Correlation Question

C

CFA_NOVA

New member
Joined
Jun 18, 2026
Messages
0
Reaction score
0
Does anyone know of a reasonable way to explain/interpret/understand the correlation formula highlighted on page 176 of Schweser Book 1? I recall this from portfolio mgt. course several years ago and applied it to final term paper... but professor's note: "Know this formula" is a bit daunting. Any thoughts???
 
I don't have the books on me.
You talking about:
corr(X,Y)=cov(X.Y)/(sdx*sdy)?
 
Cov(X,Y) could be from negative infinity to infinity, sdx and sdy are non negative, and corr is from -1 to 1. So if you cannot remember if you should use * or /, you can judge by the value.
 
Covariance is the degree at which the two variables move together, since it's a squared figure, it only give the sense of a positive or negative relationship. This is similar to variance, in that variance gives us the positive values. We square the deviations to eliminate positive or negative values and it gives weight to larger variations. In the same way covariance is squared to give weight to larger variation of the same direction, this is then totaled and divided by DF.

Correlation on the other hand "Standardizes" the covariance in the same way Standard Dev. is rooted (^1/2). The reason for this is because if you had a CovA = 1200 and CovB = 12, and STDev X (A)* STDev Y (A) = 1200 and a STDev X (B)* STDev Y (B) = 12, the two Correlation would be equal to 1. If you just interperted the CovA and CovB, you may conclude that CovA has a stronger positive relationship than B. This standardization allows you to compare relationships between two variable on the same footing.
 
Just think about it like "Correlation is to Covariance as STDev is to variance; Corr. is standardized Cov. and STDeV is Standardized Variance".
 
James, I respect you very much, but I don't like that analogy at all. St.Dev is the square root of the variance, but correlation is not the square root of covariance.
 
I agree that the analogy is general and not binding overall, the analogy is phrased and only refers to magnitudes in which they are relatively presented in. As stated I'm only saying they are both standardized for the relative magitudes of the data sets (in terms of the units) nothing more. The functions to arrive at the relative magnitude differ, the end result is the same.

*EDIT* I do agree it's not the best analogy though in terms of phrasing.



Edited 3 time(s). Last edit at Saturday, August 12, 2006 at 09:19PM by jamespucyk.
 
Based upon how I put my other post, it seems my phrasing is a little off today, especially when it comes to describing abstract statistical things, eh Joey ;).
 
you both offered some good info... and from James' analogy, I got the idea... would be about the same way I'd explain it too... now that I understand it.

Much appreciated...
 
I dont want to be a prick, but i think this is rather basic....

But well done guys for explaining the concept.
 
You comment was not necessary, but if you are so smart, please stick around we made you!

Respectfully,

jbnjc
 
jbnjc Wrote:
-------------------------------------------------------
> You comment was not necessary, but if you are so
> smart, please stick around we made you!
>
> Respectfully,
>
> jbnjc


Sorry, I mean "we may need you"
 
You must log in or register to reply here.
Back
Top