Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Suppose $X_1,\ldots,X_n$ are independent random variables with $E(X_i)=\mu$ and $\operatorname{Var}(X_i)=\sigma^2$ for all $i=1,\ldots,n$. Let $S_k=X_1+\cdots+X_k$. Find $\rho(S_k,S_n)$.

So I know that $\rho(S_k,S_n)=\operatorname{Cov}(S_k,S_n)/\sigma_k \sigma_n$ and $\operatorname{Cov}(S_k,S_n)=E(S_kS_n)-E(S_k)E(S_n)$. Since $S_k=X_1+\cdots+X_k$ I assume that $S_n=X_1,\ldots,X_n$. I'm not sure how to start this.

share|cite|improve this question
Do you mean $\operatorname{Cov}(S_k,S_n)=E(S_kS_n)-E(S_k)E(S_n)$? You have a comma in $E(S_k,S_n)$ – Thomas Andrews Nov 16 '12 at 18:28
Yes that was a typo. – Sprock Nov 16 '12 at 18:40
up vote 1 down vote accepted

There are significantly more efficient ways to go, but let's see how to complete your nearly complete calculation.

I assume you know how to find $\sigma_k$ and $\sigma_n$. Also $E(S_k)$ and $E(S_n)$ are no problem. So the only issue is computing $E(S_kS_n)$.

Note that $S_n=S_k+(X_{k+1}+X_{k+2}+\cdots +X_n)$. Let $Y=X_{k+1}+X_{k+2}+\cdots +X_n$. Then $S_kS_n=S_k^2+ S_kY$. Because $S_k$ and $Y$ are independent, we have $$E(S_kS_n)=E(S_k^2)+E(S_k)E(Y).$$ The term $E(S_k^2)$ is easy to compute, it is closely related to the variance of $S_k$. And there is no difficulty finding $E(Y)$. Now put the pieces togther.

share|cite|improve this answer

Given independence, and using bi-linearity: $$ \mathbb{Cov}\left(S_n, S_m\right) = \sum_{p=1}^n \sum_{q=1}^m \mathbb{Cov}\left(X_p, X_q\right) = \sum_{p=1}^n \sum_{q=1}^m \delta_{p,q} \mathbb{Var}(X_p) = \sum_{p=1}^{\min(n,m)} \mathbb{Var}(X_p) = \sigma^2 \min(n,m) $$

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.