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I asked I question here trying to obtain clarification about how to follow a hint. In spite of the fine answers I received there, the hint doesn't look very helpful. I'd like to know a hint for the following problem or a way to use the hint I already have.

The probability $p_{m}(r,n)$ of finding exactly $m$ cells empty placing $r$ balls into $n$ cells is $$p_{m}(r,n)=\frac{1}{n^{r}}\binom{n}{m}A(r,n-m)=\binom{n}{m}\sum_{\nu=0}^{n-m}(-1)^{\nu}\binom{n-m}{\nu}\left(1-\frac{m+\nu}{n}\right)^{r}$$

From such probability, conclude that the probability $x_{m}(r,n)$ of finding $m$ or more cells empty equals

$$\binom{n}{m}\sum_{\nu=0}^{n-m}(-1)^{\nu}\binom{n-m}{\nu}\left(1-\frac{m+\nu}{n}\right)^{r}\frac{m}{m+v}$$

HINT: Evaluate $x_{m}(r,n)-p_{m}(r,n)$.

Without using the hint, I tried to find the pattern for $p_{m}$, $p_{m+1}$, etc with the goal of factoring each term and sum over something recognizable, but I couldn't find a way to factor the last term (for example, to factor $(n-m-1-\nu)^{r}$ into some terms including the original expression $(n-m-\nu)^{r}$...I don't think it is possible. Furthermore, I don't think that strategy is going to produce the answer I'm looking for.

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r_31415
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  • What is $p_m(r,n)$? – Aryabhata Dec 28 '10 at 06:32
  • @Moron: I believe its the probability of finding exactly $m$ cells empty. – NebulousReveal Dec 28 '10 at 06:35
  • @Trevor: I see, it seems to match the page you linked in your answer too. – Aryabhata Dec 28 '10 at 06:48
  • @Moron: Trevor is right, however, the question is to find $x_{m}(r,n)$, that is, the probability of finding $m$ or more empty cells. – r_31415 Dec 28 '10 at 06:55
  • @Robert: You also have to specify what $m$, $r$ and $n$ are, along with what the experiment is. Also, If $m=0$, the formula you have implies $x_m(r,n)=0$ and so the probability of $0$ or more cells being empty is $0$. That does not look right. – Aryabhata Dec 28 '10 at 07:03
  • @Moron: I added what $m$, $r$ and $n$ are. You're right. It doesn't look right. To calculate the probability of finding $m$ or more cells empty should be $\sum_{m=0}^{n} p_{m}(r,n)$, right? – r_31415 Dec 28 '10 at 19:19
  • @Robert: Yes, you need to add the $p_m$s, that looks right. – Aryabhata Dec 28 '10 at 19:44
  • @Moron: Alright, but that doesn't look (to me, at least) as a closed form. Almost every term depends on $m$ and most of these terms depend on $\nu$ also. – r_31415 Dec 28 '10 at 20:18
  • @Robert: Yes, I was only confirming your $\sum_{m=0}^{n} p_m(r,n)$. Looks like your source (I am guessing a book) has the formula for $x_m(r,n)$ wrong, as it seems to give incorrect result for $m=0$. You are right, that is not closed form. – Aryabhata Dec 28 '10 at 20:20
  • @Moron: Yes, it could be wrong. The source is Feller's book 'An Introduction to Probability Theory and its applications'. Do you think I should ask for a closed form to such a problem? (taking for granted that the proposed answer is wrong). – r_31415 Dec 28 '10 at 20:42
  • @Robert: Maybe it is just a typo. Does the book have an errata? Asking for a closed might get you no answers, as it might be a hard problem. – Aryabhata Dec 28 '10 at 20:44
  • @Moron: To my knowledge, no. There is just an errata for the second edition, but that doesn't help. The one I'm reading is the third edition. Oh, damn. I hate to leave problems unresolved. – r_31415 Dec 28 '10 at 21:07

2 Answers2

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See the following (Theorem 2). This gives the probability that exactly $m$ cells are empty. You could probably modify this to get the multiplicative factor $\frac{m}{m+v}$.

NebulousReveal
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    Curiously, I already knew that link. However, the problem asks for the probability of finding $m$ or more empty cells. The final expression from the link you mention is basically the starting point $p_{m}(r,n)$ in the original post. – r_31415 Dec 28 '10 at 07:00
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To use Feller's original notation. Let $E_m(r,n)$ be the number of distributions leaving exactly $m$ cells empty, then we have $$p_m(r,n)=n^{-r}\binom{n}{k}A(r,n-m)$$ as the corresponding probability ($A$ is to be defined)

Now, we need compute the probability $x_m(r,n)$ of finding $m$ or more cells empty.

Now, the probability that $m$ or more cells are empty is given by $x_m = p_m + p_{m+1} + \cdots + p_{n}$. Now, observe the reason why the hint was given: $x_{m+1} = x_m - p_m$ (i.e., $m+1$ or more cells being empty is the same as $\ge m$ cells being empty minus exactly $m$ cells being empty).

So for a sanity test, you could plug in the value of $x_m(r,n)$ given, and verify whether $x_{m}(r,n)-x_{m+1}(r,n)=p_m(r,n)$ holds (with the appropriate boundary cases).

The deeper underlying reason for this hint is that Feller probably wants you to figure out an inclusion-exclusion type argument. If I get time, I will write out the details. But the above should be a sufficiently detailed hint?

  • Thanks for your answered. Yes, I get the part $x_{m+1}=x_{m}-p_{m}$ (that's what I would have expected to obtain when I asked the simplification question in the other post). However, I don't know about 'inclusion-exclusion type' of argument (related to http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle ?). By the way, Moron noticed that $m=0$ produces $x_{0}(r,n)=0$ which is contradictory. – r_31415 Dec 28 '10 at 22:42