Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: MC estimators of the variance and standard deviation for an arbitrary distribution
Replies: 1   Last Post: Mar 26, 2013 5:20 PM

 Messages: [ Previous | Next ]
 deltaquattro@gmail.com Posts: 77 Registered: 7/21/06
MC estimators of the variance and standard deviation for an arbitrary distribution
Posted: Mar 26, 2013 11:19 AM

Hi,

the MC estimator and corresponding interval estimate, for the standard deviation of a normal distribution of unknown mean and variance, are well-known. I would like to find something also for the case of an arbitrary distribution of unknown mean and variance, which is much more often the case when propagating uncertainties through fluid dynamic codes (my case). That's what I came up with so far:

Let X_i be a sequence of real RVs i.i.d with mean mu and variance sigma^2. We form the RVs

Y_i=(X_i-mu)^2

They are also i.i.d., clearly, with E[Y_1]=sigma^2 and Var[Y_1]=M_4-sigma^4 where

M_4=E[(X_1-mu)^4]

which I assume to be finite. I introduce the estimator of the mean of the Y_i,

Y^ = 1/N * sum[i=1,N] (Y_i)

Applying the Central Limit Theorem, I get

Y^ - z_(1-delta/2)*sqrt((M_4-sigma^4)/N) <= sigma^2 <= Y^ + z_(1-delta/2)*sqrt((M_4-sigma^4)/N)

where z_(1-delta/2) is the 1-delta/2 percentile of the standard gaussian distribution. Of course, this expression is useless, because mu and sigma, which are unknown, appear in it. With a bit of hand-waving, I would simply substitute them by the sample mean and the sample variance of the X_i, i.e.

X^ = 1/N * sum[i=1,N] (X_i)
V^ = 1/(N-1) * sum[i=1,N] (X_i-X^)^2

Is this correct? In that case, the above interval estimate would become asymptotic, i.e., valid for sufficiently large N. Is there a better way?

Thanks

Best Regards

deltaquattro

Date Subject Author
3/26/13 deltaquattro@gmail.com
3/26/13 mathman1