Sign up ×
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.

I am not sure how to determine the variance and mean for the following equation system. I'd greatly appreciate your help!

$ f_\xi(x) = \left\{ \begin{array}{l l} 10k_1x & \quad \text{for $0<x<1$}\\ 0 & \quad \text{for other}\\ \end{array} \right. $

$ f_\eta(x) = \left\{ \begin{array}{l l} 4k_2x & \quad \text{for $0<x<1$}\\ 4k_2(2-x) & \quad \text{for $1\leq x<2$}\\ \end{array} \right. $

I have determined the following:

$k_1 = \frac{1}{5}$


... and I believe to have found their individual variance and mean:





Now to the gist of my question: how do I calculate the variance and the mean for $10\xi+2\eta$? Thank you!

share|cite|improve this question
Since it is homework, I will not calculate mean, variance apart from giving the odd hint. But for the second, it should be obvious by symmetry that the mean is $1$. If you did it by integration, there was a slip. For the variance, often the easiest thing is to use $E(X^2)-(E(X))^2$. – André Nicolas May 20 '12 at 18:15

1 Answer 1

up vote 4 down vote accepted

Let $X$ and $Y$ be random variables, and let $a$ and $b$ be constants.

We have $E(aX+bY)=aE(X)+bE(Y)$.

If $X$ and $Y$ are independent then $\text{Var}(aX+bY)=a^2\text{Var}(X)+b^2\text{Var}(Y)$.

(If $X$ and $Y$ are not independent, one can say little about the variance of the linear combination.)

share|cite|improve this answer
Not so Andre Var(X+Y)=Var(X)+Var(Y)+2 Cov(X,Y) and so Var(aX+bY)=a^2 Var(X) + b^2 Var(Y) + 2ab Cov(X,Y). Of course he needs to calculate the covariance in his example. – Michael Chernick May 20 '12 at 16:30
@MichaelChernick: Agreed! However, the density function information provided in the problem tells us nothing much about the covariance, apart from the very weak bounds of Cauchy-Schwartz. – André Nicolas May 20 '12 at 18:12
You are right I thought he specified the joint distribution and not just the two marginals. – Michael Chernick May 20 '12 at 19:40

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.