Tracking Error

[coquin22]

pls confirm if it is not right methodology to calculate the TE


David posted
http://www.bionicturtle.com/how-to/...r_the_information_ratio_and_the_sortino_ratio

 

[David Harper CFA FRM]

Hi coquin,

No, 2.00 is the average active return, as shown by the formula.
Tracking error is the standard deviation (volatility) of the active return; hint: it is given as another answer ... i don't want to deny you the pleasure of calculating it, i am here to support you

 

[coquin22]

thanks David,
i think the question might be for active return, tracking error should be (2^2+(-4)^2+4^2+(-2)^2)^0.5=6.32

 

[David Harper CFA FRM]

sure, but i think you want [(2^2+(-4)^2+4^2+(-2)^2)/4]^0.5 for the desired answer, I actually think given the small sample that a sample standard deviation would also be acceptable (if not better) but, I guess, that's less conventional: sample volatility = [(2^2+(-4)^2+4^2+(-2)^2)/(n-1)]^0.5

 

[alexemil5]

I have check it many times, but couldn't find the actual reason to say it is wrong in calculations. Let me clear, David. I am waiting for your answer.



http://plagiarismsoftware.org/

 

[ABFRM]

sure, but i think you want [(2^2+(-4)^2+4^2+(-2)^2)/4]^0.5 for the desired answer, I actually think given the small sample that a sample standard deviation would also be acceptable (if not better) but, I guess, that's less conventional: sample volatility = [(2^2+(-4)^2+4^2+(-2)^2)/(n-1)]^0.5

it shud be 3 if it is n-1. because it punishes sample variance becuase of its biased estimator.

 

[David Harper CFA FRM]

I totally agree with ABFRM :[(2^2+(-4)^2+4^2+(-2)^2)/4]^0.5 gives the desired answer. 2 is only the average active return, but TE = standard deviation (active return). I also 100% agree with (n-1) as arguably better (i.e., unbiased variance if not standard deviation) but also unconventional. Thanks,

 

[ABFRM]

@David it makes huge difference in small samples so wat we shud use because a lot of references use n as an denominator.

 

[David Harper CFA FRM]

ABFRM I was just agreeing with you on the theory of (n-1), but I can't recall ever seeing it employed for TE. I have always encountered the population sigma /n; i.e., the MLE estimate if you will, NOT (n-1) the "sample" unbiased estimator. And in the FRM readings, the applications (to my knowledge) have always similarly employed a population standard deviation for TE (pretty sure even Bodie Kane uses population here, although they confuse their terms and call TE --> benchmark risk). I think this is because at small samples sizes, where as you say it makes a difference, the problem is that TE tends not to be significant, just like IR. Significance tends to require a large sample anyway; i.e., (n-1) doesn't do enough to get a useful t-ratio anyway, is basically Grinolds point. For what it's worth,