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I have a probability mass function: $f(q) = w^q*\frac{w-1}{w^R-1}$

Where $\sum_{q=0}^{R-1} f(q) = 1$

But also where $$\int_{q=0}^{R-1} f(q) dq = \frac{1}{\frac{w Log[w]}{(w-1)} + \frac{w Log[w]}{w^R-w}}$$

Why is the sum greater than the integral? How would I find the mean of $f(q)$ or $U$ so that $\sum_{q=0}^{U} f(q) = V$? Is it ok to normalize to the value of the integral over the full range of the PMF?

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up vote 1 down vote accepted

A probability mass function is defined for discrete random variables. It must also sum to one over its support, in this case $\Omega = \{0,1,2,\ldots, R-1\}$.

Integrating the function $f(q)$ over a continuous domain suggests there is some positive probability to the outcome, for example, q = 1.1, though this is not possible because q = 1.1 is not in $\Omega$. Ultimately, we're not interested in describing a continuum of probabilities, just those for the $R$ points in $\Omega$.

The reason integrating $f(q)$ over a continuous support yields a value less than integrating $f(q)$ over discrete support is because the latter overestimates the area beneath $f(Q)$. However, since we are only interested in the probability of discrete outcomes, and not the exact area under the curve, this is not an issue.

To find the expectation of q, compute

\begin{align*} \sum_{q=0}^{R-1} q f(q) &= \sum_{q=0}^{R-1} q w^q\left(\frac{w-1}{w^R-1}\right) \\ &= \left(\frac{w-1}{w^R-1}\right) \sum_{q=0}^{R-1} q w^q \\ &= \left(\frac{w-1}{w^R-1}\right) \left( \frac{(R-1)w^{R}}{1-w} + \frac{w(1-w^{R-1})}{(1 - w)^2} \right) \\ \end{align*}

And from there you can play with the algebra to see what comes out.

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