# A Proofless Introduction to Information Theory

There are two basic problems in information theory that are very easy to explain. Two people, Alice and Bob, want to communicate over a digital channel over some long period of time, and they know the probability that certain messages will be sent ahead of time. For example, English language sentences are more likely than gibberish, and “Hi” is much more likely than “asphyxiation.” The problems are:

1. Say communication is very expensive. Then the problem is to come up with an encoding scheme for the messages which minimizes the expected length of an encoded message and guarantees the ability to unambiguously decode a message. This is called the noiseless coding problem.
2. Say communication is not expensive, but error prone. In particular, each bit $i$ of your message is erroneously flipped with some known probably $p$, and all the errors are independent. Then the question is, how can one encode their messages to as to guarantee (with high probability) the ability to decode any sent message? This is called the noisy coding problem.

There are actually many models of “communication with noise” that generalize (2), such as models based on Markov chains. We are not going to cover them here.

Here is a simple example for the noiseless problem. Say you are just sending binary digits as your messages, and you know that the string “00000000” (eight zeros) occurs half the time, and all other eight-bit strings occur equally likely in the other half. It would make sense, then, to encode the “eight zeros” string as a 0, and prefix all other strings with a 1 to distinguish them from zero. You would save on average $7 \cdot 1/2 + (-1) \cdot 1/2 = 3$ bits in every message.

One amazing thing about these two problems is that they were posed and solved in the same paper by Claude Shannon in 1948. One byproduct of his work was the notion of entropy, which in this context measures the “information content” of a message, or the expected “compressibility” of a single bit under the best encoding. For the extremely dedicated reader of this blog, note this differs from Kolmogorov complexity in that we’re not analyzing the compressibility of a string by itself, but rather when compared to a distribution. So really we should think of (the domain of) the distribution as being compressed, not the string.

Claude Shannon. Image credit: Wikipedia

## Entropy and noiseless encoding

Before we can state Shannon’s theorems we have to define entropy.

Definition: Suppose $D$ is a distribution on a finite set $X$, and I’ll use $D(x)$ to denote the probability of drawing $x$ from $D$. The entropy of $D$, denoted $H(D)$ is defined as

$H(D) = \sum_{x \in X} D(x) \log \frac{1}{D(x)}$

It is strange to think about this sum in abstract, so let’s suppose $D$ is a biased coin flip with bias $0 \leq p \leq 1$ of landing heads. Then we can plot the entropy as follows

Image source: Wikipedia

The horizontal axis is the bias $p$, and the vertical axis is the value of $H(D)$, which with some algebra is $- p \log p - (1-p) \log (1-p)$. From the graph above we can see that the entropy is maximized when $p=1/2$ and minimized at $p=0, 1$. You can verify all of this with calculus, and you can prove that the uniform distribution maximizes entropy in general as well.

So what is this saying? A high entropy measures how incompressible something is, and low entropy gives us lots of compressibility. Indeed, if our message consisted of the results of 10 such coin flips, and $p$ was close to 1, we could be able to compress a lot by encoding strings with lots of 1’s using few bits. On the other hand, if $p=1/2$ we couldn’t get any compression at all. All strings would be equally likely.

Shannon’s famous theorem shows that the entropy of the distribution is actually all that matters. Some quick notation: $\{ 0,1 \}^*$ is the set of all binary strings.

Theorem (Noiseless Coding Theorem) [Shannon 1948]: For every finite set $X$ and distribution $D$ over $X$, there are encoding and decoding functions $\textup{Enc}: X \to \{0,1 \}^*, \textup{Dec}: \{ 0,1 \}^* \to X$ such that

1. The encoding/decoding actually works, i.e. $\textup{Dec}(\textup{Enc}(x)) = x$ for all $x$.
2. The expected length of an encoded message is between $H(D)$ and $H(D) + 1$.

Moreover, no encoding scheme can do better.

Item 2 and the last sentence are the magical parts. In other words, if you know your distribution over messages, you precisely know how long to expect your messages to be. And you know that you can’t hope to do any better!

As the title of this post says, we aren’t going to give a proof here. Wikipedia has a proof if you’re really interested in the details.

## Noisy Coding

The noisy coding problem is more interesting because in a certain sense (that was not solved by Shannon) it is still being studied today in the field of coding theory. The interpretation of the noisy coding problem is that you want to be able to recover from white noise errors introduced during transmission. The concept is called error correction. To restate what we said earlier, we want to recover from error with probability asymptotically close to 1, where the probability is over the errors.

It should be intuitively clear that you can’t do so without your encoding “blowing up” the length of the messages. Indeed, if your encoding does not blow up the message length then a single error will confound you since many valid messages would differ by only a single bit. So the question is does such an encoding exist, and if so how much do we need to blow up the message length? Shannon’s second theorem answers both questions.

Theorem (Noisy Coding Theorem) [Shannon 1948]: For any constant noise rate $p < 1/2$, there is an encoding scheme $\textup{Enc} : \{ 0,1 \}^k \to \{0,1\}^{ck}, \textup{Dec} : \{ 0,1 \}^{ck} \to \{ 0,1\}^k$ with the following property. If $x$ is the message sent by Alice, and $y$ is the message received by Bob (i.e. $\textup{Enc}(x)$ with random noise), then $\Pr[\textup{Dec}(y) = x] \to 1$ as a function of $n=ck$. In addition, if we denote by $H(p)$ the entropy of the distribution of an error on a single bit, then choosing any $c > \frac{1}{1-H(p)}$ guarantees the existence of such an encoding scheme, and no scheme exists for any smaller $c$.

This theorem formalizes a “yes” answer to the noisy coding problem, but moreover it characterizes the blowup needed for such a scheme to exist. The deep fact is that it only depends on the noise rate.

A word about the proof: it’s probabilistic. That is, Shannon proved such an encoding scheme exists by picking $\textup{Enc}$ to be a random function (!). Then $\textup{Dec}(y)$ finds (nonconstructively) the string $x$ such that the number of bits different between $\textup{Enc}(x)$ and $y$ is minimized. This “number of bits that differ” measure is called the Hamming distance. Then he showed using relatively standard probability tools that this scheme has the needed properties with high probability, the implication being that some scheme has to exist for such a probability to even be positive. The sharp threshold for $c$ takes a bit more work. If you want the details, check out the first few lectures of Madhu Sudan’s MIT class.

The non-algorithmic nature of his solution is what opened the door to more research. The question has surpassed, “Are there any encodings that work?” to the more interesting, “What is the algorithmic cost of constructing such an encoding?” It became a question of complexity, not computability. Moreover, the guarantees people wanted were strengthened to worst case guarantees. In other words, if I can guarantee at most 12 errors, is there an encoding scheme that will allow me to always recover the original message, and not just with high probability? One can imagine that if your message contains nuclear codes or your bank balance, you’d definitely want to have 100% recovery ability.

Indeed, two years later Richard Hamming spawned the theory of error correcting codes and defined codes that can always correct a single error. This theory has expanded and grown over the last sixty years, and these days the algorithmic problems of coding theory have deep connections to most areas of computer science, including learning theory, cryptography, and quantum computing.

We’ll cover Hamming’s basic codes next time, and then move on to Reed-Solomon codes and others. Until then!

Posts in this series:

# Probabilistic Bounds — A Primer

Probabilistic arguments are a key tool for the analysis of algorithms in machine learning theory and probability theory. They also assume a prominent role in the analysis of randomized and streaming algorithms, where one imposes a restriction on the amount of storage space an algorithm is allowed to use for its computations (usually sublinear in the size of the input).

While a whole host of probabilistic arguments are used, one theorem in particular (or family of theorems) is ubiquitous: the Chernoff bound. In its simplest form, the Chernoff bound gives an exponential bound on the deviation of sums of random variables from their expected value.

This is perhaps most important to algorithm analysis in the following mindset. Say we have a program whose output is a random variable $X$. Moreover suppose that the expected value of $X$ is the correct output of the algorithm. Then we can run the algorithm multiple times and take a median (or some sort of average) across all runs. The probability that the algorithm gives a wildly incorrect answer is the probability that more than half of the runs give values which are wildly far from their expected value. Chernoff’s bound ensures this will happen with small probability.

So this post is dedicated to presenting the main versions of the Chernoff bound that are used in learning theory and randomized algorithms. Unfortunately the proof of the Chernoff bound in its full glory is beyond the scope of this blog. However, we will give short proofs of weaker, simpler bounds as a straightforward application of this blog’s previous work laying down the theory.

If the reader has not yet intuited it, this post will rely heavily on the mathematical formalisms of probability theory. We will assume our reader is familiar with the material from our first probability theory primer, and it certainly wouldn’t hurt to have read our conditional probability theory primer, though we won’t use conditional probability directly. We will refrain from using measure-theoretic probability theory entirely (some day my colleagues in analysis will like me, but not today).

## Two Easy Bounds of Markov and Chebyshev

The first bound we’ll investigate is almost trivial in nature, but comes in handy. Suppose we have a random variable $X$ which is non-negative (as a function). Markov’s inequality is the statement that, for any constant $a > 0$,

$\displaystyle \textup{P}(X \geq a) \leq \frac{\textup{E}(X)}{a}$

In words, the probability that $X$ grows larger than some fixed constant is bounded by a quantity that is inversely proportional to the constant.

The proof is quite simple. Let $\chi_a$ be the indicator random variable for the event that $X \geq a$ ($\chi_a = 1$ when $X \geq a$ and zero otherwise). As with all indicator random variables, the expected value of $\chi_a$ is the probability that the event happens (if this is mysterious, use the definition of expected value). So $\textup{E}(\chi_a) = \textup{P}(X \geq a)$, and linearity of expectation allows us to include a factor of $a$:

$\textup{E}(a \chi_a) = a \textup{P}(X \geq a)$

The rest of the proof is simply the observation that $\textup{E}(a \chi_a) \leq \textup{E}(X)$. Indeed, as random variables we have the inequality $a \chi_a \leq X$. Whenever $X < a$, the value of $a \chi_a = 0$ while $X$ is nonnegative by definition. And whenever $a \leq X$,the value of $a \chi_a = a$ while $X$ is by assumption at least $a$. It follows that $\textup{E}(a \chi_a) \leq \textup{E}(X)$.

This last point is a simple property of expectation we omitted from our first primer. It usually goes by monotonicity of expectation, and we prove it here. First, if $X \geq 0$ then $\textup{E}(X) \geq 0$ (this is trivial). Second, if $0 \leq X \leq Y$, then define a new random variable $Z = Y-X$. Since $Z \geq 0$ and using linearity of expectation, it must be that $\textup{E}(Z) = \textup{E}(Y) - \textup{E}(X) \geq 0$. Hence $\textup{E}(X) \leq \textup{E}(Y)$. Note that we do require that $X$ has a finite expected value for this argument to work, but if this is not the case then Markov’s inequality is nonsensical anyway.

Markov’s inequality by itself is not particularly impressive or useful. For example, if $X$ is the number of heads in a hundred coin flips, Markov’s inequality ensures us that the probability of getting at least 99 heads is at most 50/99, which is about 1/2. Shocking. We know that the true probability is much closer to $2^{-100}$, so Markov’s inequality is a bust.

However, it does give us a more useful bound as a corollary. This bound is known as Chebyshev’s inequality, and its use is sometimes referred to as the second moment method because it gives a bound based on the variance of a random variable (instead of the expected value, the “first moment”).

The statement is as follows.

Chebyshev’s Inequality: Let $X$ be a random variable with finite expected value and positive variance. Then we can bound the probability that $X$ deviates from its expected value by a quantity that is proportional to the variance of $X$. In particular, for any $\lambda > 0$,

$\displaystyle \textup{P}(|X - \textup{E}(X)| \geq \lambda) \leq \frac{\textup{Var}(X)}{\lambda^2}$

And without any additional assumptions on $X$, this bound is sharp.

Proof. The proof is a simple application of Markov’s inequality. Let $Y = (X - \textup{E}(X))^2$, so that $\textup{E}(Y) = \textup{Var}(X)$. Then by Markov’s inequality

$\textup{P}(Y \geq \lambda^2) \leq \frac{\textup{E}(Y)}{\lambda^2}$

Since $Y$ is nonnegative $|X - \textup{E}(X)| = \sqrt(Y)$, and $\textup{P}(Y \geq \lambda^2) = \textup{P}(|X - \textup{E}(X)| \geq \lambda)$. The theorem is proved. $\square$

Chebyshev’s inequality shows up in so many different places (and usually in rather dry, technical bits), that it’s difficult to give a good example application.  Here is one that shows up somewhat often.

Say $X$ is a nonnegative integer-valued random variable, and we want to argue about when $X = 0$ versus when $X > 0$, given that we know $\textup{E}(X)$. No matter how large $\textup{E}(X)$ is, it can still be possible that $\textup{P}(X = 0)$ is arbitrarily close to 1. As a colorful example, let $X$ is the number of alien lifeforms discovered in the next ten years. We might debate that $\textup{E}(X)$ can arbitrarily large: if some unexpected scientific and technological breakthroughs occur tomorrow, we could discover an unbounded number of lifeforms. On the other hand, we are very likely not to discover any, and probability theory allows for such a random variable to exist.

If we know everything about $\textup{Var}(X)$, however, we can get more informed bounds.

Theorem: If $\textup{E}(X) \neq 0$, then $\displaystyle \textup{P}(X = 0) \leq \frac{\textup{Var}(X)}{\textup{E}(X)^2}$.

Proof. Simply choose $\lambda = \textup{E}(X)$ and apply Chebyshev’s inequality.

$\displaystyle \textup{P}(X = 0) \leq \textup{P}(|X - \textup{E}(X)| \geq \textup{E}(X)) \leq \frac{\textup{Var}(X)}{\textup{E}(X)^2}$

The first inequality follows from the fact that the only time $X$ can ever be zero is when $|X - \textup{E}(X)| = \textup{E}(X)$, and $X=0$ only accounts for one such possibility. $\square$

This theorem says more. If we know that $\textup{Var}(X)$ is significantly smaller than $\textup{E}(X)^2$, then $X > 0$ is more certain to occur. More precisely, and more computationally minded, suppose we have a sequence of random variables $X_n$ so that $\textup{E}(X_n) \to \infty$ as $n \to \infty$. Then the theorem says that if $\textup{Var}(X_n) = o(\textup{E}(X_n)^2)$, then $\textup{P}(X_n > 0) \to 1$. Remembering one of our very early primers on asymptotic notation, $f = o(g)$ means that $f$ grows asymptotically slower than $g$, and in terms of this fraction $\textup{Var}(X) / \textup{E}(X)^2$, this means that the denominator dominates the fraction so that the whole thing tends to zero.

## The Chernoff Bound

The Chernoff bound takes advantage of an additional hypothesis: our random variable is a sum of independent coin flips. We can use this to get exponential bounds on the deviation of the sum. More rigorously,

Theorem: Let $X_1 , \dots, X_n$ be independent random $\left \{ 0,1 \right \}$-valued variables, and let $X = \sum X_i$. Suppose that $\mu = \textup{E}(X)$. Then the probability that $X$ deviates from $\mu$ by more than a factor of $\lambda > 0$ is bounded from above:

$\displaystyle \textup{P}(X > (1+\lambda)\mu) \leq \frac{e^{\lambda \mu}}{(1+\lambda)^{(1+\lambda)\mu}}$

The proof is beyond the scope of this post, but we point the interested reader to these lecture notes.

We can apply the Chernoff bound in an easy example. Say all $X_i$ are fair coin flips, and we’re interested in the probability of getting more than 3/4 of the coins heads. Here $\mu = n/2$ and $\lambda = 1/2$, so the probability is bounded from above by

$\displaystyle \left ( \frac{e}{(3/2)^3} \right )^{n/4} \approx \frac{1}{5^n}$

So as the number of coin flips grows, the probability of seeing such an occurrence diminishes extremely quickly to zero. This is important because if we want to test to see if, say, the coins are biased toward flipping heads, we can simply run an experiment with $n$ sufficiently large. If we observe that more than 3/4 of the flips give heads, then we proclaim the coins are biased and we can be assured we are correct with high probability. Of course, after seeing 3/4 of more heads we’d be really confident that the coin is biased. A more realistic approach is to define some $\varepsilon$ that is small enough so as to say, “if some event occurs whose probability is smaller than $\varepsilon$, then I call shenanigans.” Then decide how many coins and what bound one would need to make the bad event have probability approximately $\varepsilon$. Finding this balance is one of the more difficult aspects of probabilistic algorithms, and as we’ll see later all of these quantities are left as variables and the correct values are discovered in the course of the proof.

## Chernoff-Hoeffding Inequality

The Hoeffding inequality (named after the Finnish statistician, Wassily Høffding) is a variant of the Chernoff bound, but often the bounds are collectively known as Chernoff-Hoeffding inequalities. The form that Hoeffding is known for can be thought of as a simplification and a slight generalization of Chernoff’s bound above.

Theorem: Let $X_1, \dots, X_n$ be independent random variables whose values are within some range $[a,b]$. Call $\mu_i = \textup{E}(X_i)$, $X = \sum_i X_i$, and $\mu = \textup{E}(X) = \sum_i \mu_i$. Then for all $t > 0$,

$\displaystyle \textup{P}(|X - \mu| > t) \leq 2e^{-2t^2 / n(b-a)^2}$

For example, if we are interested in the sum of $n$ rolls of a fair six-sided die, then the probability that we deviate from $(7/2)n$ by more than $5 \sqrt{n \log n}$ is bounded by $2e^{(-2 \log n)} = 2/n^2$. Supposing we want to know how many rolls we need to guarantee with probability 0.01 that we don’t deviate too much, we just do the algebra:

$2n^{-2} < 0.01$
$n^2 > 200$
$n > \sqrt{200} \approx 14$

So with 15 rolls we can be confident that the sum of the rolls will lie between 20 and 85. It’s not the best possible bound we could come up with, because we’re completely ignoring the known structure on dice rolls (that they follow a uniform distribution!). The benefit is that it’s a quick and easy bound that works for any kind of random variable with that expected value.

Another version of this theorem concerns the average of the $X_i$, and is only a minor modification of the above.

Theorem: If $X_1, \dots, X_n$ are as above, and $X = \frac{1}{n} \sum_i X_i$, with $\mu = \frac{1}{n}(\sum_i \mu_i)$, then for all $t > 0$, we get the following bound

$\displaystyle \textup{P}(|X - \mu| > t) \leq 2e^{-2nt^2/(b-a)^2}$

The only difference here is the extra factor of $n$ in the exponent. So the deviation is exponential both in the amount of deviation ($t^2$), and in the number of trials.

This theorem comes up very often in learning theory, in particular to prove Boosting works. Mathematicians will joke about how all theorems in learning theory are just applications of Chernoff-Hoeffding-type bounds. We’ll of course be seeing it again as we investigate boosting and the PAC-learning model in future posts, so we’ll see the theorems applied to their fullest extent then.

Until next time!

# Ramsey Number Lower Bound

Define the Ramsey number $R(k,m)$ to be the minimum number $n$ of vertices required of the complete graph $K_n$ so that for any two-coloring (red, blue) of the edges of $K_n$ one of two things will happen:

• There is a red $k$-clique; that is, a complete subgraph of $k$ vertices for which all edges are red.
• There is a blue $m$-clique.

It is known that these numbers are always finite, but it is very difficult to compute them exactly.

Problem: Prove that the Ramsey number $R(m,m) > n$ whenever $n,m$ satisfy

$\displaystyle \binom{n}{m}2^{1-\binom{m}{2}} < 1$

Solution: Color the edges of $K_n$ uniformly at random (that is, each edge has probability 1/2 of being colored red). For any complete subgraph $G = K_m$, define by event $A_G$ the event that $G$ is monochromatic (its edges are either all red or all blue).

Now the probability that $A_G$ occurs (where $G$ is fixed ahead of time) is easy to compute:

$\displaystyle \textup{Pr}(A_G) = \left (\frac{1}{2} \right)^{\binom{m}{2} - 1} = 2^{1-\binom{m}{2}}$

Since there are $\binom{n}{m}$ possible subgraphs with $m$ vertices, The probability that for some $G$ the event $A_G$ occurs is at most

$\displaystyle \binom{n}{m}2^{1-\binom{m}{2}}$

Whenever this quantity is strictly less than 1 (by assumption) then there is a positive probability that no event $A_G$ will occur. That is, there is a positive probability that a random coloring will have no monochromatic subgraph $K_m$. So there must exist such a coloring, and the Ramsey number $R(m,m)$ must be larger than $n$. $\square$

Discussion: This proof (originally due to Erdős) is a classic example of the so-called probabilistic method. In particular, we create a probability space from the object we wish to study, and then we make claims about the probability of joint events.

While it seems quite simple in nature, the probabilistic method has been successfully applied to a wide variety of problems in mathematics. For instance, there is an elegant proof in complexity theory that $\textup{BPP} \subset \textup{P/poly}$ which uses this same method. The probabilistic method has been applied to loads of problems in combinatorics, number theory, and graph theory, and it forms the foundation of the area of random graph theory (which is the setting in which one studies social networks). Perhaps unsurprisingly, there is also a proof of the fundamental theorem of algebra that uses the probabilistic method.