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Hi: I know it's true but I don't know how to prove that exponentially weighted moving average when view ed as a function of $\lambda$, is strictly convex. The exponentially weighted moving average for a time series X_t is defined as $(1-\lambda) \sum_{i=0}^{t} \lambda^{i}X_{t-i}$ for any $t > 0$. Note that $t$ is allowed to go to infinity and $0 < \lambda < 1$. Thanks for the proof or even a reference. I couldn't find much on convexity of the EWMA. Also, my apologies if this is not the correct tag for this question.

                                                               Mark
mark leeds
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  • I'm curious how you know it to be convex. It certainly does not appear to be true to me. For instance, $(1-\lambda)\lambda$ is concave; and $(1-\lambda)\lambda^k$ for $k\geq 2$ is either convex nor concave. – Michael Grant Nov 21 '14 at 22:46
  • Hi Michael: I read the statement in someone's thesis but it could be incorrect. My apologies. Also, some define it without the $1-\lambda$ term so if that's causing the lack of convexity, that would be why. – mark leeds Nov 22 '14 at 13:49
  • Even without the $1-\lambda$ term it is not convex. What if all of the $X_t$ values are negative? – Michael Grant Nov 22 '14 at 14:12
  • Hi Michael: Thank you for your input. The link of the paper I refererred to is below. The statement is on page 7 and the paragraph starts with the word : "Nevertheless". I don't doubt that you are correct but I am going to simulate the process and use the definition of convexity with different values of t_1 and t_2 to see it in action. Thanks again and I'll let you know what comes from that exercise.http://www.er.ethz.ch/publications/MAS_Johan_Boissard_Dec12.pdf – mark leeds Nov 22 '14 at 22:55
  • All this paper is saying is that $(1-\lambda)\lambda^k$ is a strictly convex function of $k$. This is evidenced by the introduction of an alternate weight set with an inflection point; see figure 1.3. – Michael Grant Nov 22 '14 at 23:08
  • hi michael: I'm sorry for the dumb question but, if the sum of convex functions is convex, then isn't the sum convex which is the function I referred to in my initial email ? I guess it doesn't but if you could explain that ( my brain and math don't always see eye to eye :) ) , it's appreciated. Mark – mark leeds Nov 23 '14 at 00:45
  • michael: I think I understand now. The sum of convex functions is convex but you're not saying that $(1-\lambda)\lambda X_t$ is a convex function of k for ALL $X_t$ but rather Just for $X_t = 1$. Is that the correct understanding ? Thanks again and I do appreciate your help. – mark leeds Nov 23 '14 at 00:49
  • The important thing is that the paper you are citing is not claiming that the EWMA is a convex function of $\lambda$. Rather, it is saying that $h(k)=(1-\lambda)\lambda^k$ is a convex function of $k$. – Michael Grant Nov 23 '14 at 01:17
  • But going back to the question you originally asked: $f(\lambda)=(1-\lambda)\lambda^k$ is neither a convex nor concave function. Drop the $(1-\lambda)$ term and it is. However, the sum of those terms is only convex if the $X_t$ values are nonnegative. No, they do not have to be equal to 1, just positive or zero. – Michael Grant Nov 23 '14 at 01:20
  • thanks michael. I get it now. I looked quickly at the paper without reading it carefully. and your explanation was really great. unfortunately, I can't assume that my $X_t$ are zero or positive which means it's not convex which means that there can be local minimums which is not good when doing optimization over $\lambda$. – mark leeds Nov 23 '14 at 01:36

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