Asymptotics for the sums from the inclusion-exclusion principle

Question

What is a method to compute the asymptotics of a sum resulting from the inclusion-exclusion principle? Each term of the sum can be approximated perhaps by Stirling's formula or the Gaussian distribution. However the alternating sign should effect some cancellation. As an example, the answer to this combinatorial problem generates a probability $$p=\frac c{n\choose j},$$ where $$c=\sum_{k=w}^j(-1)^{k-w}(j-k+1){n-k+1\choose j-k+1}.$$ What is the asymptote of $p$ for $\big|\frac jn-a\big|=o(n)$ and $\big|\frac wn-b\big|=o(n)$ for some positive numbers $a$ and $b$ as $n\rightarrow\infty$?

Answer 1

Many unsolved asymptotics problems can be written as inclusion-exclusion sums. There is no general method for solving them.

However, in your case the terms decrease in absolute value and nearly form a geometric series. This is an easy case.

If $t_k$ is the $k$-term, then $t_{w+i}\approx (-1)^i (j/n)^i t_w$, so $$c \sim \frac{t_w}{1+j/n}.$$ To get rigorous bounds, you can use the fact that the partial sums of an alternating series with absolutely decreasing terms alternate above and below the full sum.

Answer 2

Another way to approach this from the generating function perspective. Notice that $$\sum_{k=l}^u (-1)^k [x^k]\, f(x) = [x^u]\,\frac{f(-x)}{1-x} - [x^{l-1}]\,\frac{f(-x)}{1-x},$$ and so the question reduces to studying the asymptotic of the coefficients of $\frac{f(-x)}{1-x}$.

In your example, we have $$(j-k+1)\binom{n-k+1}{j-k+1} = (n-j+1)\,[x^{j-k}]\,(1-x)^{-(n+2-j)}$$ and thus \begin{split} c &=\sum_{k=w}^{j} (-1)^{k-w} (j-k+1)\binom{n-k+1}{j-k+1} \\ &= (n-j+1)(-1)^{j-w} \sum_{t=0}^{j-w} (-1)^t\,[x^t]\,(1-x)^{-(n+2-j)} \\ &= (n-j+1)(-1)^{j-w}\,[x^{j-w}]\,\frac{1}{(1+x)^{n+2-j}(1-x)} \\ &= (n-j+1)\,[x^{j-w}]\,\frac{1}{(1-x)^{n+2-j}(1+x)}. \end{split}

Now, noticing the partial fraction decomposition $$\frac{1}{(1-x)^m(1+x)} = \frac{2^{-m}}{1+x} + \sum_{i=0}^{m-1} \frac{2^{-(i+1)}}{(1-x)^{m-i}},$$ we obtain the formula: $$c=(n-j+1)\left(\frac{(-1)^{j-w}}{2^{n+2-j}} + \sum_{i=0}^{n+1-j} \frac{1}{2^{i+1}}\binom{n+1-i-w}{j-w}\right).$$ Here all terms in the sum are positive and thus are more amenable to asymptotic analysis. For example, we can deduce $$\frac{n-j+1}{2}\binom{n+1-w}{j-w} \lesssim c \lesssim (n-j+1)\binom{n+1-w}{j-w}.$$

Answer 3

Your question seems extremely general! Perhaps the most famous result of this type states that the number of fixed points of the $N$-letter permutations $$\chi:S_N\to\mathbb N,\quad,\quad\chi(\sigma)=\#\{i|\sigma(i)=i\}$$ becomes Poisson (1) in the $N\to\infty$ limit. The proof uses the inclusion-exclusion principle at fixed $N\in\mathbb N$, and then some simple asymptotics in the limit. All this generalizes the fact that $$\mathbb P(\chi(\sigma)=0)\simeq\frac{1}{e}$$ in other words that "a random permutation is a derangement with probability about $1/e$''. All beautiful mathematics all this, you can take a look at papers by Diaconis et al. for more.

Answer 4

The limiting probability is $0$ and it goes there pretty rapidly if we fix the ratios $\frac{j}n \approx a \lt 1$ and $\frac{w}n \approx b \gt 0$ and then let $n$ grow. In fact $p\lt n a^{w} \approx n{a}^{bn}$

This is a kind of cheap way out of the interesting problem of estimating $c$ but that is how it is.

The convergence to $0$ is pretty apparent from numerical experiments, however here is a calculation:

$$c\lt j\binom{n-w+1}{j-w+1}=j\frac{(n-w+1)!}{(j-w+1)!(n-j)!}$$

so $$\frac{c}{\binom{n}{j}} \lt j\frac{(n-w+1)!j!}{n!(j-w+1)!}=j\frac{j(j-1)(j-2)\cdots(j-w+2)}{n(n-1)(n-2)\cdots(n-w+2)} $$

since $a=\frac{j}{n} \gt\frac{j-k}{n-k}$

$$\frac{c}{\binom{n}{j}} \lt j(a)^{w-1}\approx na^w. $$