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.

I saw this expression in a book and I cannot understand how did he get this expression. Suppose $Z_t$ and $D_t$ are some stochastic processes and we have these expressions, $Z_{t_k} - Z_{t_{k-1}} = (t_k - t_{k-1})(D_1 + ... + D_{n-k+1}) - t_{k-1}D_{n-k+2}$

$Z_{t_i} - Z_{t_{i-1}} = (t_i - t_{i-1})(D_1 + ... + D_{n-i+1}) - t_{i-1}D_{n-i+2}$

How did they get this covariance equation?

$Cov(Z_{t_k} - Z_{t_{k-1}}, Z_{t_i} - Z_{t_{i-1}}) = (t_k - t_{k-1})(t_i - t_{i-1}) \sum_{m=1}^{n-i+1} Var(D_m) - t_{i-1}((t_k - t_{k-1})Var(D_{n-i+2})$

Any hint appreciated as I cannot figure out which formula they are using here.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Assume $n-i+1 < n-k+1$ and that $D_i$ are independent. Substitute for $Z_{t_k}-Z_{t_{k-1}}$ and $Z_{t_i}-Z_{t_{i-1}}$ in ${\rm Cov}(Z_{t_k}-Z_{t_{k-1}},Z_{t_i}-Z_{t_{i-1}})$. Use linearity of covariance. Use the fact that independent random variables are uncorrelated. Use that fact that ${\rm Cov}(X,X) = {\rm Var}(X)$.

share|improve this answer
    
Of course, the same holds if $D_i$ are only assumed uncorrelated. –  Shai Covo Feb 25 '11 at 8:38
    
got it. linearity of covariance was new to me. –  user957 Feb 25 '11 at 10:07

Your Answer

 
discard

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.