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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

A French nobleman, Chevalier de M茅r茅, had asked a famous mathematician, Pascal, to explain why the following two probabilities were different (the difference had been noted from playing the game many times): (1) at least one six in 4 independent casts of a six-sided die; (2) at least a pair of sixes in 24 independent casts of a pair of dice. From proportions it seemed to de M茅r茅 that the probabilities should be the same. Compute the probabilities of \((1)\) and \((2)\).

A die is cast independently until the first 6 appears. If the casting stops on an odd number of times, Bob wins; otherwise, Joe wins. (a) Assuming the die is fair, what is the probability that Bob wins? (b) Let \(p\) denote the probability of a \(6 .\) Show that the game favors Bob, for all \(p\), \(0

Consider the events \(C_{1}, C_{2}, C_{3}\). (a) Suppose \(C_{1}, C_{2}, C_{3}\) are mutually exclusive events. If \(P\left(C_{i}\right)=p_{i}, i=1,2,3\), what is the restriction on the sum \(p_{1}+p_{2}+p_{3} ?\) (b) In the notation of Part (a), if \(p_{1}=4 / 10, p_{2}=3 / 10\), and \(p_{3}=5 / 10\) are \(C_{1}, C_{2}, C_{3}\) mutually exclusive?

For each of the following, find the constant \(c\) so that \(p(x)\) satisfies the condition of being a pmf of one random variable \(X\). (a) \(p(x)=c\left(\frac{2}{3}\right)^{x}, x=1,2,3, \ldots\), zero elsewhere. (b) \(p(x)=c x, x=1,2,3,4,5,6\), zero elsewhere.

Hunters \(\mathrm{A}\) and \(\mathrm{B}\) shoot at a target; the probabilities of hitting the target are \(p_{1}\) and \(p_{2}\), respectively. Assuming independence, can \(p_{1}\) and \(p_{2}\) be selected so that \(P(\) zero hits \()=P(\) one hit \()=P(\) two hits \() ?\)
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