Answer:
To prove the inequality P(X ≥ a) ≤ E[X] / a for a > 0, where X is a non-negative random variable, we can use Markov's inequality.
Markov's inequality states that for any non-negative random variable Y and any constant c > 0, we have P(Y ≥ c) ≤ E[Y] / c.
Let's apply Markov's inequality to the random variable X - a, where a > 0:
P(X - a ≥ 0) ≤ E[X - a] / 0
Simplifying the expression:
P(X ≥ a) ≤ E[X - a] / a
Since X is a non-negative random variable, E[X - a] = E[X] - a (the expectation of a constant is equal to the constant itself).
Substituting this into the inequality:
P(X ≥ a) ≤ (E[X] - a) / a
Rearranging the terms:
P(X ≥ a) ≤ E[X] / a - 1
Adding 1 to both sides of the inequality:
P(X ≥ a) + 1 ≤ E[X] / a
Since the probability cannot exceed 1:
P(X ≥ a) ≤ E[X] / a
Therefore, we have proved that P(X ≥ a) ≤ E[X] / a for a > 0, based on Markov's inequality.