Jim_Garrett@bd.com wrote:
> I'm a newcomer to Maxima. I'm really delighted to find this, and I intend
> to learn to use it well because, since it's GPL'd, no one will ever be able
> to take it away from me! Thanks to everyone who's contributed to it.
>
> I'm a statistician, so run into summations (over observed data points)
> often. I seem to be running into some walls regarding manipulating
> summations.
>
> I've been practicing with a well-known, simple example: demonstrating that
> the average is the maximum likelihood estimator for the mean of a Gaussian
> series of observations (independent and having common parameters mu and
> sigma). I'll use Latex notation. Here are a few questions:
>
> (If you wish to send me information off-list, please send it to this
> address and to my home address, jim_garrett@comcast.net.)
>
> 1. Suppose I define the log-likelihood function for a single data point
> $x_i$ (the log-likelihood is the log of the probability density function
> for the observations, regarding the observed data as fixed and the
> parameters--mu and sigma here--as varying):
>
> loglik_1 : log(1/sqrt(2 * %pi) * exp(-1/2 * (x[i] - mu)^2));
try loglik_1(i) := log(1/sqrt(2 * %pi) * exp(-1/2 * (x[i] - mu)^2));
>
> Now if I see $x_1, x_2, \ldots, x_n$ data observations, the log-likelihood
> for the sequence is obtained by adding loglik_1, with a difference $i$ each
> time. I tried
>
> loglik : sum(loglik_1, i, 1, 10);
try
loglik : sum(loglik_1(i),i,1,10);
>
> but that gave me 10 * loglik_1 for a single arbitrary $i$. How can I
> communicate that each term in the summation is different?
>
> 2. Moving on, I meet something of this form:
>
> \[ \sum_{i = 1}^n (x_i - \mu) = 0. \]
>
> In order to solve for mu, I want to make this into two summations:
>
> \[ \sum_{i = 1}^n (x_i - \mu) = \sum_{i = 1}^n x_i - \sum_{i = 1}^n \mu.
> \]
>
> How can I do that?
There may be some simplification flag to do this, but
here's another way..
Maybe use some other operator for sum(*, i,1,n), say s. Declare s
to be a linear operator. then replace by sum.
>
> 3. Can Maxima realize that for one summation it's summing a constant,
> i.e., $\sum_{i = 1}^n \mu = n \mu$?
I think so.
>
> I'm sure I'll have more questions later on. These are standard things that
> arise in statistics, so once I figure this out I'll be able to advise other
> statisticians with a "design pattern", if you will.
>
> Thanks for any advice.
>
> Jim Garrett
> Becton Dickinson Diagnostic Systems
> Baltimore, Maryland, USA
>
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