Trending Content
Career Resources
WSO Free Modeling Series - Now Open Through
May 31, 2024
9 Free Financial Modeling Lessons
Get instant access to lessons taught by experienced private equity pros and bulge bracket investment bankers including financial statement modeling, DCF, M&A, LBO, Comps and Excel Modeling.
+Bonus: Get 27 financial modeling templates in swipe file
+Bonus: Get 27 financial modeling templates in swipe file
Corr(A,B)=0 -> Cov(A,B)=0 Corr(B,C)=0 -> Cov(B,C)=0
Cov(A,C)=E(AC)-E(A)E(C) Cov(A,B)=E(AB)-E(A)E(B)=0 -> E(AB)=E(A)E(B) Cov(B,C)=E(BC)-E(B)E(C)=0 -> E(BC)=E(B)E(C)
To prove that Corr(A,C)=0 the variables should be independent. While independence implies uncorrelatedness, uncorrelatedness does not imply independence.
If the variables are independent, then it naturally follows that Corr(A,C)=0
So assuming the variables are independent, can you still prove it mathematically?
I understand the formulas and logic you provide, but isnt there a way to prove it using the formulas?
If X and Y are independent, then the following holds for expectations
E(XY) = E(X)E(Y)
Hence, it follows directly from the covariance formula that Cov(A,C)=0.
Alternatively, Cov(A,B)=E(AB)-E(A)E(B)=0 -> E(AB)=E(A)E(B) -> E(A) = E(AB)/E(B) Cov(B,C)=E(BC)-E(B)E(C)=0 -> E(BC)=E(B)E(C) -> E(C) = E(BC)/E(B)
Cov(A,C)=E(AC)-E(A)E(C)=E(AC) - [E(AB)/E(B)]*[E(BC)/E(B)]=E(AC)-E(AB)E(BC)/(E(B))^2
Given that that the variables are independent, the expression reduces to Cov(A,C)=E(AC)-E(AC)=0
Since Corr(A,C) = Cov(A,C)/st.dev(A)*st.dev(C), it follows that Corr(A,C)=0
The alternative route is redundant, though, once you know that the variables are independent.
Hope it helps.
...how are you coming to the conclusion that A is independent of C? The fact that cov(A,B) = 0 and cov(B,C)=0 only tells you that A is independent of B and C is independent of B but you don't necessarily know the relationship between A and C. The relationship could actually be A=C. You're assuming that step it seems, and as a result you can't really say whether cov(AC) would or wouldn't be 0
Come on, NY, read the math! You can derive the result from Cov(A,B)=0 and Cov(B,C)=0 (see the alternative route). However, you need to make the assumption of independence.
The fact that Cov(A,B)=0 does not mean that A is independent of B! However, if two variables are independent, it implies zero covariance.
corr(a,b)=corr(b,c)=0 Set correlation equations equal to each other (A-E(A))(B-E(B))/(s.d(A))(s.d(B))=((B-E(B))(C-E(C))/(s.d(b))(s.d(c))) Multiply both sides by s.d(B)/(B-E(B)) and get: (A-E(A))/s.d(A)=(C-E(C))/s.d(C) Multiply both sides by (A-E(A))/s.d(A)) and you get (A-E(A))^2 / (s.d. (A))^2 =((A-E(A))(C-E(C))/(s.d(A))(s.d(C))) Which is equal to Var(A)/Var(A) = corr(A,C) Therefore corr(A,C)=1
Someone please tell me if you see a flaw, but it looks good to me
Nostrum ipsa aliquid placeat ex deserunt. Ducimus dolores velit a porro rerum est omnis. Aut occaecati molestiae et expedita.
Eos explicabo qui non sit iure eveniet et. Architecto sit laboriosam provident non consequatur. Velit dolorem aperiam et molestiae. Sit dolorem sit doloribus illo iure et. Et minima perspiciatis quis natus nemo.
Temporibus dolore saepe corporis nostrum reiciendis. Modi eos quasi quis a sed enim numquam. Et non ut maxime magnam veniam quia asperiores. Suscipit numquam officia quis omnis possimus nobis omnis.
Explicabo commodi consequuntur qui aut reprehenderit. Possimus voluptatum accusantium architecto tempora consequuntur eius.
See All Comments - 100% Free
WSO depends on everyone being able to pitch in when they know something. Unlock with your email and get bonus: 6 financial modeling lessons free ($199 value)
or Unlock with your social account...