the normalization constant

Question

On page 4 of this article, the authors wants to find the normalizing constant $c$ but it is very hard to compute so I used the same formula to bound $c$.

Take the non-negative integers $~0 \leq x_1,x_2,...,x_r \leq X$ where $\sum_{i=1}^{r}x_i=X$. Using the the Multinomial distribution to find the probability of $x_1,x_2,...,x_r $ and assuming that $ p(x_i)=p=\frac{1}{r}$; where $r$ is a constant, then:

$$P(x_1,x_2,...,x_r;\frac{1}{r})= \frac{1}{c}\frac{X!}{x_1!x_2!...x_r!}\prod_{i=1}^{r} p^{x_i}$$. $$= \frac{1}{c}\frac{X!}{x_1!x_2!...x_r!}p^{\sum_{i=1}^{r}x_i}$$

To prove hat $c\leq1$, I did:

$$ c= \sum_{x_1+x_2+...x_r=X}\frac{X!}{x_1!x_2!...x_r!}p^{X}$$ $$c \leq \sum_{x_1+x_2+...x_r=X}\frac{1}{r^{X}}$$. let $y=\frac{1}{r}$, $$c \leq \sum_{0}^{X}{y^{X}}$$ $$c \leq\frac{1-\frac{1}{r^{X+1}}}{1-\frac{1}{r}}$$ $$c \leq \frac{r(r^{X+1}-1)}{r(r^{X+1}-r^{X})}$$ $$c \leq \frac{r^{X+1}-1}{r^{X+1}-r^{X}}$$

Is this correct or not? and if not, which part I missed?

P.S: Please write a comment if there is something unclear in the question so that I can clarify it or edit the question if needed.

Thanks for help.

Answer 1

The distribution $$ \mathbb{P}_{\mathbf{p}}(X_1=x_1,X_2=x_2,...,X_r=x_r)= \frac{1}{c}\frac{X!}{x_1!x_2!...x_r!}\prod_{i=1}^{r} p_i^{x_i} $$ is a standard Multinomial distribution. The normalising constant is thus $c=1$. Since $$ \sum_{\substack{x_1,\ldots,x_r\\\sum\limits_{i=1}^r x_i=X}} \frac{X!}{x_1!x_2!...x_r!}\prod_{i=1}^{r} p_i^{x_i} = \left\{\displaystyle{\sum_{i=1}^r} p_i\right\}^{X}=1\,. $$