This problem asks us to show that if \(X\) is a non-negative random variable and \(a > 0\), then the probability that \(X\) is at least \(a\) is less than or equal to the expected value of \(X\) divided by \(a\). We can proof this statement using Markov’s inequality.
Markov’s inequality states that for any non-negative random variable \(X\) and any \(a > 0\),
\[ P(X \geq a) \leq \frac{E(X)}{a} \]
where \(E(X)\) is the expected value of \(X\).
The expected value \(E(X)\) is:
For discrete random variables, \[ E(X) = \sum_{\text{all } x} x \cdot P(X = x) \]
For continuous random variables, \[ E(X) = \int_{-\infty}^{\infty} x \cdot f(x) \, dx \]
where \(f(x)\) is the probability density function of \(X\).
When considering only the values of \(X\) that are at least \(a\), any such value contributes at least \(a \cdot P(X = x)\) (in the discrete case) or \(a \cdot f(x) \, dx\) (in the continuous case) to the expected value. Therefore, if we consider the part of \(E(X)\) coming from \(X \geq a\), it must be that:
\[ E(X) \geq a \cdot P(X \geq a) \]
Rearranging this inequality gives:
\[ P(X \geq a) \leq \frac{E(X)}{a} \]
This is the statement of Markov’s inequality, which we wanted to show.