Proof of Theorem 1.1. Let X be a ${\mathcal B}(m,p)$ -distributed random variable. Then

(2.1) \begin{equation}{\mathbb E}\, \exp\!(\gamma X) = (1+p(\textrm{e}^\gamma-1))^m, \qquad \gamma\in {\mathbb R}.\end{equation}

Now let $X_j$ , $1\le j\le n$ , be ${\mathcal B}(m,p)$ -distributed random variables. We do not assume any independence. Then, for every $\gamma>0$ ,

(2.2) \begin{equation}{\mathbb E}\, \exp\Big(\gamma \max_{1\le j\le n} X_j\Big) \le {\mathbb E}\, \sum_{j=1}^n \exp\!(\gamma X_j) =n (1 + p(\textrm{e}^\gamma - 1))^m.\end{equation}

By Jensen’s inequality,

\begin{align*}\exp \Big(\gamma {\mathbb E}\, \max_{1\le j\le n} X_j \Big) \le {\mathbb E}\, \exp\!(\gamma \max_{1\le j\le n} X_j) \le n (1 + p(\textrm{e}^\gamma - 1))^m.\end{align*}

It follows that

\begin{equation*}{\mathbb E}\, \max_{1\le j\le n} X_j \le \gamma^{-1}(\!\log n + m\log\!(1 + p(\textrm{e}^\gamma - 1))) \le\gamma^{-1}(\!\log n + mp(\textrm{e}^\gamma - 1)).\end{equation*}

We choose $\gamma \;:\!=\; (({2\log n})/{mp})^{1/2}$ . By (1.2), $\gamma\to 0$ . Using the expansion $\textrm{e}^\gamma - 1 = \gamma + \gamma^2(1+o(1))/2$ , we obtain

\begin{align*}{\mathbb E}\, \max_{1\le j\le n} X_j& \le \gamma^{-1}(\!\log n + mp[\gamma +\gamma^2(1+o(1))/2]) \\[5pt] & = mp + \gamma^{-1} \log n + mp\,\gamma(1+o(1))/2 \\[5pt] & = mp + (2 mp \log n)^{1/2} (1+o(1)).\end{align*}

Furthermore, by (1.2) the second term is negligible and we obtain ${\mathbb E}\, \max_{1\le j\le n} X_j \le m p (1+o(1))$ .

The same approach applies to the minima. With the same notation we have, for every $\gamma>0$ ,

\begin{align*}{\mathbb E}\, \exp\!(\!-\! \gamma \min_{1\le j\le n} X_j) \le {\mathbb E}\, \sum_{j=1}^n \exp\!(\!-\!\gamma X_j) =n(1 + p(\textrm{e}^{-\gamma} - 1))^m.\end{align*}

By Jensen’s inequality,

\begin{align*}\exp \Big({-} \gamma {\mathbb E}\, \min_{1\le j\le n} X_j\Big) \le{\mathbb E}\, \exp\Big({-} \gamma \min_{1\le j\le n} X_j\Big) \le n (1 + p(\textrm{e}^{-\gamma} - 1))^m.\end{align*}

It follows that ${\mathbb E}\, \min_{1\le j\le n} X_j \ge - \gamma^{-1}(\!\log n + m \log\!(1 + p(\textrm{e}^{-\gamma} - 1)))$ . We still use $\gamma \;:\!=\; (({2\log n})/{mp})^{1/2}\to 0$ . The expansion $\textrm{e}^{-\gamma}-1= -\gamma +\gamma^2(1+o(1))/2$ yields

\begin{align*}\log\!(1 + p(\textrm{e}^{-\gamma} - 1)) = p (\textrm{e}^{-\gamma} - 1) (1+o(1)) =- p \gamma (1+o(1)) + p \gamma^2 (1+o(1)) / 2.\end{align*}

From this we get

\begin{align*}{\mathbb E}\, \min_{1\le j\le n} X_j& \ge -\gamma^{-1}(\!\log n + mp[\!-\!\gamma (1+o(1)) + \gamma^2(1+o(1))/2]) \\[5pt] & = mp (1+o(1)) - \gamma^{-1} \log n - mp\gamma(1+o(1))/2 \\[5pt] & = mp (1+o(1)) - (2 mp \log n)^{1/2} (1+o(1)).\end{align*}

By (1.2), the second term is negligible and we obtain ${\mathbb E}\, \min_{1\le j\le n} X_j \ge mp (1+o(1))$ .

We now apply these results to the multinomial assignment process. Here, the joint law of the entries $X_{ij}$ is ${\mathcal M}(m, n^{2})$ and every $X_{ij}$ follows the binomial law ${\mathcal B}(m,p)$ with $p = n^{-2}$ . Our bound for the maxima yields

\begin{align*}{\mathbb E}\, \max_\sigma S(\sigma) \le \sum_{i=1}^n {\mathbb E}\, \max_{1\le j\le n} X_{ij} =n \cdot {\mathbb E}\, \max_{1\le j\le n} X_{1j} \le \frac{m}{n} (1+o(1)),\end{align*}

while the bound for the minima yields

\begin{align*}{\mathbb E}\, \min_\sigma S(\sigma) \ge \sum_{i=1}^n {\mathbb E}\, \min_{1\le j\le n} X_{ij} =n \cdot {\mathbb E}\, \min_{1\le j\le n} X_{1j} \ge \frac{m}{n} (1+o(1)).\end{align*}

It follows that ${\mathbb E}\, \max_\sigma S(\sigma) = ({m}/{n}) (1+o(1))$ and ${\mathbb E}\, \min_\sigma S(\sigma) = ({m}/{n}) (1+o(1))$ , as required.

Proof of Proposition 1.1. The claim will follow by squeezing once we prove that, for all ${\varepsilon}>0$ , ${\mathbb P} \big(\!\max_\sigma S(\sigma) \geq (1 + {\varepsilon}) {m}/{n}\big) \to 0$ and ${\mathbb P} \big(\!\min_\sigma S(\sigma) \leq (1 - {\varepsilon}) {m}/{n}\big) \to 0$ .

Again, let $X_j$ , $1\le j\le n$ , be ${\mathcal B}(m,p)$ -distributed random variables. Note that, by (2.2), for $\gamma \to 0^+$ ,

\begin{align*}{\mathbb E}\, \exp \Big\{\gamma\max_{1 \le j \le n} X_j \Big\} \le \exp\{\log n + mp\gamma (1+o(1))\}.\end{align*}

Using Markov’s inequality, we get

\begin{align*}{\mathbb P} \Big( \max_{1 \le j \le n} X_j \geq (1+{\varepsilon}) mp \Big) \le \exp\{\log n + mp\gamma (o(1)-{\varepsilon})\} ,\end{align*}

in which we set $\gamma = (\!\log n/ mp)^{1/2} $ to get

\begin{align*}{\mathbb P} \Big(\!\max_{1 \le j \le n} X_j \geq (1+{\varepsilon}) mp \Big) \le \exp\{\log n - (\!\log n\,mp)^{1/2} (o(1)+{\varepsilon})\}.\end{align*}

By (1.2), there exists some $n_0$ such that, for all $n\geq n_0$ , $mp \geq 10 \log n/ {\varepsilon}^2$ , and thus

\begin{align*}{\mathbb P} \Big( \max_{1 \le j \le n} X_j \geq (1+{\varepsilon}) mp \Big) \le n^{-2} \end{align*}

for sufficiently large n. We can then conclude that

\begin{align*}{\mathbb P} \bigg(\!\max_\sigma S(\sigma) \geq (1 + {\varepsilon}) \frac{m}{n}\bigg)& \leq {\mathbb P}\bigg(\!\max_{1\le i\le n}\max_{1\le j\le n}X_{ij} \geq (1 + {\varepsilon})\frac{m}{n^2} = (1 + {\varepsilon})mp\bigg) \\[5pt] & \leq \sum_{i=1}^n {\mathbb P} \Big(\!\max_{1\le j\le n} X_{ij} \geq(1 + {\varepsilon}) mp\Big) \to 0.\end{align*}

The claim for the minimum follows along the same lines.

Proof. The proof is split into two blocks dealing with the lower and upper bounds required to establish both claims.

For the upper bound in (2.5) and the lower bound in (2.6), let $H>H_*$ . Then

(2.7) \begin{equation}H\log H-(H-1)>\frac 1c.\end{equation}

Let $r\;:\!=\;Hp$ . Then, by (2.3), $m r=H m p = c H \log \eta (1+o(1))$ .

Applying the exponential Chebyshev inequality for every j and every $v>0$ , we obtain

(2.8) \begin{align}{\mathbb P}(X_j \ge c H \log \eta + v)& = {\mathbb P}(X_j\ge mr(1+o(1))^{-1} + v) \nonumber \\[5pt] & \le \frac{{\mathbb E}\, \textrm{e}^{\gamma X_j}}{\exp\{\gamma {m r}/({1+o(1)}+v)\}} =\bigg[\frac{1 + p(\textrm{e}^\gamma - 1)}{\exp\{{\gamma r}/({1+o(1)})\}}\bigg]^m \textrm{e}^{-\gamma v}. \end{align}

By choosing the optimal $\gamma\;:\!=\; \log\!([{(1-p)r}]/[{p(1-r)}])$ and setting $r^\textrm{o} \;:\!=\; {r}/({1+o(1)})$ , we have

\begin{align*}\frac{1 + p(\textrm{e}^\gamma - 1)}{\textrm{e}^{\gamma r}}&= \bigg(\frac pr\bigg)^{{r}/({1+o(1)})} \exp\!((1 - r^\textrm{o})\log\!(1-p)- (1-r^\textrm{o})\log\!(1-r)) \\[5pt] & = H^{-{Hp}/({1+o(1)})} \exp\!(\!-\!p + r + O(p^2)) \\[5pt] &= \exp\!(\!-\!((1+o(1))H\log H - (H-1)\big) p + O(p^2)).\end{align*}

Hence,

\begin{align*} \bigg[\frac{1 + p(\textrm{e}^\gamma - 1)}{\textrm{e}^{\gamma r}}\bigg]^m& = \exp\!(\!-\!(H\log H - (H-1) + o(1)) mp) \\[5pt] &= \exp\!(\!-\!(H\log H - (H-1)) c (\!\log \eta) (1+o(1))) \;:\!=\; \eta^{-\beta+o(1)},\end{align*}

where, by (2.7), $\beta = (H\log H -(H-1)) c > 1$ .

Substituting the above results in (2.8) we obtain ${\mathbb P}(X_j \ge c H \log \eta + v) \le \eta^{-\beta+o(1)} \textrm{e}^{-\gamma v}$ . It is now trivial that

\begin{align*}{\mathbb P}\Big(\!\max_{1\le j\le \eta} X_j \ge c H \log\eta + v\Big) \le \eta^{-(\beta-1)+o(1)} \textrm{e}^{-\gamma v}.\end{align*}

It follows that

\begin{align*}{\mathbb E}\, \max_{1\le j\le \eta} X_j - cH\log \eta& = {\mathbb E}\, \Big(\!\max_{1\le j\le \eta} X_j - cH\log \eta\Big) \\[5pt] & \le {\mathbb E}\, \Big( \max_{1\le j\le \eta} X_j - cH\log \eta\Big)_+ \\[5pt] & = \int_0^\infty {\mathbb P}\Big(\!\max_{1\le j\le \eta} X_j \ge c H \log \eta + v\Big) \, \textrm{d} v \\[5pt] & \le \eta^{-(\beta-1)+o(1)} \int_0^\infty \textrm{e}^{-\gamma v} \, \textrm{d} v \\[5pt] & = \eta^{-(\beta-1)+o(1)}\frac 1\gamma = \eta^{-(\beta-1)+o(1)} \frac{1}{\log H} (1+o(1)) \to 0.\end{align*}

Therefore, ${\mathbb E}\, \max_{1\le j\le \eta} X_j \le c H\log \eta + o(1)$ . By letting $H\searrow H_*$ we obtain the upper bound in (2.5).

The lower bound in (2.6) is obtained in exactly the same way through the Chebyshev inequality for the lower tails.

We now consider the converse bounds. The lower bound in (2.5) is reached in four steps: we give a Poissonian approximation of binomial laws, then provide a lower bound for this Poissonian approximation. This bound provides a lower bound for the maximum’s expectation of independent binomial i.i.d. random variables. Finally, using the negative association argument, we reduce the claim to the independence case.

First, let X be a binomial ${\mathcal B}(m,p)$ -distributed random variable. Elementary calculations show that Poissonian approximation

\begin{align*}{\mathbb P}(X=k) = \textrm{e}^{-mp} \frac{(mp)^k}{k!} (1+o(1))\end{align*}

is valid if $p^2m\to 0$ , $pk\to 0$ , and ${k^2}/{m}\to 0$ . Indeed, we have

\begin{align*}{\mathbb P}(X=k) = \textrm{e}^{-mp} \frac{m !}{k!(m-k)!} p^k (1-p)^{m-k}\end{align*}

and, with the limiting behaviour as above,

\begin{align*}& (1-p)^{m} = \exp\!(m \log\!(1-p)) = \exp\!(\!-\!mp + m\operatorname{O}\!(p^2)) = \exp\!(\!-\!mp)(1+\operatorname{o}\!(1)) \text{ as } p^2m\to 0; \\[5pt] & (1-p)^{k} = \exp\!(k \log\!(1-p)) = \exp\!(k\operatorname{O}\!(p)) = 1 + \operatorname{o}\!(1) \text{ as } kp\to 0; \\[5pt] & \frac{m!}{(m-k)!} \le m^k; \\[5pt] & \frac{m!}{(m-k)!} \geq m^k\bigg(1 - \frac{k}{m}\bigg)^k =m^k \exp\bigg(k \operatorname{O}\bigg(\frac{k}{m}\bigg)\bigg) = m^k(1+\operatorname{o}\!(1)) \text{ as } k^2/m\to 0.\end{align*}

Second, let $c>0$ and $H>1$ . Let $k=\lfloor cH \log \eta +1 \rfloor $ and $\lambda = c (\!\log \eta) (1+o(1))$ .

Then, Stirling’s approximation and basic computations yield $\textrm{e}^{-\lambda} {\lambda^k}/{k!} = \eta^{-\beta+o(1)}$ , where

(2.9) \begin{equation}\beta\;:\!=\; c (H\log H - (H-1)).\end{equation}

Now we combine the results of the two steps. Note that with (2.3), (2.4), and, for $k = cH (\!\log \eta) (1+o(1))$ , all three assumptions of the first step are verified and, with $\lambda=mp$ , we obtain ${\mathbb P}(X \ge cH \log \eta) \ge {\mathbb P}(X=k) = \eta^{-\beta+o(1)}$ . If $1 < H < H_*$ , then $\beta < 1$ .

Third, let $(\widetilde{X}_j)_{1\le j\le \eta}$ be independent copies of X. Then

(2.10) \begin{align}{\mathbb P}\Big(\!\max_{1\le j\le \eta} \widetilde X_j \le c H \log \eta\Big)& = {\mathbb P}(X\le c H \log \eta)^\eta \le (1 - \eta^{-\beta+o(1)})^\eta \nonumber \\[5pt] & \le \exp\!(\!-\!\eta^{1-\beta+o(1)}) \to 0. \end{align}

It follows that

(2.11) \begin{align}{\mathbb E}\, \max\tilde X_j& \geq {\mathbb E}\, \big[(\!\max \tilde X_j){\mathbf 1}_{ \{\max \tilde X_j\geq cH\log \eta\}}\big] \nonumber \\[5pt] & \geq (cH \log \eta){\mathbb P}(\!\max \tilde X_j \geq cH \log \eta) = cH(\!\log \eta)(1-o(1)). \end{align}

Fourth, from the disintegration theorem for negatively associated variables [Reference Christofides and Vaggelatou4] (see also [Reference Bulinski and Shashkin3, Chapter 2, Theorem 2.6, and Lemma 2.2]), we have

(2.12) \begin{equation} {\mathbb E}\, \max_{1\le j\le \eta} X_j\ge {\mathbb E}\, \max_{1\le j\le \eta} \widetilde X_j.\end{equation}

Combining this estimate with the result of the third step, for every $H<H_*$ we obtain

\begin{align*}{\mathbb E}\, \max_{1\le j\le \eta} X_j \ge cH \log \eta (1+o(1)).\end{align*}

Letting $H\nearrow H_*$ , we obtain the lower bound in (2.5).

The upper bound in (2.6) follows in a similar way. Now let $k\;:\!=\;[cH \log \eta]$ . By using Poissonian approximation and Poissonian asymptotics we obtain

\begin{align*}{\mathbb P}(X\le cH \log \eta) \ge {\mathbb P}(X=k) = \eta^{-\beta+o(1)}\end{align*}

with the same $\beta$ as in (2.9). If $\widetilde{H}_*<H<1$ , then $\beta<1$ .

As before, for independent variables we obtain

\begin{align*}{\mathbb P}\Big(\!\min_{1\le j\le \eta} \widetilde X_j \ge cH \log \eta\Big) \le \exp\big({-}\eta^{1-\beta+o(1)}\big).\end{align*}

It follows that

\begin{align*}{\mathbb E}\, \min_{1\le j\le \eta} \widetilde X_j& = {\mathbb E}\Big[\min_{1\le j\le \eta}\widetilde X_j {\mathbf 1}_{ \{\min_{1\le j\le \eta}\widetilde X_j \le cH \log \eta\}}\Big]+ {\mathbb E}\Big[\min_{1\le j\le \eta}\widetilde X_j {\mathbf 1}_{ \{\min_{1\le j\le \eta}\widetilde X_j > cH \log \eta\}}\Big] \\[5pt] & \le c H \log \eta +\sum_{j=1}^\eta {\mathbb E}\big[\widetilde X_j {\mathbf 1}_{ \{\min_{1\le i\le \eta, i\not=j}\widetilde X_i > cH \log \eta\}}\big] \\[5pt] & = c H \log \eta + \eta {\mathbb E}\, \widetilde X_1 {\mathbb P}\Big(\!\min_{2\le i\le \eta}\widetilde X_i > c H \log \eta\Big) \\[5pt] & \le c H \log \eta + \eta \cdot c \log \eta (1+o(1)) \exp\big({-}\eta^{1-\beta+o(1)}\big) \\[5pt] & = c H \log \eta + o(1).\end{align*}

The final negative association argument reads as follows. Since $(X_j)$ are negatively associated, so are $(\!-\!X_j)$ . From the disintegration theorem cited above, it follows that

\begin{align*}{\mathbb E}\, \max_{1\le j\le \eta} (\!-\!X_j) \ge {\mathbb E}\, \max_{1\le j\le \eta} (\!-\!\widetilde X_j),\end{align*}

which is equivalent to ${\mathbb E}\, \min_{1\le j\le \eta} X_j \le {\mathbb E}\, \min_{1\le j\le \eta} \widetilde X_j$ . By combining the obtained results, we have ${\mathbb E}\, \min_{1\le j\le \eta} X_j \le c H \log \eta (1+o(1))$ . Finally, letting $H\searrow \widetilde H_*$ we obtain the upper bound in (2.6).

Proof of Theorem 1.2. Recall that a multinomial distribution is negatively associated, see [Reference Joag-Dev and Proschan8] and [Reference Bulinski and Shashkin3, Chapter 1, Theorem 1.27]. Furthermore, with $p=n^{-2}$ , the assumption (2.4) is also valid.

Therefore, the bounds (2.5) and (2.6) apply to the sums of the entries $X_{ij}$ . They yield, respectively,

\begin{align*}{\mathbb E}\, \max_\sigma S(\sigma)& \le \sum_{i=1}^{n} {\mathbb E}\, \max_{1\le j\le n} X_{ij} \le c H_* n \log n (1+o(1)), \\[5pt] {\mathbb E}\, \min_\sigma S(\sigma)& \ge \sum_{i=1}^{n} {\mathbb E}\, \min_{1\le j\le n} X_{ij} \ge c \widetilde H_* n \log n (1+o(1)).\end{align*}

The opposite bounds follow by the ‘greedy method’ dating back to [Reference Steele15], for instance, and introduced for Gaussian random variables in [Reference Mordant and Segers12] (and used in [Reference Lifshits and Tadevosian9]) that we recall now. This method allows the construction of a quasi-optimal permutation $\sigma^*$ that provides a sufficiently large value or sufficiently small value of the assignment process. Recall that $[1..i] \;:\!=\; \{1, 2, \dots, i\}$ . Define $\sigma^*(1) \;:\!=\; \arg \max_{j\in[1..n]} X_{1j}$ , and let, for all $i=2,\dots, n$ , $\sigma^*(i) \;:\!=\; \arg \max_{j \not\in \sigma^*([1..i-1])} X_{ij}$ . It is natural to call this strategy ‘greedy’, because at every step we consider row i, take the maximum of its available elements (without considering the influence of this choice on subsequent steps), and then forget row i and the corresponding column $\sigma^*(i)$ . The number of variables used at consequent steps is decreasing from n to 1.

By using the greedy method, we have

(2.13) \begin{equation} {\mathbb E}\, \max_\sigma S(\sigma) \ge {\mathbb E}\, \sum_{i=1}^{n} X_{i\,\sigma^*(i)} =\sum_{i=1}^{n} {\mathbb E}\, \max_{j \not\in \sigma^*([1..i-1])} X_{ij} =\sum_{i=1}^{n} {\mathbb E}\, \max_{1\le j \le n-i+1} X_{ij}.\end{equation}

The latter equality may seem surprising because the index sets $[n]\backslash \sigma^*([1..i-1])$ are random and depend on the matrix X. However, it is justified by the following lemma.

Proof of Lemma 2.2. Let $S = S(X^{(1)}) \;:\!=\; \sum_{j=1}^{N_1} X_j$ . Recall that the conditional distribution of $X^{(2)}$ with respect to $X^{(1)}$ is ${\mathcal M}_{m-S,N_2}$ . This means that, for all $x_1\in {\mathbb N}^{N_1}$ and $x_2\in {\mathbb N}^{N_2}$ ,

\begin{align*}{\mathbb P}(X^{(2)}=x_2, X^{(1)}=x_1) = {\mathbb P}(X^{(1)}=x_1) {\mathcal M}_{m-S(x_1), N_2} (x_2).\end{align*}

For every fixed set $J\subset (N_1,N_1+N_2]$ of size q,

\begin{align*}{\mathbb P}(X^{(2)}=x_2, {\mathcal J}=J ) = \sum_{s=0}^m {\mathbb P}({\mathcal J}=J, S=s) {\mathcal M}_{m-s, N_2} (x_2) \end{align*}

by summing over $x_1\in {\mathcal J}^{-1}(J)$ , where ${\mathcal J}^{-1}(J)$ is the preimage of the set J under ${\mathcal J}$ . Now, for every nonnegative integer $\mu$ , by summing over $x_2$ such that $\max_{j\in J} x_{2j} = \mu$ , we obtain

\begin{align*}{\mathbb P}\Big(\!\max_{j\in J}X_{j} = \mu, {\mathcal J} = J\Big) =\sum_{s=0}^m{\mathbb P}({\mathcal J} = J, S = s){\mathcal M}_{m-s, N_2}\Big(x_2 \colon \max_{j\in J} x_{2j} = \mu\Big).\end{align*}

The latter factor does not depend on a particular set J due to the exchangeability property of the multinomial law. We may thus write ${\mathcal M}_{m-s, N_2}(x_2 \colon \max_{j\in J} x_{2j} = \mu) \;=\!:\; F(m-s,N_2,q,\mu)$ and obtain

\begin{align*}{\mathbb P}\Big(\!\max_{j\in J} X_{j} = \mu, {\mathcal J} = J\Big) = \sum_{s=0}^m {\mathbb P}({\mathcal J}=J, S=s) F(m-s,N_2,q,\mu).\end{align*}

By summing over all sets J of size q, we see that

\begin{align*}{\mathbb P}\!\left(\!\max_{j\in {\mathcal J}} X_{j}=\mu\right) = \sum_{s=0}^m {\mathbb P}(S=s) F(m-s,N_2,q,\mu)\end{align*}

does not depend on the specific choice of ${\mathcal J}$ , and the claim of the lemma follows.

Proof of Theorem 1.3. We are going to use an old result from [Reference Erdös and Rényi7] about the existence of perfect matching in a random bipartite graph. Let G be a uniformly distributed $n+n$ bipartite graph with $m=m(n)$ edges. If

(2.14) \begin{equation}\lim_{n\to\infty} \bigg(\frac{m}{n} - \log n\bigg) = \infty,\end{equation}

then with probability tending to 1, as $n\to \infty$ , G has a perfect matching.

In matrix form, this result asserts the following. Let $Y=Y(n,m)=\{Y_{ij}\}_{1\le i,j\le n}$ be a uniformly distributed random $n \times n$ matrix with entries taking values in $\{0,1\}$ and satisfying $\sum_{i,j=1}^n Y_{ij} =m$ . If (2.14) holds, then

(2.15) \begin{equation}\lim_{n\to\infty} {\mathbb P}\Bigg(\!\max_\sigma \sum_{i=1}^n Y_{i\sigma(i)} = n\Bigg) = 1.\end{equation}

Now let $X=(X_{ij})$ be our matrix following the multinomial law ${\mathcal M}(m, n^{2})$ . Introduce the matrix ${\widetilde Y}$ as

\begin{align*}{\widetilde Y}_{ij} \;:\!=\;\begin{cases}0, & X_{ij}>0,\\[5pt] 1, & X_{ij}=0.\end{cases}\end{align*}

Note that ${\mathbb P}({\widetilde Y}_{ij}=1) = {\mathbb P}(X_{ij} = 0) = (1 -p)^m = \exp\!(\!-\! m p (1+o(1)))$ . Let $T\;:\!=\; \sum_{i,j=1}^n {\widetilde Y}_{ij}$ be the number of empty cells in our matrix X. Observe that, conditioned on T, the matrix ${\widetilde Y}$ has the same distribution as Y(n, T). Taking into account that the probability in (2.15) is nondecreasing as a function of m, we have, for every positive integer M,

(2.16) \begin{align}& {\mathbb P}\Big(\!\min_\sigma S(\sigma) = 0\Big) ={\mathbb P}\Bigg(\!\max_\sigma\sum_{i=1}^n{\widetilde Y}_{i\sigma(i)} = n\Bigg) \nonumber \\[5pt] & = \sum_{k\geq 0} {\mathbb P}\Bigg(\!\max_\sigma\sum_i\tilde Y_{i\sigma(i)} = n \mid T=k \Bigg){\mathbb P}(T=k) \nonumber \\[5pt] & \geq {\mathbb P}\Bigg(\!\max_\sigma\sum_{i=1}^n Y(n,M)_{i\sigma(i)} = n\Bigg)\sum_{k\geq M}{\mathbb P}\Bigg(\!\max_\sigma\sum_i\tilde Y_{i\sigma(i)} = n \mid T=k\Bigg){\mathbb P}(T=k) \nonumber \\[5pt] & \ge {\mathbb P} (T\ge M) {\mathbb P} \Bigg(\!\max_\sigma\sum_{i=1}^n Y(n,M)_{i\sigma(i)} = n \Bigg). \end{align}

We choose $M=n^{\beta}$ with $\beta\in(1,2-c)$ and show that both probabilities in the latter product tend to 1 as $n\to\infty$ .

For the first one, using (1.4), we have ${\mathbb E}\, T = n^2 {\mathbb E}\,{\widetilde Y}_{11} = n^2 \exp\!(\!-\! m p (1+o(1))) \ge n^{2-c(1+o(1))}$ . Furthermore, since the variables ${\widetilde Y}_{ij}$ are negatively correlated, we have $\textrm{Var}\, T \le n^2 \textrm{Var}\,{\widetilde Y}_{11} \le n^2 {\mathbb E}\,{\widetilde Y}_{11} = {\mathbb E}\, T$ . Finally, using $\beta < 2-c$ , by Chebyshev’s inequality,

\begin{align*}{\mathbb P}(T\le n^\beta)& \le {\mathbb P}(|T-{\mathbb E}\, T|\ge {\mathbb E}\, T-n^\beta) = {\mathbb P}(|T-{\mathbb E}\, T|\ge {\mathbb E}\, T (1+o(1))) \\[5pt] & \le \frac{\textrm{Var}\, T}{({\mathbb E}\, T)^2 (1+o(1))} \le \frac{{\mathbb E}\, T}{({\mathbb E}\, T)^2 (1+o(1))} \to 0.\end{align*}

On the other hand, since $\beta>1$ , the assumption (2.14) with $m\;:\!=\;M=n^\beta$ is true. Therefore, the second probability in the product (2.16) tends to 1 by the result in [Reference Erdös and Rényi7]. We obtain from (2.16) that $\lim_{n\to\infty} {\mathbb P}(\!\min_\sigma S(\sigma) = 0) = 1$ , which is the desired claim.

Proof of Theorem 1.4. The proof goes along the same lines as for Theorem 1.2. Instead of the key relation (2.5), we prove the following claim. Let $(X_j)$ be negatively associated random variables following the binomial law ${\mathcal B}(m,p)$ . Then, under assumptions (1.5) and (1.6),

(2.17) \begin{equation}{\mathbb E}\, \max_{1\le j\le n}X_j = \frac{\log n}{\log\!(({\log n})/{mp})} (1+o(1)) \qquad \textrm{as } n\to\infty.\end{equation}

For the upper bound in (2.17) that we shall prove now, no lower bound on mp is needed; we only use (1.5).

Let $\beta>1$ , $y \;:\!=\; ({\beta \log n})/{mp}$ , $r \;:\!=\; {y}/({\log y})$ . Notice that under (1.5) we have $y,r\to \infty$ . Next, for a binomial ${\mathcal B}(m,p)$ random variable X and for every $v>0$ ,

\begin{align*}{\mathbb P} \bigg(X\ge \frac{\beta\log n}{\log\!(({\log n})/{mp})} + v\bigg)& \le {\mathbb P}\bigg(X \ge \frac{\beta \log n}{\log\!(({\beta \log n})/{mp})} + v\bigg) \\[5pt] & = {\mathbb P}\bigg(X \ge \frac{(\beta\log n)/{mp}}{\log\!(({\beta \log n})/{mp})} mp+ v\bigg) \\[5pt] & = {\mathbb P}\bigg(X \ge \frac{y}{\log y} mp + v\bigg) = {\mathbb P}(X \ge r mp + v).\end{align*}

In the next calculation we use the Poisson version of the bound for exponential moment, ${\mathbb E}\, \exp\!(\gamma X) \le \exp\!(mp (\textrm{e}^\gamma-1))$ , that immediately follows from the exact formula (2.1). By applying Chebyshev’s inequality with Poisson-optimal parameter $\gamma=\log r$ we obtain

\begin{align*}{\mathbb P}(X \ge rmp + v)& \le {\mathbb E}\, \exp\!(\gamma X) \exp\!(\!-\!\gamma (rmp+v)) \\[5pt] & \le \exp\!(\!-\!mp(\gamma r - \textrm{e}^\gamma + 1) - \gamma v) \\[5pt] & = \exp\!(\!-\! m p (r\log r - r + 1) - \gamma v).\end{align*}

Since $r\to\infty$ , $r \log r - r + 1 \sim r \log r \sim y = ({\beta \log n})/{mp}$ . It follows that

\begin{align*}{\mathbb P}(X \ge rmp + v)& \le \exp\!(\!-\!\beta\log n (1+o(1)) - \gamma v) = n^{-\beta(1+o(1))} \exp\!(\!-\!\gamma v), \\[5pt] {\mathbb P}\Big(\!\max_{1\le j\le n} X_j \ge rmp + v\Big)& \le n{\mathbb P}(X \ge rmp + v) \le n^{-(\beta-1)(1+o(1))}\exp\!(\!-\!\gamma v).\end{align*}

Hence,

\begin{equation*}{\mathbb E}\,\max_{1\le j\le n} X_j \le rmp + n^{-(\beta-1)(1+o(1))}\int_0^\infty \exp\!(\!-\!\gamma v)\,\textrm{d} v =rmp + n^{-(\beta-1)(1+o(1))} \gamma^{-1}.\end{equation*}

Note that $rmp\gamma = r\log r\, mp \sim y mp = \beta \log n \to \infty$ , and hence we conclude that $n^{-(1-\beta)(1+o(1))}\gamma^{-1}$ is negligible compared to rmp; thus,

\begin{equation*}{\mathbb E}\, \max_{1\le j\le n} X_j \le rmp (1+o(1)) \sim \frac{\beta\log n}{\log\!(({\log n})/{mp})}\end{equation*}

and the required upper bound follows by letting $\beta \searrow 1$ .

For the lower bound, let $\beta\in (0,1)$ , $y \;:\!=\; ({\beta \log n})/{mp}$ , $r \;:\!=\; {y}/({\log y})$ , and

\begin{align*}k\;:\!=\; r mp = \frac{y}{\log y} mp = \frac {\beta \log n}{\log y} .\end{align*}

Assumption (1.5) yields $y\to\infty$ , $k=o(\!\log n)$ , $\textrm{e}^k = n^{o(1)}$ , and $\textrm{e}^{mp}=n^{o(1)}$ .

On the other hand, under assumption (1.6) we have $|\log\!(mp)|\ll \log n$ , which yields $\log y \ll \log n$ , hence $k\to \infty$ .

Rewriting the binomial probability mass function as

\begin{align*}{\mathbb P}(X=k) = \textrm{e}^{-mp} \frac{(mp)^k}{k!}\bigg[\frac{m(m-1)\cdots(m-k+1)}{m^k} \times\bigg(1-\frac{pm}{m}\bigg)^{m-k} \textrm{e}^{pm} \bigg],\end{align*}

and observing that the product between square brackets converges to 1 for m, p, k as above, we can make a Poissonian approximation. Using the latter, we obtain

\begin{align*}{\mathbb P}(X \ge k)& \ge {\mathbb P}(X=k) \sim \textrm{e}^{-mp}\frac{(mp)^k}{k!} \sim \textrm{e}^{-mp}\textrm{e}^k(2\pi k)^{-1/2}\bigg(\frac{mp}{k}\bigg)^k \\[5pt] & = n^{o(1)} r^{-k} = n^{o(1)} r^{-rmp} = n^{o(1)} \exp\!(\!-\!r \log r\, mp) \\[5pt] & = n^{o(1)}\exp\!(\!-\!y (1+o(1)) mp) = n^{-\beta+o(1)}.\end{align*}

By repeating the arguments from (2.10), (2.11), and (2.12), we obtain

\begin{align*}{\mathbb E}\,\max_{1\le j\le n} X_j \ge k(1+o(1)) = \frac{y}{\log y} mp (1+o(1)) =\frac{\beta \log n}{\log y} (1+o(1)) ;\end{align*}

letting $\beta\nearrow 1$ provides the required lower bound in (2.17).

Once (2.17) is proved, the proof of Theorem 1.4 is completed by the same simple arguments (including the greedy method) as for Theorem 1.2. Indeed, combining (2.17) with (2.13) implies that

\begin{equation*} {\mathbb E}\,\max_\sigma S(\sigma) \geq \sum_{i=1}^{n} {\mathbb E}\,\max_{1\le j \le n-i+1} X_{ij} \geq\sum_{i=1}^{n} \frac{\log\!(n-i+1)}{\log[{\log\!(n-i+1)}/{mp}]} (1+o(1)),\end{equation*}

from which the claim easily follows.

Proof of Theorem 1.5. We first establish the upper bound. Let $X_j$ , $1\le j\le n$ , be ${\mathcal B}(m,p)$ -binomial random variables. We have

\begin{align*}{\mathbb E}\, \max_{1\le j\le n} X_j& = {\mathbb E}\,\Big[\max_{1\le j\le n} X_j{\mathbf 1}_{ \{\max_{1\le j\le n} X_j \le k\}}\Big] +{\mathbb E}\,\Big[\max_{1\le j\le n} X_j{\mathbf 1}_{ \{\max_{1\le j\le n} X_j > k\}}\Big] \\[5pt] & \le k + \sum_{j=1}^n {\mathbb E}\,\big[X_j{\mathbf 1}_{ \{X_j > k\}}\big] = k + n {\mathbb E}\,\big[X_1 {\mathbf 1}_{ \{X_1 > k\}}\big].\end{align*}

Furthermore, since the law of $X_1$ is ${\mathcal B}(m,p)$ ,

\begin{align*}{\mathbb P}(X_1=\ell) = \frac{m!}{(m-\ell)!} \frac{p^\ell}{\ell!} (1-p)^{m-\ell} \le\frac{m^\ell p^\ell}{\ell!}, \qquad 0\le \ell \le m.\end{align*}

Hence,

\begin{equation*}{\mathbb E}\,\big[X_1 {\mathbf 1}_{ \{X_1 > k\}}\big] \le \sum_{\ell=k+1}^\infty \frac{(mp)^\ell}{(\ell-1)!} =\sum_{q=0}^\infty \frac{(mp)^{k+1+q}}{(k+q)!} \le (mp)^{k+1} \textrm{e}^{mp} = (mp)^{k+1} (1+o(1)).\end{equation*}

Therefore, ${\mathbb E}\,\max_{1\le j\le n}X_j \le k + c^{k+1}n^{1-a(k+1)}(1+o(1)) = k + o(1)$ , where we used the lower bound in (1.8) at the last step.

It follows immediately that

(2.18) \begin{equation}{\mathbb E}\, \max_\sigma S(\sigma) \le n k (1+\operatorname{O}\!(1)).\end{equation}

Turning to the lower bound, for every positive integer v in the independent case, we have

\begin{align*}{\mathbb P}\Big(\!\max_{1\le j\le v} X_j < k\Big) = {\mathbb P}(X_1 < k)^{v}& = (1 - {\mathbb P}(X_1 \ge k))^{v} \\[5pt] & \le (1 - {\mathbb P}(X_1 = k))^{v} \\[5pt] & \le \exp\{-v {\mathbb P}(X_1 = k)\} \\[5pt] & = \exp\bigg\{{-}v \frac{c^k n^{-a k}}{k!} (1+o(1))\bigg\},\end{align*}

where the second last inequality follows from the inequality $1 + x \le \exp x$ , and the last one follows from

\begin{align*}{\mathbb P}(X_1 = k) = \frac{(mp)^k}{k!}\bigg[\frac{m(m-1)\cdots(m-k+1)}{m^k} (1-p)^{m-k}\bigg] =\frac{(mp)^k}{k!}(1+o(1))\end{align*}

applied to this case; recall (1.7). Let us choose some arbitrary $\delta\in (0,1)$ . By letting $v=[\delta n]$ and using the upper bound in (1.8) we obtain ${\mathbb P}\big(\!\max_{1\le j\le [\delta n] } X_j < k\big) \to 0$ . It follows that

\begin{align*}{\mathbb E}\,\max_{1\le j\le [\delta n]}X_j \ge k{\mathbb P}\Big(\!\max_{1\le j\le [\delta n]} X_j \ge k\Big) = k(1+o(1)).\end{align*}

By using the negative association argument (2.12), we also obtain ${\mathbb E}\,\max_{1\le j\le [\delta n]} X_j \ge k(1+o(1))$ in the multinomial setting. Combining this with the greedy method (2.13) yields

\begin{equation*}{\mathbb E}\,\max_\sigma S(\sigma) \geq \sum_{i=1}^{n} {\mathbb E}\, \max_{1\le j \le n-i+1} X_{ij} \geq\sum_{i=1}^{n-[\delta n]+1} k (1+o(1)) = (1-\delta) n k (1+o(1)).\end{equation*}

By letting $\delta\searrow 0$ we obtain ${\mathbb E}\, \max_\sigma S(\sigma) \geq n k (1+o(1))$ . We then conclude from this and (2.18) that ${\mathbb E}\, \max_\sigma S(\sigma) =k n (1+o(1))$ .

Proof of Theorem 1.6. The upper bound $\max_\sigma S(\sigma) \le m$ is trivial; it remains to prove the lower bound.

Let us denote by $(u_i,v_i)_{1\le i\le m}$ the coordinates of the particles thrown on the square table. All $u_i$ and all $v_i$ are i.i.d. random variables uniformly distributed on the integers $[1..n]$ . Let $U_0=V_0=\emptyset$ , $U_k \;:\!=\; \{u_i, 1 \le i \le k\}$ , $V_k \;:\!=\; \{v_i, 1 \le i \le k\}$ , $1 \le k \le m$ , and introduce the events $A_k \;:\!=\; \{u_{k} \not\in U_{k-1}, v_{k} \not\in V_{k-1}\}$ , $1\le k\le m$ . Note that the sequence of events $A_k$ describes the possibility of constructing a matching of size k iteratively by keeping track of the rows and columns already used in the construction of a permutation matrix and ‘locking’ them. This understanding underlies the validity of (2.19) below. It is obvious that, for each k, ${\mathbb P}(A_k) \ge 1 - {2m}/{n}$ ; hence, by $m \ll n$ ,

\begin{align*}{\mathbb E}\,\Bigg(\sum_{k=1}^m{\mathbf 1}_{ \{A_k\}}\Bigg) \ge m\bigg(1 - \frac{2m}{n}\bigg) = m (1+o(1)).\end{align*}

On the other hand, we have

(2.19) \begin{equation}\max_\sigma S(\sigma) \ge \sum_{k=1}^m {\mathbf 1}_{ \{A_k\}},\end{equation}

which entails the desired ${\mathbb E}\, \max_\sigma S(\sigma) \ge m (1+o(1))$ .