Question
Can you give an example (similar to that for Markov's inequality in Exercise 3.16) that shows that Chebyshev's inequality is tight? If not, explain why not.
Step 1
For any random variable X with mean μ and variance σ^2, and any k > 0, we have: P(|X - μ| ≥ kσ) ≤ 1/k^2 Now, let's consider a discrete random variable X with the following probability distribution: - X = μ - kσ with probability 1/2 - X = μ + kσ with probability Show more…
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One-Sided Chebyshev : Using the Markov Inequality, one can also show that for any random variable with mean μ and variance σ^2, and any positive number a > 0, the following one-sided Chebyshev inequalities hold: P(X ≤ μ - a) ≤ σ^2 / (σ^2 + a^2)
This problem shows that Markov's inequality is as tight as it could possibly be. Given a positive integer $k$, describe a random variable $X$ that assumes only nonnegative values such that $$ \operatorname{Pr}(\boldsymbol{X} \geq k \mathbf{E}[X])=\frac{1}{k} $$
Let $X$ be a random variable on a sample space $S$ such that $X(s) \geq 0$ for all $s \in S .$ Show that $p(X(s) \geq a) \leq E(X) / a$ for every positive real number $a$. This inequality is called Markov's inequality.
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Expected Value and Variance
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