# Power Spectral Density from Probability Density Function

The samples of a signal $$x[n]$$ are i.i.d. and follow a triangular pdf with $$a = 0,\ b = 2,\ c = 1$$:

The DC-power of the signal is

$$\mu_x^2 = \big(\mathbb{E}(X)\big)^2 = \left(\int_{-\infty}^{\infty} x f_x(x)dx\right)^2 = 1$$

the total power of the signal

$$\mathbb{E}(X^2) = \int_{-\infty}^{\infty} x^2 f_x(x)dx = \frac{7}{6}$$

and the variance (AC-power) as

$$\sigma^2 = \mathbb{E}\big((X-\mu_x)^2\big)= \mathbb{E}\left(X^2\right)-\mu_x^2 = \frac{1}{6}$$.

Thus I thought, that the PSD of the signal has the following form:

$$S_{xx}(e^{j\omega}) = 2\pi \delta(\omega) + \frac{1}{6}$$

as the AC-power results in a constant value over the whole spectrum and the DC power in a Dirac-Delta at $$\omega = 0$$. Furthermore, the integration over the PSD should return the total power of the signal: $$\frac{1}{2\pi} \int_{-\pi}^{\pi} S_{xx} dx = \mathbb{E}(X^2)$$

Is my solution correct or did I miss something? Can this procedure be done for every probability density function under the assumption that all samples of $$x[n]$$ are iid?

If the $$x[n]$$ are i.i.d. random variables with mean $$\mu$$ and variance $$\sigma^2$$, then the power spectral density of this discrete-time random process (which is effectively white noise plus a possibly nonzero mean $$\mu$$) is what you have calculated. The shape of the common pdf of the random variables is irrelevant in all this except insofar as the shape determines the mean and variance; for example. you would get the same PSD if the $$x[n]$$ were Gaussian random variables with mean $$\mu$$ and variance $$\sigma^2$$.