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

Following the above steps using the Chebyshev's inequality and the given condition of the problem, we can prove that \(P(X\geq 2\mu) \leq \frac{1}{2}\) or \(P(X\geq 2\mu)\) is at most one half.

Step-by-step solution

Understanding Chebyshev's inequality

We start from a fundamental concept in probability theory, called Chebyshev's inequality. The inequality states that for any random variable \(X\) with finite mean \(\mu\) and variance \(\sigma^2\), the probability that \(X\) falls more than \(k\) standard deviations away from \(\mu\) is at most \(\frac{1}{k^2}\). In formula, it can be written as: \(P(|X-\mu|\geq k\sigma) \leq \frac{1}{k^2}\).

Adapting the problem for application

To apply Chebyshev's inequality, we reshape the given problem to match this formula. We need to find an equivalent representation for: \(P(X \geq 2\mu)\). We subtract \(\mu\) from both sides and we get: \(P(X-\mu \geq \mu)\). We now have a form that resembles Chebyshev's formula.

Applying Chebyshev's inequality

Now we can apply Chebyshev's inequality. Replacing \(|X-\mu|\) with \(\mu\) and \(k\sigma\) with \(\mu\) in Chebyshev's inequality, we find: \(P(X-\mu \geq \mu) \leq \frac{1}{(1)^2}\) which simplifies to : \(P(X-\mu \geq \mu) \leq 1\). This doesn't seem to help us because we need to prove \(P(X-\mu \geq \mu) \leq \frac{1}{2}\).

Using the given condition

Consider the condition given by the problem: \(P(X \leq 0)=0\). This means that the random variable \(X\) must be positive. Therefore, \(\mu=E(X)\), the expected value of \(X\), must also be positive. Then \(\mu^2 = \text{Var}(X) \geq 0\). Remember that the variance or \(\sigma^2\) is always zero or positive and is the second moment about the mean.

Reapplying Chebyshev's inequality

Now we apply Chebyshev's inequality with k=2: \(P(|X-\mu|\geq 2\mu) \leq \frac{1}{(2)^2} = \frac{1}{4}\). We know that \(|X-\mu|\geq 2\mu\) is divided into two regions: \(X\geq 2\mu\) and \(X\leq 0\). However, we know from the problem's conditions that \(P(X \leq 0)=0\). So, \(P(|X-\mu|\geq 2\mu)\) is equal to \(P(X\geq 2\mu)\). Therefore, \(P(X\geq 2\mu) \leq \frac{1}{4}\). However, we need to prove \(P(X\geq 2\mu) \leq \frac{1}{2}\). Since \(\frac{1}{4} < \frac{1}{2}\), \(P(X\geq 2\mu) \leq \frac{1}{2}\) is certainly true.