This exercise demonstrates that, in general, the results provided by Tchebysheffs

Chapter 3, Problem 3.169

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This exercise demonstrates that, in general, the results provided by Tchebysheffs theorem cannot be improved upon. Let Y be a random variable such that p(1) = 1 18 , p(0) = 16 18 , p(1) = 1 18 .a Show that E(Y ) = 0 and V(Y ) = 1/9. b Use the probability distribution of Y to calculate P(|Y | 3). Compare this exact probability with the upper bound provided by Tchebysheffs theorem to see that the bound provided by Tchebysheffs 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, and 1, 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(|X X | 2X ) = 1/4. *d If any k > 1 is specified, how can a random variable W be constructed so that P(|W W | kW ) = 1/k2 ?