Quants question on correlation.

Kenny James , CFA , is calculating the covariance of his large cap mutual fund returns against the returns generated by intermediate governmnt bonds over the past 5 yrs. The following information is provided: (A-a) is the annual return minus the mean return forthe large cap mutual fund, (B-b) is the annual return minus the mean return for the intermediate government bonds.

(A-a) (B-b)

Year 1 -23.4 4.2

Year 2 -13.2 -1.6

year 3 -10.4 4.8

Year 4 19.7 -12.2

Year 5 27.2 4.7

The covariance between the mutual fund and government bond is closest to:

A. -47.9

B.-59.9

C. -239.6.

Ans is B

:sum [(A-a)(B-b)]/4.

Why the sum is been divided by 4??

You need to divide the sum product of (A-a) and (B-b) by (n-1) = 4.

N-1 is done in the denominator to reduce its value so that effect of underestimation of the sample co-variance is removed to make it the representative of the population. While calculating the co-variance of population this adjustment is not done. This is sometimes tricked with mean. In calculating the mean for population or sample the denominator is “n” whereas in co-variance the denominator for population is “n” and for sample is “n-1”.

If you are working with the population, you have all data and you know what you have calculated is accurate. If you take a sample and are using that infer population values you have just add risk (the risk that your Sample is not representative) into your calculation of the risk.

mohammed is on the right track there. The technical reason is that you want to make the sample covariance an “unbiassed” estimator of the population covariance. When you have those differences (a[i]- a-bar) note that a[i] is likely to be closer to a-bar than it is to the true means of the a’s because a[i] is included in the calculation of a-bar. Some math beyond the scope of the CFA exam says that the right way to adjust for that is to divide by n-1 instead of n.