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 $t$?

Also note that your last statement is incorrect. Weak stationarity requires more than just a constant mean.

• Ah thank you, I get it now, but just an afterthought, since A and B are RV, is it possible that lets' say at t=0, A=1, giving X(1) = 1 and then at say t=2, A=1/cos(2) to give back X(2) = 1 and then for all t, A and B changes such that X(t) = 1, which would still fulfill E(X) = constant? In such a case, A and B doesn't have to be 0 all the time? – John Tan Apr 11 '12 at 12:58
• $A$ 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