# Lecture 17: Chebychev's inequality and the weak law of large numbers

• Reading: MCS 20.2 20.4

• Chebychev's inequality
• statement, proof, example
• Weak law of large numbers
• statement, proof, setting up sample space / random variables

## Chebychev's inequality

Claim (Chebychev's inequality): For any random variable $$X$$, $Pr(|X - E(X)| \geq a) \leq \frac{Var(X)}{a^2}$

Proof: Note that $$|X - E(X)| \geq a$$ if and only if $$(X - E(X))^2 \geq a^2$$. Therefore $$Pr(|X - E(X)| \geq a) = Pr((X - E(X))^2 \geq a^2)$$. Applying Markov's inequality to the variable $$(X - E(X))^2$$ gives

\begin{aligned} Pr(|X - E(X)| \geq a) &= Pr((X - E(X))^2 \geq a^2) \\ &\leq \frac{E((X - E(X))^2)}{a^2} \\ &= \frac{Var(X)}{a^2} \end{aligned}

by definition.

Example: Last time we used Markov's inequality and the fact that the average height is 5.5 feet to show that if a door is 55 feet high, then we are guaranteed that at least 90% of people can fit through it.

If we also know that the standard deviation of height is $$σ = 0.2$$ feet, we can use Chebychev's inequality to build a smaller door. Let $$X$$ be the height random variable. $$Var(X) = σ^2 = 0.04$$.

If $$x - E(X) \geq a$$ then $$|x - E(X)| \geq a$$. Therefore, the event $$(X - E(X))$$ is a subset of the event $$(|X - E(X)| \geq a)$$, and thus $$Pr(X - E(X) \geq a) \leq Pr(|X - E(X)| \geq a)$$. This lets us apply Chebychev's inequality to conclude $$Pr(X - E(X) \geq a) \leq \frac{Var(X)}{a^2}$$.

Solving for $$a$$, we see that if $$a \geq .6$$, then $$Pr(X -E(X) \geq a) \leq 0.10$$. This in turn gives us $$Pr(X \lt a + E(X)) = Pr(X - E(X) \lt a) \geq 0.9$$. Thus, if the door is at least $$6.1$$ feet tall, then 90% of the people can fit through.

## Weak law of large numbers

Suppose we wish to estimate the average value of the height of a population by sampling $$n$$ people from the population and averaging their height. The weak law of large numbers says that this will give us a good estimate of the "real" average.

Formally, we can model this experiment by letting our outcomes be sequences of $$n$$ people. We can define several random variables: $$X_1$$ is the height of the first person sampled; $$X_2$$ is the height of the second person sampled, $$X_3$$ is the height of the third and so forth.

Since these are all measures of height, $$E(X_1) = E(X_2) = \cdots = E(X_n)$$ (let's call this value $$\mu$$) and $$Var(X_1) = \cdots = Var(X_n)$$ (let's call this value $$\sigma^2$$). The result of our sampled average is given by the random variable $$(X_1 + X_2 + \cdots + X_n)/n$$. The weak law of large numbers says that this variable is likely to be close to the real expected value:

Claim (weak law of large numbers): If $$X_1, X_2, \dots, X_n$$ are independent random variables with the same expected value $$\mu$$ and the same variance $$σ^2$$, then $Pr\left(\left|\frac{X_1 + X_2 + \cdots + X_n}{n} - μ\right| \geq a\right) \leq \frac{σ^2}{na^2}$

Proof: By Chebychev's inequality, we have $Pr\left(\left|\sum X_i/n - E(\sum X_i/n)\right| \geq a\right) \leq \frac{Var(\sum X_i/n)}{a^2}$

Now, by linearity of the expectation, we have $E(\sum X_i/n) = \sum E(X_i)/n = nμ/n = μ$

As was shown in homework 5, $$Var(cX) = c^2Var(X)$$, and we also know that if $$X$$ and $$Y$$ are independent, that $$Var(X + Y) = Var(X) + Var(Y)$$. Therefore, we have $Var(\sum X_i/n) = \sum Var(X_i)/n^2 = nσ^2/n^2 = σ^2/n$

Plugging these into the result from Chebychev's, we have $Pr\left(\left|\sum X_i/n - μ\right| \geq a\right) \leq \frac{σ}{na^2}$

which is what we were trying to show.