Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

From the book A First Course in Probability 8th ed by Ross, expected value linearity is defined as $ E[aX + b] = aE[X] + b $ where a and b are both constants and X is a random variable.

However a random process defined $ X(t) = Kcos(wt) $ where K is a random variable has an expected value of $ E[X(t)] = E[K]cos(wt)$. My question is how is this true when cosine is a function of t and not a constant.


share|improve this question
@AndréNicolas. Even if the distribution of $K$ involved $t$ as a parameter, you would still be able to get $\cos(w t)$ outside of the expectation. –  Learner Dec 8 '12 at 4:17
Yes, thank you, it was very imprecisely phrased. –  André Nicolas Dec 8 '12 at 4:38

1 Answer 1

up vote 1 down vote accepted

At this stage of your study, heuristically, you could consider constants everything that is not random. In the question, $X(t)=K \cos(w t)$ is a random function of $t$ only because $K$ is random. When you take the expectation, you could ignore everything that is not random. Were either $w$ or $t$ or both random, you would not be able to get $\cos( w t)$ outside of the expectation.

share|improve this answer
I am assuming this is the same case for variance? $ Var(X(t)) = cos(wt)^2Var(K)$ ? –  wi1 Dec 8 '12 at 6:14
Yes. Remember that $Var(X)=E[X^2]-E[X]^2$. So the rules that apply to the expectation follow for the variance from that formula. –  Learner Dec 8 '12 at 6:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.