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I have generated a random series of returns drawn from a normal distribution and generated a random price series by compounding these returns (X) so $P_i = P_1(1+X)^i$. I want to show the analytic comparison between the variance of this price series to the variance of a k-period moving average of the price.

What I am finding is that the variance of the price series (not price returns) converges to the variance of k period average price for $all$ values of k (so the variance of moving average price = price variance regardless of the averaging period) for large data sets. This surprised me as I would have expected the variance of the moving average to be scaled down by k regardless of the period.

I have tried to derive this result mathematically but without success. Here is the starting point of my calculation: $$MA_k = 1/k . {\sum^k_{i=1}P_i} = $$ $$=1/k . (P_1 + P_1(1+x) + ... + P_1(1+x)^{k-1})$$ $$P_1/k.{\sum^{k-1}_{i=0}}(1+x)^i$$

From here I use a power series expansion, ignore higher order terms, and take the variance of the expression. I think I should find this expression $var(MA_k) = P_1^2.var(x)$ but that's a guess. I can see I am not incorporating the dependency on the size of the data because there appears to convergence in variances between the rolling average of the price and the price, but I am not sure how to do that.

Any assistance in filling in the missing steps would be very welcome here.

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  • $\begingroup$ Shouldn't it be $P_i=P_{i-1}(1+x_i)$ ? $\endgroup$ – noob2 Jun 25 '17 at 20:52
  • $\begingroup$ Hi Noob, yes I believe that would make the formula recursive. I did not know how to solve the recursive formula. Instead I made P a constant and tried to solve for the variance as function of x through an expansion (which I treat as a random variable) and by ignoring $O(x^2)$ terms. $\endgroup$ – markm Jun 25 '17 at 21:03
  • $\begingroup$ Hi - I have edited the question to show my approach in more details and reduced to a single goal as opposed to the dual goals originally stated in the question $\endgroup$ – markm Jun 26 '17 at 8:42
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    $\begingroup$ I'm voting to close this question as off-topic because it belongs on the Cross Validated site. $\endgroup$ – Richard Hardy Jun 26 '17 at 19:06
  • $\begingroup$ Good suggestion Richard. The question does have a quantitative finance application in mind. However I can see that those with statistical/mathematical bent might find it more interesting (and easier to solve). $\endgroup$ – markm Jun 27 '17 at 8:40