# 104.3 Pages 6 - 7, PQ Set Ch 23 - Hull

[Dr. Jayanthi Sankaran]

Hi David,

In your question below:

104.3 Assume that instead of an infinite historical window, we want to apply EWMA to only the last ten daily returns (i.e., including T-10 but no further back). What is the weight assigned to the seventh prior daily squared-return if we want the sum of the weights to equal 1.0?
a. 5.97%
b. 6.97%
c. 7.97%
d. 8.97%

and the answer below:

104.3 D. 8.97% Weight (d,N) = (1-lambda)/(1-0.94^N)*lambda^(d-1), where N = total days in truncated series
Weight (7,10) = (1-0.94)/(1-0.94^10)*0.94^6 = 8.97%

Linda Allen discusses the issue of a historical window that is isn't long/infinite: the problem is that the weights given by Weight (d) = (1-lambda)*lambda^(d-1) will not sum to 1.0 as the series is not infinite.
So she gives two solutions: 1, increase the days so the remainder is small enough not to matter, or
2. this solution in 104.3: basically distributed the reminder to all the weights, thereby forcing them to sum to 1.0. These weights are given by

Weight (d,N) = (1-lambda)/(1-0.94^N)*lambda^(d-1)
i.e., the original weight is multiplied by 1/(1-lambda^N) where N is the total number of days in the series.

I don't understand how you get the following - please explain:

So, the difference is, here are the 10 weights assuming an infinite series:
6.00%
5.64%
5.30%
4.98%
4.68%
4.40%
4.14%
3.89%
3.66%
3.44%

but they only sum to 46.14%
In solution #2, the weight are (effectively) multiplied by 1/46.14% = 2.16, and this produces an exponentially declining series but where the sum of the weights is 1.0:
13.00%
12.22%
11.49%
10.80%
10.15%
9.54%
8.97%
8.43%
7.93%
7.45%

Thanks a tonne!
Jayanthi

 

[ShaktiRathore]

Hi,
Previous m days weights are exponentially decreasing in order lambda^0*k,lambda^1*k,lambda^2*k,..,lambda^m-1*k for m days right from previous day to mth day. Now all weights should sum to 1 implies lambda^0*k+lambda^1*k+lambda^2*k+....+lambda^m-1*k=1=>k*(1-lambda^m)/(1-lambda)=1=>k=(1-lambda)/(1-lambda^m)
So the weights for 1st,2nd, mth previous day becomes lambda^0*(1-lambda)/(1-lambda^m),lambda^1*(1-lambda)/(1-lambda^m),..in general ith day weight is lambda^i-1*(1-lambda)/(1-lambda^m) weight for i=7th day=.94^6*(1-.94)/(1-.94^10)=8.97%
Thanks

 

[Dr. Jayanthi Sankaran]

Hi Shakti,

Thanks for the detailed answer - appreciate it! Although, computing 8.97% is a piece of cake, I am having trouble getting 6% and so on.....I am missing something here.....can you show me the illustration for computing 6% at the start of the tenth day series.

Jayanthi

 

[ShaktiRathore]

Hi jayanti,
For infinite series weights just put limit m>>infinity in formula w(i)= lambda^i-1*(1-lambda)/(1-lambda^m) so that lambda ^m=0 as m tends to infinity. Thus for infinite series our formula for ith day weight is w(i)=lambda^i-1*(1-lambda), so w(1)=.94^0*(1-.94)=.06 or 6%,w(2)=.94^1*(1-.94)=.94*.06=.0564 or 5.64% and so on.
Thanks

 

[Dr. Jayanthi Sankaran]

Hi Shakti,

Got it! Thanks for your detailed and simply reply - apparently, I was relegating a lot of terms to the exponential power.

Jayanthi