How Does Time Shifting Convert a WSCS Process to a WSS Process?

[MajorGrubert]

Hello everybody

I have a bit of a problem with understanding the conversion from a WSCS process X(t) to a WSS process Y(t) = X(t - \Delta). With \Delta the time shift being a uniform random variable on (0,T), independent of X(t) and T being the period of the mean function of X(t)

The problem begins with the method to find the mean function of Y(t) :

m_{Y} = E\{X(t - \Delta)\} = E\{E[X(t - \Delta)|\Delta]\} = E\{m_{X}(t - \Delta)\}

First, and it might seem very basic, I don't get the syntax E[X(t - \Delta)|\Delta]

And second, why by averaging the mean of the WSCS process over its period T would we get the mean function of the WSS process ?

If I understand that I could understand the same kind of process used to find the autocorrelation function of Y(t) from the autocorrelation function of X(t)

Please help me !

 

[Stephen Tashi]

On this forum, the LaTex will appear more gracefully if you use the "itex" tag when you want the expression to appear inline with the text.

the conversion from a WSCS process [itex] X(t) [/itex] to a WSS process [itex] Y(t) = X(t - \Delta) [/itex].

the conversion from a WSCS process X(t) to a WSS process Y(t) = X(t - \Delta).

I don't get the syntax E[X(t - \Delta)|\Delta]

On my screen the square brackets are hard to distinguish from the vertical bars. My interpretation is that it is a "conditional expectation". Roughly speaking, compute the expected value of X(t - \Delta) for one particular value of \Delta.

So:

E( E\{X(t-\Delta)| \Delta\}) = \int_{y=0}^{y=T} \bigg( \int_{-\infty}^{\infty} (x) p(X(t-y)=x)| \Delta=y) dx \bigg) p(\Delta=y) dy