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Discrete.Math.in.Computer.Science,2002

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5.7. PROBABILITY DISTRIBUTIONS AND VARIANCE

213

15.This problem develops an important law of probability known as Chebyshev’s law. Suppose we are given a real number r > 0 and we want to estimate the probability that the di erence |X(x) − E(X)| of a random variable from its expected value is more than r.

(a)Let S = {x1, x2, . . . , xn} be the sample space, and let E = {x1, x2, . . . , xk } be the set of all x such that |X(x) − E(X)| > r. By using the formula that defines V (X), show

that

k

V (X) > P (xi)r2 = P (E)r2

i=1

(b)Show that the probability that |X(x) − E(X)| ≥ r is no more than V (X)/r2. This is called Chebyshev’s law.

16.Use Exercise 5.7-14 show that in n independent trials with probability p of success,

P

# of successes − np

r

1

 

n

 

 

4nr2

17.This problem derives an intuitive law of probability known as the law of large numbers from Chebyshev’s law. Informally, the law of large numbers says if you repeat an experiment many times, the fraction of the time that an event occurs is very likely to be close to the probability of the event. In particular, we shall prove that for any positive number s, no matter how small, by making the number n independent trials in a sequence of independent trials large enough, we can make the probability that the number X of successes is between np − ns and np + ns as close to 1 as we choose. For example, we can make the probability that the number of successes is within 1% (or 0.1 per cent) of the expected number as close to 1 as we wish.

(a)Show that the probability that |X(x) − np| ≥ sn is no more than p(1 − p)/s2n.

(b)Explain why this means that we can make the probability that X(x) is between np−sn and np + sn as close to 1 as we want by making n large.

18.On a true-false test, the score is often computed by subtracting the number of wrong answers from the number of right ones and converting that number to a percentage of the number of questions. What is the expected score on a true-false test graded this way of someone who knows 80% of the material in a course? How does this scheme change the standard deviation in comparison with an objective test? What must you do to the number of questions to be able to be a certain percent sure that someone who knows 80% gets a grade within 5 points of the expected percentage score?

19.Another way to bound the deviance from the expectation is known as Markov’s inequality. This inequality says that is X is a random variable taking only non-negative values, then, for any k ≥ 1,

P (X > kE(X)) k1 .

Prove this inequality.