# Approximating Binomial Coefficients with Entropy

This post recounts the story of one of my favorite “probabilistic method” proofs,

$\log \binom{n}{k} \leq n H(k/n)$

where $H(p)$ is the binary entropy function i.e. the entropy of a coin flipped with bias $p$. Maybe more properly the proof is just counting in two ways?

By a tiny tweak to our entropy argument as done here, one can get a better bound

$\log\displaystyle\sum_{i = 0}^{k} \binom{n}{k} \leq nH(k / n)$

This is what’s typically used in practice to bound e.g. the sum of the first $1/3$ of the binomial coefficients by $2^{n H(1/3)}$. The proof here gives all the essential ideas for that proof. With Stirling’s approximation used to give error bounds, both approximations are tight when $k = \Theta(n)$,

$\binom{n}{k} = (1+o(1))2^{n H(k / n)}$

The experiment: Let the random variable $X$ denote the result of flipping $n$ fair coins i.e. $X$ is binomially distributed. Now suppose we know $X$ has exactly $k$ Heads.

The question: How much entropy is in the conditional distribution of $X$ given $k$ Heads? I.e. what is $H(X \mid k\text{ Heads})$?

The left side: The conditional distribution of $X$ is uniform over $\binom{n}{k}$ possible results. The entropy of the uniform distribution on $N$ objects is $\log N$, therefore

$H(X \mid k\text{ Heads}) = \log \binom{n}{k}$.

The right side: On the other hand, $X$ is specified by its bits $X_i$, so

$H(X \mid k \text{ Heads}) = H(X_1, X_2, \dots, X_n \mid k \text{ Heads})$

$H(X_1, X_2, \dots, X_n \mid k \text{ Heads}) \leq H(X_1 \mid k \text{ Heads}) + H(X_2 \mid k \text{ Heads}) + \cdots + H(X_n \mid k \text{ Heads})$

This expression is easy to compute: when flipping $k$ Heads of $n$ flips, each $X_i$ looks like a coin with bias $k /n$, therefore its entropy is $H(k/n)$. So

$H(X_1 \mid k \text{ Heads}) + H(X_2 \mid k \text{ Heads}) + \cdots + H(X_n \mid k \text{ Heads})= n H(k/n)$

giving the right-hand side.

## Further Generalization: Multinomial Coefficients

We can naturally generalize the above argument to provide an approximation for multinomial coefficients. Fix natural numbers $k_1, k_2, \dots, k_m$ with $\sum_i k_i = n$. Let $K$ be a distribution over $[m]$ taking $i$ with probability $k_i / n$. Then

$\log \binom{n}{k_1, k_2, \dots, k_m} \leq nH(K)$

The proof generalizes perfectly: the underlying set is $[m]^{n}$ (place a letter from $\{1, 2, \dots, m\}$ at each of $n$ indices), and we consider the entropy of a uniform choice of a string constrained to have the fixed letter histogram $k_1, k_2, \dots, k_m$. This is exactly the left-hand side, and subadditivity yields the right-hand side.

Moreover, by Stirling again the approximation is tight when $k_i = \Theta(n)$.

# How I Remember the Chernoff Bound

The Chernoff bound is of critical importance in computer science and probability. When I need it, though, I almost always have to look it up to make sure I get all the parameters correct. Not any more, though. I’ll explain how the Chernoff bound is essentially the PDF of the “Gaussian you would expect” from the Central Limit Theorem.

The most common form of the Chernoff bound is the following: suppose you have $n$ independent and identically distributed coin flips $X_1, X_2, \dots, X_n$, say the result of repeatedly flipping a coin that comes up Heads with probability $p$. The number of Heads is by definition a Binomial distribution; let $S$ denote this random variable. Then with all but exponentially small probability, $S$ is within $O(\sqrt{n})$ of its mean:

$\text{Pr}[\left|S - np\right| > \delta \sqrt{n}] \leq 2\exp(-\delta^2 / 2p(1-p))$

The Central Limit Theorem (CLT) tells us that as $n \to \infty$, the distribution $(S - \mathbb{E}(S)) / \sqrt{\text{Var}(S)}$ approaches a standard normal distribution. Philosophically, $S$ approaches a normal distribution with the parameters you would expect: mean $\mathbb{E}(S)$ and variance $\text{Var}(S)$ (though the “scaled up” distributions may not necessarily converge from the CLT alone).

In our case, $\mathbb{E}(S) = np$ and $\text{Var}(S) = np(1-p)$. The PDF of the “Gaussian you would expect” is

$p(S = x) \approx C \cdot \exp(-(x - np)^2 / 2np(1-p))$

for an appropriate normalizing constant $C = 1/\sqrt{2\pi n p(1-p)}$. Let’s rewrite that with $x = np + \delta\sqrt{n}$:

$p(S = np + \delta\sqrt{n}) \approx C \cdot \exp(-(\delta \sqrt{n})^2 / 2np(1-p)) =C \cdot \exp(-\delta^2/2p(1-p))$

This is (up to the factor of $C$) the expression appearing in the Chernoff bound! Thus we can think of the Chernoff bound as expressing an “even when $n$ is small” version of the CLT, with a little bit of loss from $1/\sqrt{2\pi}$ to $2$ in the multiplicative factor.

### Using the CDF instead of the PDF

We lost some credibility at one point in the technique above: we actually should have been looking at the cumulative probability

$\text{Pr}[\left|S - np \right| \geq \delta \sqrt{n}]$

Instead we noticed that the Chernoff bound can be remembered by looking at the PDF (and ignoring a nonconstant factor)

$p(S = np+\delta\sqrt{n})$

But using the CDF instead of the PDF actually gives the same expression, and with a truly constant factor. Let’s compute:

$\text{Pr}[\left|S - np \right| \geq \delta \sqrt{n}] = 2\int_{np+\delta\sqrt{n}}^\infty p(S = x) d x$

$= 2\int_{\delta\sqrt{n}}^\infty 1/ \sqrt{2\pi np(1-p)} \cdot \exp(-x^2 / 2np(1-p)) d x$

If we break up the integral into little $\sqrt{n}$ size pieces from $\delta \sqrt{n}$ to $\infty$, the integral on a piece $[k\sqrt{n}, (k+1)\sqrt{n}$ looks like

$\int_{k\sqrt{n}}^{(k+1)\sqrt{n}} 1/ \sqrt{2\pi np(1-p)} \cdot \exp(-x^2 / 2np(1-p)) d x$

$\leq1/ \sqrt{2\pi np(1-p)} \int_{k\sqrt{n}}^{(k+1)\sqrt{n}} \exp(-k^2 / 2p(1-p)) d x$

$= 1/ \sqrt{2\pi p(1-p)} \exp(-k^2 / 2p(1-p))$

The exponent increases faster than linearly,  and it is the only thing that changes in $k$, so this contribution to the integral goes to 0 faster than geometrically.  Therefore the integral is bounded by an infinite geometric series with initial term proportional to $\exp( -\delta^2/2p(1-p))$. Therefore the cumulative probability is (up to a constant) the expression in the Chernoff bound

$\exp(-\delta^2/2p(1-p))$

and this time the constant is actually constant!

# How to Easily Compute Matrix Derivatives, Fréchet Style

The routine, traditional interpretation of a derivative is as a slope of the tangent line. This is great for a function of a single variable, but if you have a multivariate function $f: \mathbb{R}^m \to \mathbb{R}^n$, how can you talk about slopes? One approach is to make a conceptual leap: a derivative is a linear approximation to your function; the tangent line itself is the derivative, rather than the slope of the tangent line. This is called the Fréchet derivative.

Suppose you have a multivariate function $f: \mathbb{R}^m \to \mathbb{R}^n$. At some point in the domain $x_0 \in \mathbb{R}^m$, we want to say that if you make a small change $\epsilon \in \mathbb{R}^m$ around $x_0$, the change in $f$ is approximately a linear function in $\epsilon$:

$f(x_0 + \epsilon) \approx f(x_0) + Df_{x_0}(\epsilon)$

Here $Df_{x_0} : \mathbb{R}^m \to \mathbb{R}^n$ is our notation for the derivative function (which is linear, and can be represented as a matrix applied to $\epsilon$). $Df_{x_0}$ could depend highly on $x_0$; after all, $f$ will probably have different slopes at different points. More formally, we want a function $Df_{x_0}$ such that

$f(x_0 + \epsilon) = f(x_0) + Df_{x_0}(\epsilon) + o(\epsilon)$

where $o(\epsilon)$ is a function that “is tiny compared to $\epsilon$” e.g. every entry of the vector is the square of the largest entry of $\epsilon$.

### Computing Fréchet Derivatives is Incredibly Elegant

To compute $Df_{x_0}(\epsilon)$, just plug in $x_0 + \epsilon$ to $f$, collect all the terms that are linear in $\epsilon$, and drop all the terms that “get small as $\epsilon \to 0$“. In practice, this means we drop all “higher-order terms” in which we multiply two coordinates of $\epsilon$. The best way to illustrate is with examples.

### Example 1: f(x) = x^Tx

If we change $x$ a little, how does $x^\top x$ change? Here $f: \mathbb{R}^n \to \mathbb{R}$ is $f(x) = x^\top x$.

$f(x + \epsilon) = (x + \epsilon)^\top (x + \epsilon) = x^\top x + x^\top \epsilon + \epsilon^\top x + \epsilon^\top \epsilon$

$= f(x) +x^\top \epsilon + \epsilon^\top x + \epsilon^\top \epsilon$

The part linear in $\epsilon$ here is $x^\top \epsilon + \epsilon^\top x$, and $\epsilon^\top \epsilon$ is tiny compared to $\epsilon$. So

$Df_x(\epsilon) = x^\top \epsilon + \epsilon^\top x = 2x^\top \epsilon$

$f(x+\epsilon) \approx f(x) + 2x^\top \epsilon$

If we want to see where $x^\top x$ is minimized, the critical points are found by finding where the function $Df_x$ is 0; that is, where we can’t move a little bit to go up or down. In this case,

$\forall \epsilon.\; 2x^\top \epsilon = 0 \iff x = 0$

### Example 2: Derivative of the Matrix Inverse

Let’s look at a function of a matrix,

$f: \mathbb{R}^{n \times n} \to \mathbb{R}^{n \times n}$

$f(M) = M^{-1}$

As we change $M$ by a matrix $X$ with small entries (in particular, small enough to ensure $M+X$ is still invertible), how does $(M+X)^{-1}$ change from $M^{-1}$? Just thinking about dimensions, the answer to that question should be by adding a small matrix to $M^{-1}$.

$f(M+X) = (M+X)^{-1} \approx M^{-1} + Df_M(X)$

The hard part is massaging $(M+X)^{-1}$ to look like $M^{-1}$ plus a linear function of $X$. To start,

$(M+X)^{-1} = M^{-1}(I + XM^{-1})^{-1}$

The matrix $I + XM^{-1}$ can be inverted using the Neumann series, which is just a fancy name for the geometric series summation formula

${1 \over I + XM^{-1}} = I - XM^{-1} + (XM^{-1})^2 - (XM^{-1})^3 + \cdots$

Note that the series converges for small enough $X$ because $XM^{-1}$ has all eigenvalues less than 1.

Because $X$ is small, $X^2$ is negligibly tiny. For our linear approximation, we can drop the terms that are too tiny to be noticed by the linear approximation, which is all but the first two terms of this series.

${1 \over I + XM^{-1}} \approx I - XM^{-1}$

$M^{-1}(I + XM^{-1})^{-1} \approx M^{-1}(I - XM^{-1}) = M^{-1} - M^{-1}XM^{-1}$

Remembering $f(M) = M^{-1}$, altogether we just computed

$f(M+X) \approx f(M) - M^{-1}XM^{-1}$

Noting that this second term is linear in $X$, we have the derivative

$Df_{M}(X) = -M^{-1}XM^{-1}$

As a remark: whenever we wrote $\approx$, we could have written $+ O(X^2)$ to be a little more formal about carrying the higher-order terms through the computation.

### Example 3: Derivative of the Log of the Determinant

Now let’s do another example of a matrix function, the log of the determinant

$f: \mathbb{R}^{n \times n} \to \mathbb{R}$

$f(M) = \log |M|$

This function arises frequently in machine learning and statistics when computing log likelihoods.

Change $M$ by a matrix $X$ which has small entries

$f(M + X) = \log |M + X| = \log |M| |I + M^{-1}X| = \log |M| + \log |I + M^{-1}X|$

Now let’s look at $|I + M^{-1}X|$. Each of the entries of $M^{-1}X$ is “small”, though they’re essentially the same size as elements of $X$ (just differing by a few constants depending on $M^{-1}$). The determinant of a matrix $A$ is a sum over all permutations

$|A| = \displaystyle\sum_{\sigma \in S_n} \prod_{i = 1}^n A_{i, \sigma(i)}$

When performing this sum for $I + M^{-1}X$, if a product contains two off-diagonal terms, those will be two “small” terms from $M^{-1}X$ multiplied together, and multiplying two small terms makes a negligible term that we drop. So we only need to sum up $\sigma$ that include at least all but 1 diagonal elements. The only permutation that does this is the identity permutation.

$|I + M^{_1}X| \approx \prod_{i = 1}^n (I + M^{-1}X)_{i,i} =\prod_{i = 1}^n (1 + M^{-1}X_{i,i})$

Same trick: this sum has $2^n$ terms, but any term that multiples an $M^{-1}X_{i,i}$ with a $M^{-1}X_{j,j}$ is negligibly small. To get a term with at most one of the $M^{-1}X_{j,j}$, either we (1) take 1 from every factor, or (2) take 1 from all but one factor

$\prod_{i = 1}^n (1 + M^{-1}X_{i,i}) \approx 1^n + 1^{n-1}M^{-1}X_{1,1} +1^{n-1}M^{-1}X_{2,2} + \cdots + 1^{n-1}M^{-1}X_{n,n}$

$= 1 + \text{tr}(M^{-1}X)$

This is our linear approximation $|I + M^{-1}X| \approx 1 +\text{tr}(M^{-1}X)$. To incorporate the log, note that we’re taking the log of something that is pretty much 1. The normal derivative of log at 1 is 1, and in our Fréchet derivative formulation

$\log (1 + \epsilon) = \epsilon + O(\epsilon^2)$

$\log |I + M^{-1}X| \approx \log(1 + \text{tr}(M^{-1}X)) \approx\text{tr}(M^{-1}X)$

And so the derivative is $Df_{M}(X) = \text{tr}(M^{-1}X)$.

As a remark: incorporating log illustrates the composability of Fréchet derivatives. This makes sense when spoken aloud: if we have a linear approximation $Df$ to $f$ and a linear approximation $Dg$ to $g$, a linear approximation of $f \circ g$ is $Df \circ Dg$.

### Other Viewpoints

Actually, $Df_{x_0}(\epsilon) = J\epsilon$ where $J$ is the Jacobian matrix of partial derivatives ${\partial f_i \over \partial x_j}$. But it’s often easier to not compute the Jacobian entirely, and keep $Df_{x_0}$ in an “implicit” form like in Example 2.