Let $X_1,X_2,X_3$ be uniform random variables on the interval $(0,1)$ with $$\newcommand{\cov}{\text{cov}} \newcommand{\var}{\text{var}}\cov(X_i,X_j)=\frac{1}{24} \text{ for } i,j\in\{1,2,3\}, i\ne j$$

Calculate the variance of $X_1+2X_2-X_3$

I did the below approach:

\begin{align*}\var(x_1+2x_2-x_3)&=\cov(x_1+2x_2-x_3,x_1+2x_2-x_3)\\&=\cov(x_1,x_1)+2\cov(x_1,x_2)-\cov(x_1,x_3)\\&+2\cov(x_2,x_1)+4\cov(x_2,x_2)-2\cov(x_2,x_3)\\&-\cov(x_3,x_1)-2\cov(x_3,x_2)+\cov(x_3,x_3)\\&= \frac{2}{24} -\frac{1}{24}+\frac{2}{24}-\frac{2}{24}-\frac{1}{24}-\frac{2}{24}=-\frac{2}{24}\end{align*} but the answer is $\dfrac{5}{12}$???

As $\cov(x_i,x_j)=\frac{1}{24} \text{ for } i\ne j$ so, $\cov(x_1,x_1), \cov(x_2,x_2),\cov(x_3,x_3)$ I assumed to be $0$. Is my assumption is wrong and I need to calculate the $\cov(x_1,x_1)$ from the UDF given.

  • $\begingroup$ i kind of did that didn't i? $\endgroup$ – Prabir Acharya Jul 10 '16 at 6:53

For any $X$, $\text{Cov}(X,X)=\text{Var}(X)$. So you for that part of the calculation you need to calculate the variance of the uniform distribution on $(0,1)$.

That variance is $\frac{1}{12}$, so you will be adding $\frac{1}{12}+\frac{4}{12}+\frac{1}{12}$ to the number you calculated. Note that $-\frac{2}{24}$ cannot be right: variance is always non-negative.

  • $\begingroup$ ok!! so my assumption was wrong $\endgroup$ – Prabir Acharya Jul 10 '16 at 6:54
  • $\begingroup$ got the answer as 5/12, thanks for the help $\endgroup$ – Prabir Acharya Jul 10 '16 at 6:57
  • $\begingroup$ @PrabirAcharya: You are welcome. $\endgroup$ – André Nicolas Jul 10 '16 at 7:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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