# What are the requirments for the sequence term in a power series distribution? Why is the binomial distribution one?

Given that a power series distribution is a distribution that can be expressed as

$$f_{\theta}(x) = {{a\theta^x}\over{C(\theta)}}$$

where $a$ is a sequence of non-negative real numbers, I've also seen it expressed as $a(x)$. I understand that you can re-parametrize a function so that there are separate terms for $\theta^x$ and some function of $\theta$ but what are the limits on what $a$ could be?

For example, according to this site, the binomial distribution is a type of power series distribution, such that $a(x) = {n \choose x}$ but $n$ is a parameter, and the sequence $a(x)$ is a function of a parameter and the random variable. What makes it okay to treat $n\choose x$ as part of $a$ but not, for example, $\theta^x$?

• Hint: the only additional restrictions on $(a)$ come from the need for $C(\theta)$ to be defined. By considering how $C(\theta)$ is computed you will obtain the answer.
– whuber
Oct 1, 2013 at 17:06
• So, it's up to me whether to define $a$ as $n\choose x$ and $C(\theta)$ as $(1-p)^n$ or the other way around, or any other rearrangement as long as $\theta^x$ is by itself? Then why express this in terms $a\over {C(\theta)}$ at all? Why not just have $a$ be part of $C(.)$? Oct 1, 2013 at 18:26
• $C$ alone cannot determine the countable sequence of $a$'s. It is up to you to stipulate what $a$ is. If you want to define $a(x)=\binom{n}{x}$ for $0\le x\le n$ and $a(x)=0$ otherwise, that's fine. In that case $C(\theta)=(1+\theta)^n.$ In general, when $C(\theta)$ (which depends on $a$) exists, it is possible to define $f_\theta$. That's the only restriction (apart from the $a$ being non-negative).
– whuber
Oct 1, 2013 at 18:52