geometric distribution on interval Is it possible to define a geometric distributed variable with parameter p on an interval $[a,b]$, such that
$$ c \cdot \sum_{j=a}^b (1-p)^j p=1$$
with a scale factor $c$.
Is that reasonable?
 A: It is possible to define it, although the name "geometric distribution" may be confusing, since it commonly refers to a distribution on a different domain. "Geometric distribution on a bounded domain" seems correct – maybe it's used somewhere. You might also call it a "geometrically varying" distribution if you like, since subsequent probabilities are in a fixed ratio, just like for the ordinary geometric distribution.
Such a distribution can in fact be rewritten in the form $\mathrm{p}(i \vert \lambda) = \frac{1}{Z(\lambda)} \exp(\lambda\ i)$ with
$$Z(\lambda):=\frac{\exp[(b+1/2)\ \lambda]-\exp[(a-1/2)\ \lambda]}{\exp(\lambda/2)-\exp(-\lambda/2)}\ ,$$
obtainable using the expression of the partial sum of a geometric series.
Note that the parameter $\lambda \equiv \ln(1-p)$ is determined by the mean of the distribution, and vice versa. I don't think there's an analytic expression giving $\lambda$ as a function of the mean, but I'm not sure to be honest.
Also, the mean (ie, first moment) is a sufficient statistics because of the special exponential form (see ref. below). So this is also a maximum-entropy distribution with the mean (ie, first moment) as constraint.
I think that the parameter $p$ in your expression ceases to have the specific meaning it has for the ordinary geometric distribution.
Whether this distribution is reasonable or not depends completely on the specific problem you're applying it to – for example, can you consider the mean to be a sufficient statistics, in your problem?
See eg Bernardo, Smith: Bayesian Theory (Wiley 2000), § 4.5.
