Re: Project Management

[Lloyd Provost <lprovost@fc.net>]

Jean-Marie Gogue wrote:

> Rudi, In your post dated 30 Jul 2000 ...
> >When the distribution is normal the sum of the squares of task variances is
> >equal to the variance of the projects completion time. This means that the
> >standard deviation of the project is much smaller than the sum of the
> >standard deviations of the tasks!
> >
> >What is the situation when the distribution is skewed? Does the the same hold
> >true? Is the standard deviation of the project still much smaller than the
> >sum of the std deviations of the tasks? Is the project even safer because of
> the
> >skew or is it less safe (vs a situation where there is no skew?)
>
> ...... Let me add my 2 cents contribution to your first question.
> The variance of the sum of normal and independent variables is equal to the
> sum of the variances. You know this theorem. There is no equivalent theorem
> applicable to any random variables. Therefore it's useless to look for a
> calculation giving the variance of the sum of random variables in general.
> .......
>

There is a general theorem from mathematical statistics about the variance of sums
of random variables (no normality assumption). This will be in most introductory
mathematical statistical books. From Feller, William, "An Introduction to
Probability Theory and Its Applications", Vol I, Third Ed, 1968, page 230:
If X1, .... Xn are random variables with finite variances (sigma1, ...sigman) and
Sn = X1 + .... + Xn, the the Variance of Sn = sum of Variances + 2 sum of
covariances (covariances summed over all pairs).

If the Xj ar mutually independent, then the variance of the sum is = to the sum of
the individual variances.

In Rudi's problem, the skewness will result in larger individual variances than
one would get from symmetrically distributed data.

Lloyd Provost
API
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