# Expectation of weakly stationary process

Hi all I have a question. Suppose

$$X(t) = A \cos(t) + B \sin(t)$$

is a weakly stationary process, where $A$ and $B$ are random variables.

Why is it that the mean of $A$ and $B$ are necessarily 0? As I understand it, weak stationarity only requires that the $E(X) = \mathrm{constant}$.

• I've changed the acronym, since this makes the internet the better place :) – mpiktas Apr 11 '12 at 6:10

Take the expectation of $X(t)$:
$$E \big( X(t) \big) =E (A) \cdot \cos (t) + E(B) \cdot \sin(t)$$
and think what you have on the left hand side (constant), and on the right hand side (real function). Now when the expression on the left hand side is constant for all $t$?
• $A$ and $B$ can't magically change with time to make $X(t)$ constant. That's not how random variables work. – Emre May 11 '12 at 6:48