Finding PDF of a converted random process

Question

Let $s(t)$ be a periodic triangle wave as illustrated in the accompanying figure. Suppose a random process is created according to $X(t) = s(t − T)$, where $T$ is a random variable uniformly distributed over $(0, 1)$.

And I need to find the probability density function of this new $X(t)$...

So I thought like this. $s(t)$ got a period of $1$. So $X(t)$ can be any waveform created by shifting $s(t)$ within it's period time frame. Since $T$ is random within the $s(t)$'s period, is it fine to get the PDF from $s(t)$ directly by integrating it like this?

$$f_X(x) = \int\limits_{0}^{x}s(t)dt$$

Answer 1

Your argument is that $s(t-T)$ and $s(T)$ have identical distributions due to periodicity and symmetry.   This is okay.   However that then gives you:

$$\begin{align}\mathsf P(X(t)\leq x) ~=&~ \mathsf P(s(t-T)\leq x) \\[1ex] ~=&~ \mathsf P(s(T)\leq x)\\[1ex] ~=&~ \int_{-1}^x f_{s(T)}(z)\operatorname d z~\mathbf 1_{x\in(-1;1)}+\mathbf 1_{x\in[1;\infty)}\end{align}$$

Then by taking derivatives: $~f_{X(t)}(x) = f_{s(T)}(x)~$ (... which we could have said directly).

In short: What you need to determine is, if $T$ is uniformly distributed over $(0,1)$, then what distribution does $s(T)$ have?   (Hence what is its probability density function?)

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The support of $X(t)$ is $(-1;1)$ and the distribution is uniform because ... the relation between $T$ and $s(T)$ is linear in either of two parts of the support of $T$, and both of these intervals are folded onto the support of $s(T)$ (a surjective mapping).   tl/dr: The density function is $f_X(x) = \tfrac 12 \mathbf 1_{x\in(-1;1)}$