2}\\4e^{-2} & 4\end{bmatrix}$$
We also need the inverse of this
$$C_X^{-1} =\frac{1}{16(1-e^{-4})}\begin{bmatrix}4 & -4e^{-2}\\-4e^{-2} & 4\end{bmatrix}$$
The two variable Gaussian PDF is given by;
$$f_{X_1X_2}(x_1, x_2)=\frac{1}{2\pi\sqrt{det(C_X)}}e^{-\left(\frac{1}{2}X^TC_X^{-1}X\right)}$$
by substituting;
$$f_{X_1X_2}(x_1, x_2;t_1, t_2)=\frac{1}{2\pi\sqrt{16(1-e^{-4})}}e^{-\left(2x_1^2 + 2x_2^2-4x_1x_2e^{-2}\right)}$$
Is this the correct approach?
Best Answer
Since the process is Gaussian and one-dimensional, its probability density function (PDF) must be
$$f_X(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\tag{1}$$
where $\mu$ is the mean, and $\sigma^2$ is the variance. From the problem statement we know that $\mu=0$, so we only need to find $\sigma^2$.
From the definition of the auto-correlation function we know that
$$R_x(0)=E[X^2(t)]=\sigma^2+\mu^2=\sigma^2\tag{2}$$
where the last equality follows from the fact that $\mu=0$. From $(2)$ and the given auto-correlation function we know that $\sigma^2=R_x(0)=4$, so we finally get for the PDF
$$f_X(x)=\frac{1}{2\sqrt{2\pi}}e^{-\frac{x^2}{8}}\tag{3}$$