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Chapter 1: Problem 2

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

Expert verified
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

01

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: \(P(|X-\mu|\geq k\sigma) \leq \frac{1}{k^2}\).
02

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.
03

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 \(P(X-\mu \geq \mu) \leq \frac{1}{2}\).
04

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.
05

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.

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Most popular questions from this chapter

Players \(A\) and \(B\) play a sequence of independent games. Player \(A\) throws a die first and wins on a "six." If he fails, \(B\) throws and wins on a "five" or "six." If he fails, \(A\) throws and wins on a "four," "five," or "six." And so on. Find the probability of each player winning the sequence.

(Monte Hall Problem). Suppose there are three curtains. Behind one curtain there is a nice prize while behind the other two there are worthless prizes. A contestant selects one curtain at random, and then Monte Hall opens one of the other two curtains to reveal a worthless prize. Hall then expresses the willingness to trade the curtain that the contestant has chosen for the other curtain that has not been opened. Should the contestant switch curtains or stick with the one that she has? If she sticks with the curtain she has then the probability of winning the prize is \(1 / 3 .\) Hence, to answer the question determine the probability that she wins the prize if she switches.

Let \(X\) be a random variable such that \(E\left[(X-b)^{2}\right]\) exists for all real \(b\). Show that \(E\left[(X-b)^{2}\right]\) is a minimum when \(b=E(X)\).

Two distinct integers are chosen at random and without replacement from the first six positive integers. Compute the expected value of the absolute value of the difference of these two numbers.

A bowl contains ten chips numbered \(1,2, \ldots, 10\), respectively. Five chips are drawn at random, one at a time, and without replacement. What is the probability that two even-numbered chips are drawn and they occur on even- numbered draws?
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