a repository of mathematical know-how

Bounding probabilities by expectations

Quick description

If X is a random variable that takes non-negative values, then the probability that X\geq t is at most \mathbb{E}X/t. This simple result is called Markov's inequality. Clever uses of Markov's inequality can have surprisingly powerful consequences. The main trick is to choose an increasing function F and apply Markov's inequality to the random variable F(X) instead. If F is rapidly growing but \mathbb{E}F(X) is not too large, then one can prove that the probability that X\geq t, which equals the probability that F(X)\geq F(t), is very small.


A familiarity with basic concepts in probability such as random variables and expectation.

Example 1

Let us begin with a proof of Markov's inequality itself. By the law of total probability we know that

\mathbb{E}X=\mathbb{E}[X|X\geq t]\mathbb{P}[X\geq t]+\mathbb{E}[X|X<t]\mathbb{P}[X<t].

Since X\geq 0 (by hypothesis), this is at least \mathbb{E}[X|X\geq t]\mathbb{P}[X\geq t], which is obviously at least t\mathbb{P}[X\geq t], and from this it follows that \mathbb{P}[X\geq t]\leq\mathbb{E}X/t, as we wanted.

Example 2

It is frequently useful in probability theory (and also in probabilistic combinatorics) to prove that a random variable is close to its mean most of the time. Of course, this may not always be true, which is why we invent concepts such as variance to give us some measure of how concentrated a random variable is about its mean. A precise relationship between the variance of a random variable X and the concentration of X about its mean \mu is given by Chebyshev's inequality, which states that if t>0, then \mathbb{P}[|X-\mu|\geq t]\leq\mathrm{Var}X/t^2.

To prove this, we first recall the definition of the variance of X, which is \mathbb{E}(X-\mu)^2. This instantly suggests a function that we can plug into Markov's inequality: instead of looking at the random variable X we shall look at (X-\mu)^2. (The function x\mapsto(x-\mu)^2 is not increasing, but this is all right because we are actually interested in the random variable |X-\mu|.)

The random variable (X-\mu)^2 is non-negative, so by Markov's inequality we find that

\mathbb{P}[(X-\mu)^2\geq t^2]\leq\mathbb{E}(X-\mu)^2/t^2=\mathrm{Var}X/t^2,

and we are done, since \mathbb{P}[(X-\mu)^2\geq t^2]=\mathbb{P}[|X-\mu|\geq t].

General discussion

Suppose that we have a random variable X with mean \mu such that \mathbb{E}|X-\mu|=r and \mathrm{Var}X=\mathbb{E}(X-\mu)^2=s^2. Then Markov's inequality implies that \mathbb{P}[|X-\mu|\geq t]\leq r/t, while Chebyshev's inequality implies that \mathbb{P}[|X-\mu|\geq t]\leq s^2/t^2. Since all probabilities are at most 1, Markov's inequality tells us nothing unless t\geq r, and Chebyshev's inequality tells us nothing unless t\geq s. Since \mathbb{E}(X-\mu)^2\geq(\mathbb{E}|X-\mu|)^2, this tells us that Markov's inequality will usually give us some information for smaller values than Chebyshev's inequality. However, if \mathbb{E}(X-\mu)^2 and (\mathbb{E}|X-\mu|)^2 are reasonably comparable, then Chebyshev's inequality is much stronger for larger values of t. This illustrates a general principle: a good bound on \mathbb{E}F(X) for a quickly growing positive function F gives us a good upper bound for the probability that X is large. In the next example, we shall see a very strong result of this kind.

Example 3

Let X_1,X_2,\dots,X_n be independent random variables, each of mean 0 and each taking values in the interval [-1,1]. What is the probability that |X_1+\dots+X_n|\geq t?

To get a preliminary feel for this question, let us use the fact that variances of independent events add together. Each X_i obviously has variance at most 1, since it cannot differ from its mean by more than 1. Therefore, the variance of |X_1+X_2+\dots+X_n| is at most n, so by Chebyshev's inequality the probability that |X_1+\dots+X_n|\geq t is at most n/t^2. This tells us that once t gets past \sqrt{n}, the probability decays fairly rapidly. However, we shall now see that it decays far more rapidly than Chebyshev's inequality implies.

To prove this, we shall estimate the probability that X_1+\dots+X_n\geq t and then, by symmetry, we will also have an estimate for the probability that X_1+\dots+X_n\leq -t. To obtain the first estimate we shall think about the expectation of \exp(\lambda(X_1+\dots+X_n)).

The exponential function has three properties that make it a particularly good choice. The first is that it is non-negative. The second is that it turns sums into products, which is very useful to us because the expectation of a product of independent random variables is equal to the product of the expectations. The third is that it is a rapidly growing function.

Using these properties, we find that

 \mathbb{E}\exp(\lambda(X_1+\dots+X_n))=\mathbb{E}\prod_{i=1}^n\exp(\lambda X_i)=\prod_{i=1}^n\mathbb{E}(\exp(\lambda X_i)).

We would like to show that this is not too large, and our problem is now reduced to showing that each \mathbb{E}(\exp(\lambda X_i)) is not too large. It is at this point that we exploit the fact that each X_i has mean 0 and takes values in [-1,1]. As long as \lambda\geq 0, this implies that

 \mathbb{E}(\exp(\lambda(X_i))=1+\lambda\mathbb{E}X_i+\frac{\lambda^2}2\mathbb{E}X_i^2+\dots\leq 1+\frac{\lambda^2}2+\frac{\lambda^3}{3!}+\dots

And now a small calculation shows that if \lambda\leq 1 then this is at most \exp(\lambda^2). Indeed, we can prove this coefficient by coefficient: \frac{\lambda^2}2+\frac{\lambda^3}{3!}\leq\lambda^2, \frac{\lambda^4}{4!}+\frac{\lambda^5}{5!}\leq\frac{\lambda^4}{2}, and so on. Therefore, \mathbb{E}\exp(\lambda(X_1+\dots+X_n))\leq\exp(n\lambda^2).

Now we are ready to apply Markov's inequality. The probability that X_1+\dots+X_n\geq t is equal to the probability that \exp(\lambda(X_1+\dots+X_n))\geq\exp(\lambda t), which by Markov's inequality is at most \exp(n\lambda^2-\lambda t). We are now free to choose \lambda so as to minimize this estimate, provided that 0\leq\lambda\leq 1. The minimum occurs when \lambda=t/2n, when it gives \exp(-t^2/4n). This estimate is valid if t\leq 2n, and therefore for all t, since the probability that X_1+\dots+X_n\geq 2n is zero.

After applying the same argument to -X_1-\dots-X_n, we find that the probability that |X_1+\dots+X_n|\geq t is at most 2\exp(-t^2/4n), so once t gets beyond \sqrt{n} the probability decays very quickly indeed.

This conclusion is far stronger than that of Chebyshev's inequality, but it should be stressed that we have also used a stronger hypothesis: here we assumed that the random variables were bounded, but we could have obtained an upper bound of n/t^2 from the weaker assumption that each X_i had variance at most 1.


a suggestion for improvement

Here is a point which I feel needs to be emphasized more in Example 3.

If we try to make the argument work without leaving \lambda to be optimized later, but by, say, initially setting \lambda=1, the argument will fail: we will get a useless bound. The reason: we need to pick a function F which will grow rapidly while E F(X) will grow slowly. In this case, E e^{X_1 + \ldots X_n} grows pretty fast with n.

What kind of functions F work for this? Well, F(X)=X^2 works: out of the n^2 terms in (X_1 + \cdots + X_n)^2, almost all have expectation 0. Similarly, F(X)=X^{2k} works fine as well: in the sum (X_1 + \cdots + X_n)^{2k}, we will be able to easily cancel many terms because they have expectation 0.

The central insight behind the proof in Example 3 is that picking F(x)=(1+t)^x for really, really small t works. The reason: the average of 1+t and 1/(1+t) is really close to 1. That is: near one, inversion is about the same as subtraction. This insight can be pushed a little further to argue that if X has mean 0, the quantity E (1+t)^X should be pretty close to 1. Thus one might hope that E[(1+t)^{X_1 + \cdots X_n}] = (E[(1+t)^{X_1}])^n grows slowly - after all, it is "approximately" 1^n = 1.

Post new comment

(Note: commenting is not possible on this snapshot.)

Before posting from this form, please consider whether it would be more appropriate to make an inline comment using the Turn commenting on link near the bottom of the window. (Simply click the link, move the cursor over the article, and click on the piece of text on which you want to comment.)