Exam #1, Afternoon Session: Question # 79 regarding Covariance between 2 assets and Correlation Coefficient. Can anyone tell me why we use 71.75 / 12 to find Covariance, but when we calculate correlation coefficient we use 71.75 / 19.49 x 11.62 ? Why don’t we use (71.75 / 12) / 19.49 x 11.62 to find correlation coefficient? I’d type the whole question in, but the way they give the info on the question is a little weird. But from what I can tell, there are 12 observations of return for 2 assets. Here is the info we are given: Covariance (I think) = 71.75 Variance for A = 379.9 Variance for B = 135.06 What is the covariance between the 2 sets of returns and what is the correlation coefficient?
correlation (A,B) = covariance(A,B)/[stdev(A)*stdev(B)] covariance is estimated as [1/(n-1)]*(sum[A_i-mean(A)]*[B\_i-mean(B)] or [1/n] if mean for A and mean for B are known.
Sorry, I am still not getting it. I’ll be a little more clear. The answers given by Schweser are: Covariance = 5.98 Correlation coefficient = 0.32 I guess my question is, if Covariance = 5.98 why does coefficient correlation not equal 5.98 / 19.49 x 11.62, or 0.026? Why are they using 71.75 as covariance to calculate correlation coefficient?
from what you told me, you are correct. if covariance (A,B) is 5.98, stdev(A) = 19.49, stdev(B) = 11.62, then correl = 0.026 either their answer is wrong or you were incorrect describing the problem.
Covariance isn’t 71.75. Its (71.75/12-1) = 6.52. 6.52/[(379.9/11)^.5][(135.06/11)^.5] = Correlation coefficient correlation coefficient is the covariance divided by the standard deviations NOT the variances. Therefore, variance^.5
I figured it out. Since we are using historical data, we need to divide each of the variances by the number of observations, and then take the square root to find the standard deviation. In this case, Covariance does equal 71.75 / 12 = 5.98. std dev of A = (379.9/12)^.5 = 5.627 and std dev of B = (135.06/12)^.5 = 3.355. So correlation coefficient = 5.98 / (5.627)(3.355) = 5.98 / 18.88 = .316 We would not use n-1 in this case b/c this is a population, not a sample.
rbford, you should review your definition of variance and you misinformed us. variances were not 379.9 and 135.06. they were 379.9/12 and 135.06/12 (or 11, if it’s a sample).
incidentally – if you used jalmy8’s formula above… bcos you consistently use 11 all over you would get the same answer of .316 as well. So I think the point is, just be consistent all over… use either 12 or 11 all over, not a combination of 11 in some places, 12 in others… CP