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I'm trying to solve the following problem:

The probability $p_{i}$ that a given cell contains exactly i balls is given by the binomial distribution $p_{i}=\displaystyle \frac{\binom{r}{i}(n-1)^{r-i}}{n^{r}}$. Prove that the most probable number is the integer $v$ such that $\displaystyle \frac{(r-n+1)}{n}< v \leq \displaystyle \frac{(r+1)}{n}$.

I've tried to get the inequality using first $\displaystyle \frac{p_{i}}{p_{i-1}}$ (which didn't work quite well) and later $\displaystyle \frac{p_{i}}{p_{i+1}}$. For example, for the first attempt the result I obtained was:

$$\frac{p_{i}}{p_{i-1}}=\frac{r-i+1}{i(n-1)}$$

In this kind of problem, I'm expecting to obtain a $1$ in the resulting expression in order to figure out for which values of $i$ there is a maximum in the function, however, I can't get the correct result.

Any help will be appreciated.

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1 Answer 1

up vote 2 down vote accepted

See this (also for a better formulation of the problem, using common notation).

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Thanks for your answer. I think that's exactly what I'm doing. In the case of $\displaystyle \frac{p_{i}}{p_{i+1}}$, we have (when the distribution is increasing) $r-i>(i+1)(n-1)$. Simplifying, we obtain: $\displaystyle \frac{r+1-n}{n}>i$. This is the opposite that the first part of the inequality. I missed something?. By the way, the notation I used is more or less standard although I understand it can be a bit odd. –  Robert Smith Dec 15 '10 at 5:29
    
The inequality $(r+1-n)/n > i$ only agrees with the first part of the desired inequality, so I don't see a problem. –  Shai Covo Dec 15 '10 at 6:05
    
The problem is that $(r+1-n)/n > i$ is saying the opposite than $\displaystyle \frac{(r-n+1)}{n}< i$. Furthermore, I'd like to get both parts of the inequality. –  Robert Smith Dec 15 '10 at 19:02
    
No, $(r+1-n)/n > i$ corresponds to $p_i < p_{i+1}$. So, $p_i$ increases for $i$ up to about $(r+1-n)/n$ (and then decreases, as you should probably be able to show analogously), which agrees with the desired inequality. –  Shai Covo Dec 15 '10 at 19:26
    
Oh, right. However, I have to suppose that $p_{i-1}<p_{i}>p_{i+1}$ or play around with those quotients to see which inequalities are mutually compatible. In this example, $\displaystyle \frac{p_{i}}{p_{i-1}}$ corresponds to $i<\frac{r+1}{n}$ (with $p_{i-1}<p_{i}$) and $\displaystyle \frac{p_{i+1}}{p_{i}}$ corresponds to $\displaystyle \frac{r+1+n}{n}<i$ (with $p_{i+1}<p_{i}$). In this way, $\displaystyle \frac{r+1+n}{n}<i^{*}<\frac{r+1}{n}$. Is this reasoning correct?. –  Robert Smith Dec 15 '10 at 22:41

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