1.5 Markov’s Inequality
If \(X\) is a random variable that takes only non-negative values, then for any \(a>0\), \[P\left (X \geq a\right ) \leq \frac {E(X)}{a}.\]
Proof. \(\begin {aligned}[t] E(X) = \int _0^{\infty } x\, f(x)\, dx & = \int _0^ax\, f(x)\, dx \, + \, \int ^{\infty }_ax\, f(x)\, dx\\ & \geq \int ^{\infty }_a x\, f(x)\, dx\\ & \geq \int _a^{\infty }a\, f(x)\, dx\\ & = a\int ^{\infty }_a f(x)\, dx\\ & = a\, P(X\geq a). \end {aligned}\)
Therefore \(\, E(X) \geq a\, P(X\geq a)\hspace {0.3cm}\implies \, \dfrac {E(X)}{a}\geq P(X\geq a)\). □
Example 1.5.1. Suppose we know that the number of items produced in a factory during
a week is a random variable with mean 500.
What can be said about the probability that this week’s production will be atleast 1000?
Solution. Let \(X\) denote the weekly production. \[P(X\geq 1000) \leq \dfrac {E(X)}{1000}\]
\[\text {i.e}\hspace {0.3cm} P(X\geq 1000) \leq \frac {500}{1000} = \frac {1}{2}.\] □
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