Hello,I am new to this forum and I have a question concerning the square-root-rule.First, concerning the application to volatility.Application to log returns is straightforward, however you can approximate log returns with simple returns and you then obtain the square-root-rule for the volatility of simple returns as well.Let's call R_01 the simple return for one period which we approximate by R_01^log (the 1 period log return).As we use a first order taylor approximation we would have a certain error. My question is, how does this error influence the error of approximating vol_01 by vol_01^log and consequently what is the error for vol_0n approximated by vol_0n^log = sqrt(n)*vol_01^log? (dependent on n).The 2nd question is about VaR.Using the approximation above I found motivations for the use of the square-root-rule for two-parameter distributions( location-scale family). But I also saw several sources claiming that it was approximately true in general. Is there a mathematical motivation for the use of the square-root-rule without a distribution assumption?

the - in (1-\mu) is supposed to be a +.but that doesn't change anything

To answer your other post, the general motivation for the square-root ruleis an iid assumption. Suppose R(i), where R(i) is the simple return in period i is drawn independently each period from an identical distribution.Then, the final wealth after N periods is W(N) = Prod_i (1 + R(i)), where Prod means "product of the terms".So log W(N) = X(1) + X(2) + ... + X(N), where X(i) = log (1 + R(i)) are also iid.Then, you should prove to yourself that Var log[W(N)] = N sig^2, where sig^2 = Var X.Take a square root of both sides and you have the square root rule for general distributions.To answer your last, post, if you insist on analyzing Var W(N) = Var Prod (1 + R(i)),then things are -completely different-. I suggest you take thecase where the R(i) are distrbuted log-normally. Do a computationand you will find that Var W(N) = e^(2 mu N) [exp(2 sig^2 N) -1],so the variance grows exponentially with N. Take a square-rootand you still have exponential growth. So, the square-root rule becomes anexponential rule.Basically, what is going on when things are compounding is that everything is growing at an exponential rate, including normal fluctuations from the trend.

I also assumed i.i.d. returns.But I focussed on simple returns whereas you use log returns.Frequently, one approximates simple returns by log returns and then claims that the square-root rule holds by your derivation.I am interested in the error we obtain by doing so. That is why I used the delta-method to obtain the asymptotical variance of the n period simple return which results in a linearly increasing volatility (assuming a zero-mean...this stems from the application on VaR)So my question now is whether there is any problem with using the delta-method. The square-root-rule is wrong for simple returns. You can compute the actual variance easily and obtain Var[R_n] = (sig^2 + (1+mu)^2)^n - (1+mu)^2n

Yes, your delta-method approximation is generally wrong in its dependence on N.Proof: it fails to reproduce the correct answer: exponential growth of variance with N when the iid returns are log-normal.

Actually I don't see your pointwe have Var(f(X))~ (f'(E[X])^2 *Var(X)which in the case of X= 1+R and f(x) = x^nyields n^2* (1+E[R]])^2(n-1) * Var[R]then I assumed that E[R] = 0 ...and then we have growth with n^2*Var[R]the assumption of E[R]= 0 also removes the compounding effect to summarize:I am NOT assuming i.i.d. log returns. the square-root rule trivially holds for those.I am looking at i.i.d. arithmetic/simple returns (not e^(log return) )and the error we get by using the square root rule (which we motivate by ln (1+R)= R^log ~ R for small R).the error obviously gets really large for a non-zero mean as the square-root-rule ignores compounding. for a zero-mean though it grows slower and I am interested in the exact rate depending on n (not depending on the return size).also...you were talking about exponential growth. this would even in the case of a non-zero mean only be obtained in the limit as I am considering n period arithmetic returns.

QuoteOriginally posted by: AlanTo answer your other post, the general motivation for the square-root ruleis an iid assumption. Suppose R(i), where R(i) is the simple return in period i is drawn independently each period from an identical distribution.Then, the final wealth after N periods is W(N) = Prod_i (1 + R(i)), where Prod means "product of the terms".So log W(N) = X(1) + X(2) + ... + X(N), where X(i) = log (1 + R(i)) are also iid.Then, you should prove to yourself that Var log[W(N)] = N sig^2, where sig^2 = Var X.Take a square root of both sides and you have the square root rule for general distributions.To answer your last, post, if you insist on analyzing Var W(N) = Var Prod (1 + R(i)),then things are -completely different-. I suggest you take thecase where the R(i) are distrbuted log-normally. Do a computationand you will find that Var W(N) = e^(2 mu N) [exp(2 sig^2 N) -1],so the variance grows exponentially with N. Take a square-rootand you still have exponential growth. So, the square-root rule becomes anexponential rule.Basically, what is going on when things are compounding is that everything is growing at an exponential rate, including normal fluctuations from the trend.Hi, can we possibly compute upper and lower bounds on root(Var(prod(1+Ri))) as some functions of "root n" and then proceed to prove the "more or less squareroot" law ?