# Why is $P(X \geq a)$ the same as $E[\mathbf{1}_{X \geq a}]$

in a proof of the markov's inequality I didn't understand the last step

the markov's inequality : $$P(X \geq a) \leq \frac{E[X]}{a}$$

let $X : \Omega \to \Bbb{R}$ be a random variable

let $$\mathbf{1}_{X \geq a} :\mathbb{R} \to \{0,1\}$$

$$\mathbf{1}_{X \geq a}(X) = \begin{cases} 0, & \text{if X<a } \\ 1, & \text{if X\geq a } \end{cases}$$

and let $$g = a \cdot \mathbf{1}_{X \geq a} :\mathbb{R} \to \{0,a\}$$

$$g(X) = \begin{cases} 0, & \text{if X<a } \\ a, & \text{if X\geq a } \end{cases}$$

then $$g(X) \leq X$$ $$E[g(X)] \leq E[X]$$ $$a\cdot E[\mathbf{1}_{X \geq a}(X)] \leq E[X]$$ $$E[\mathbf{1}_{X \geq a}(X)] \leq \frac{E[X]}{a} \implies P(X \geq a) \leq \frac{E[X]}{a}$$

I didn't understand this last step can someone please elucidate.

• Also considering the expectation of the Bernoulli random variable may help you understand a bit I feel, e.g. $\bf{1}_{X\geq a}(X)$ is a Bernoulli random variable, its expectation is the probability that X is greater and equal to a in this sense. – lzstat Mar 26 '17 at 17:01

The expectation of an indicator variable related to an event $E$ is always the probability that $E$ happens :

Let

$$I_E = \begin{cases} 1 & \text{if } E \text{ happens} \\ 0 & \text{if } E \text{ doesn't happen} \end{cases}$$ Then $$E[I_E] \overset{(*)}{=} 1\cdot Pr[E \text{ happens}] + 0 \cdot Pr[E \text{ doesn't happens}] = Pr[E \text{ happens}]$$ $(*)$ By the definition of expecation.

In your case the event is "$X \geq a$". Therefore $$E[I_{X \geq a}] = 1\cdot Pr[X \geq a \text{ happens}] + 0 \cdot Pr[X \geq a\text{ doesn't happen}] = Pr[X \geq a]$$

Taking expectation is to integrate a random variable over the outcome space w.r.t the probability measure. Recall that $\int_{\Omega} 1_{A} dP = \int_{A} 1 dP = P(A)$ by basic properties of Lebesgue integration.

Also note that the set $\{ X \geq a \}$ is the shorthand for $\{ \omega \in \Omega \mid X(\omega) \geq a \}$ which is always a subset of $\Omega$ and which is actually simply the preimage $X^{-1}([a, +\infty])$.

• Don't want to offense you but I don't think we should go into "deep" measure theory for a question related to Markov inequality in a "simple probabilistic framework". I might be wrong though :) – Zubzub Mar 24 '17 at 13:50