Consider a $n \times n$ matrix $A$ with $k$ nonzero entries. Assume every row and every column of $A$ has at most $\sqrt{k}$ nonzeros. Permute uniformly at random the rows and the columns of $A$. Divide $A$ in $k$ submatrices of size $n/\sqrt{k} \times n/\sqrt{k}$ (i.e. $\sqrt{k}$ meta-rows and meta-columns). Enumerate the $k$ nonzeros and define the following indicator random variable:

\begin{equation} X_{\ell,z} = \begin{cases} 1 & \text{if the $z$-th nonzero entry is in $A^\ell$} \\ 0 & \text{otherwise} \end{cases} \end{equation}

The expected number of nonzero entries in a generic submatrix $A^\ell$ is one. Is it possible to prove Chernoff-Hoeffding bounds on the sum $X_\ell = \sum_{z=1}^k X_{\ell,z}$?

My first guess was to prove negative association, following Dubhashi and Panconesi analysis. Unfortunately, $X_{\ell,z}$ and $X_{\ell,z^\prime}$ are not negatively associated (following the book's notation, if $z$ and $z^\prime$ are in the same row/column then $\mathbf{E}[f(X_{\ell,z})g(X_{\ell,z^\prime})] > \mathbf{E}[f(X_{\ell,z})] \mathbf{E}[g(X_{\ell,z^\prime})]$).

  • 1
    $\begingroup$ "Consider a n×n matrix $A$ with $k$ nonzero entries. Assume every row and every column of $A$ has $\sqrt{k}$ nonzeros." so $k = n\sqrt{k}$, $k=n^2$? $\endgroup$
    – John L.
    Aug 20, 2018 at 12:01
  • $\begingroup$ Edited to address the comment of @Apass.Jack. I have an upper bound on the number of nonzeros in each row/column. $\endgroup$
    – Matteo
    Aug 20, 2018 at 12:11
  • $\begingroup$ I don't understand the notation $A^\ell$. Are you taking a matrix power? $\endgroup$ Aug 21, 2018 at 14:40
  • $\begingroup$ No @ThomasAhle, the notation $A^\ell$ is just a way to enumerate the $k$ submatrices. There is no matrix manipulation except from random permutations. $\endgroup$
    – Matteo
    Aug 21, 2018 at 14:57

1 Answer 1


Okay, here is a full answer.

We will use the fact, that any bipartite graph of maximum degree $d$ can be broken into (at most) $d$ matchings.

In our case, this means that we can split $A$ into (at most) $\sqrt{k}$ disjoint sets of elements $S_i\subseteq\{(i,j)\in[n]\times[n] \mid A_{i,j}=1\}$ of size at most $n$ such that any every element in each set has a unique row and column. (This uses that a $1$ in $A$ can be seen as an edge in the $n\times n$ bipartite graph that is the matrix.)

Now let $X$ be the total number of $1$s in your $n/\sqrt k\times n/\sqrt k$ matrix, $A^\ell$, and let $X_i$ be the number of elements coming from $S_i$. Then $X\ge t\implies \exists_i\, X_i\ge t/\sqrt k$ and so $$\Pr[X\ge t]\le \sqrt k \Pr[X_1 \ge t/\sqrt k]$$ by the union bound. Now, because elements in $S_1$ don't share any rows/columns, it is easy to analyze using your original approach.

In particular, reorder the rows and columns using the matching, so that the first element in $S_1$ is $(1,1)$ the next is $(2,2)$ and so on. Let $r_i$ be the random variable that is 1 if row $i$ is chosen and $c_i$ likewise. Then we want to upper bound $\Pr[\sum_i r_ic_i \ge t]$. Notice $E[r_ic_ir_jc_j] = E[r_ir_j]E[c_ic_j] \le E[r_i]^2E[c_i]^2$ by Maclaurin’s Inequality, so $$\exp(t\sum_i r_ic_i) = \sum_k t^k/k!\,E\left(\sum_i r_ic_i\right)^k \le \sum_k t^k/k!\,E\left(\sum_i b_i\right)^k = \exp(t\sum_i b_i)$$ where $b_i$ is a normal binomial variable with $p=(n/\sqrt{k}/n)^2=1/k$.

Finally, we then get

$$\Pr[X\ge t]\le \sqrt k \Pr[X_1 \ge t/\sqrt k] \le \sqrt k \exp\left(-2\left(\frac{t-p n \sqrt{k}}{n\sqrt k}\right)^2 n\right)$$

or more simply

$$\Pr[X\ge \lambda\sqrt{nk}+n/\sqrt k]\le \sqrt k \exp(-2\lambda^2)$$

or if we use the Hoeffding bound for small values of $p$ (large $k$):

$$\Pr[X\ge \lambda\sqrt{nk}+n/\sqrt k]\le \sqrt k \exp\left(\frac{-\lambda^2}{2/k+\lambda/\sqrt n}\right)$$

Note that if we had simply assumed all elements were independent, we would have gotten the bound $\Pr[X\ge \lambda\sqrt{n\sqrt{k}}+n/\sqrt k]\le \exp(-2\lambda^2)$. Equal to ours except for a factor $k^{1/4}$ in the exponent.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.