Question
For the distribution of Exercise 2, calculate $P(|X-\mu| \geq 1.5)$ and compare this with a) the upper bound obtained from Tchebycheff's inequality and with b) the upper bound obtained from the stronger form of Tchebycheff's inequality (3.27.3).
Step 1
We need to know the mean \(\mu\) and the variance \(\sigma^2\) of the random variable \(X\) in order to proceed with the calculations. Show more…
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This exercise demonstrates that, in general, the results provided by Tchebysheff's theorem cannot be improved upon. Let $Y$ be a random variable such that $$p(-1)=\frac{1}{18}, \quad p(0)=\frac{16}{18}, \quad p(1)=\frac{1}{18}$$ a. Show that $E(Y)=0$ and $V(Y)=1 / 9$ b. Use the probability distribution of $Y$ to calculate $P(|Y-\mu| \geq 3 \sigma) .$ Compare this exact probability with the upper bound provided by Tchebysheff's theorem to see that the bound provided by Tchebysheff's theorem is actually attained when $k=3$ *.c. In part.(b) we guaranteed $E(Y)=0$ by placing all probability mass on the values $-1,0,$ and1, with $p(-1)=p(1) .$ The variance was controlled by the probabilities assigned to $p(-1)$ and $p(1) .$ Using this same basic idea, construct a probability distribution for a random variable $X$ that will yield $P\left(\left|X-\mu_{X}\right| \geq 2 \sigma_{X}\right)=1 / 4$ * d. If any $k>1$ is specified, how can a random variable $W$ be constructed so that $P\left(\left|W-\mu_{W}\right| \geq k \sigma_{W}\right)=1 / k^{2} ?$
Discrete Random Variables and Their Probability Distributions
Tehebysheff’s Theorem
A random variable X has a mean of 3 and a variance of 2. Use the Chebyshev inequality to obtain an upper bound for (a) P(|X-3| ≥ 2), (b) P(|X - 3| ≥ 1).
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