Chapter 1 Covariance

[Aisha]

Hello
I'm unable to recall the formula used here for calculating the covariance . This is the solution of Question 1.2(i) from chapter 1 practice questions ( on page 29)

 

[Calm]

Hello
I'm unable to recall the formula used here for calculating the covariance . This is the solution of Question 1.2(i) from chapter 1 practice questions ( on page 29)

These are the covariance formulae used in NTU's CT6 and CT4, respectively.

 

[Aisha]

Hey
Can you please explain how have we used these formulas in solving the above problem?

 

[Calm]

Hey
Can you please explain how have we used these formulas in solving the above problem?

 

[Aisha]

Thanks!

 

[Bill SD]

Thanks Calm and Aisha.
A related question: The solution for Q1.7 uses the fact that (0,t) and (t,t+s) are non-overlapping time periods. But why does this justify why Cov((X(t), X(t+s)-X(t)) =0?
Surely this should be broken down to Cov(X(t),-X(t)) +Cov(X(t),X(t+s)). And the first expression Cov(X(t),-X(t)) = var (X(t)) so the solution to the question should be 2var (X(t)) Tia

 

[Andrew Martin]

Hello

The process has independent increments, so the increment \( X_{t+s} - X_{t} \) is independent of \( { X_u, 0 \le u \le t } \). This means that \( X_{t+s} - X_{t} \) is independent of \( X_t \) and so \(cov(X_{t+s}- X_{t}, X_t ) \) is 0.

Hope this helps

Andy