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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need some help understanding an excercise.

Let $X_1, X_2, X_3 \sim N(-2,3).$

(right here there is an ambiguity about the second parameter: is it $\sigma$ or $\sigma^2$ ?)

First they calculate the variance

$$\sigma^2\left(\sum_{i=1}^3 i X_i\right) = \sum_{i=1}^3 i^2 \sigma^2 X_i = 14 \cdot 9 = 126$$

So far I think I understand. This result implies that $\sigma = 3$ . Am I correct?

Then they give the following line without any explanation:

$$\operatorname{cov}\left(\sum_{i=1}^3 i X_i, \sum_{i=1}^3 X_i\right) = \sum_{i=1}^3 i \cdot \sigma^2 X_i = 54$$

Can you explain what happens here? How did they derive $\sum_{i=1}^3 i \cdot \sigma^2 X_i $?

share|cite|improve this question
up vote 1 down vote accepted

Imagine expanding the product $(X_1+2X_2+3X_3)(X_1+X_2+X_3)$. By independence, or uncorrelatedness (not mentioned, but necessary) we have $\operatorname{Cov}(X_iX_j)=0$ if $i\ne j$. And of course $\operatorname{Cov}(iX_i,X_i)=i\operatorname{Var}(X_i)$.

Remark: It is distressing that $N(a,b)$ has two interpretations. If the quoted calculation is correct, $b=\sigma$ is the intended interpretation. A gppd solution of the problem is to not use the abbreviation.

share|cite|improve this answer

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.