Question

Let \(X\) be a random variable such that \(P(X \leq 0)=0\) and let \(\mu=E(X)\) exist. Show that \(P(X \geq 2 \mu) \leq \frac{1}{2}\).

Short answer

By applying Markov's Inequality on \(X\) and doing some simple algebraic manipulations, we find that indeed, the probability that \(X\) is at least twice its expected value is at most \(\frac{1}{2}\).

Step-by-step solution

Apply Markov's Inequality

Let's use Markov's Inequality on \(X\) itself, setting \(a = 2\mu\). This gives us \(P(X \geq 2\mu) \leq \frac{E(X)}{2\mu}\).

Simplify the inequality

The expected value of \(X\), \(E(X)\), is \(\mu\), so we substitute \(E(X)\) with \(\mu\), for simplification. We get \(P(X \geq 2\mu) \leq \frac{\mu}{2\mu}\).

Final Steps

By simplifying the fraction \(\frac{\mu}{2\mu}\), we see that it is equal to \(\frac{1}{2}\). So, we have \(P(X \geq 2\mu) \leq \frac{1}{2}\), which is exactly what we were asked to show.

Key concepts

Probability Theory

Probability theory is a branch of mathematics that deals with the analysis of random phenomena. It provides the foundation for quantifying the likelihood of various outcomes. In the exercise, the random variable, which is a fundamental concept in probability theory, is called upon to express the uncertainty surrounding a particular event, in this case, the event where the value of the random variable is at least twice its expected value.

The problem at hand involves the understanding of how likely a random variable is to assume values that deviate significantly from its average or expected value. This aspect of probability theory plays a vital role in fields ranging from insurance to finance, where assessing risks is crucial.

Expected Value

The expected value, often denoted as E(X), can be thought of as the long-run average value of repetitions of the experiment it represents. It is a weighted average of all possible values that a random variable can take on, each value weighted by its probability of occurring. In the context of our exercise, the expected value denotes the central or 'average' value that the random variable X is most likely to be around.

Calculating the expected value is essential in many areas of statistics and probability, as it serves as a measure of the center of the distribution of the variable. Understanding the concept of the expected value enables us to contextualize how improbable it is for the value of X to be at least double that average, providing a pathway to grasp what Markov's inequality is quantifying.

Inequalities in Statistics

Inequalities in statistics are tools that provide bounds and limitations to probabilities and expectations. They are crucial for understanding how far variables can deviate from some norm, like their mean, and are thus a way to gauge 'rare' events in a probabilistic sense.

Markov's inequality, which is applied in the problem, is one such inequality. It tells us that for a non-negative random variable, the probability that the variable exceeds a positive value is at most the expected value of the variable divided by that positive value. This inequality provides an upper bound and is particularly useful when we have limited information about the distribution of a random variable but still wish to make probabilistic statements.

Probability Inequalities

Probability inequalities like Markov's and Chebyshev's give us ways to bound probabilities without exact distributions. Markov's inequality, utilized in the problem, ensures that for any non-negative random variable, the probability that it will be greater or equal to a certain multiple of its mean is bounded by a reciprocal of that multiple.

This theoretical insight is powerful in probabilistic analysis because it allows us to make assertions about the tail behavior of the distribution — how it behaves far away from the mean — which is essential for risk assessment and management. Applying Markov's Inequality as demonstrated in the problem exemplifies how we can use these inequalities to estimate probabilities in a broad range of applications, despite having incomplete knowledge of the underlying distribution.