Lectures on Elementary Probability

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32 CHAPTER 5. CONTINUOUS RANDOM VARIABLES The mean or expectation of X is µ X = E[X] = ∫ ∞ −∞ xf(x) dx. (5.4) Sometimes for technical purposes the following theorem is useful. It expresses the expectation in terms of an integral of probabilities. It is most often used for random variables Y that have only positive values; in that case the second term is of course zero. Theorem 5.1 Let Y be a continuous random variable with probability density function f. Then ∫ ∞ ∫ 0 E[Y ] = P [Y > y] dy − P [Y < y] dy. (5.5) 0 −∞ Proof: We have E[Y ] = ∫ ∞ 0 xf(x) dx+ ∫ 0 −∞ xf(x) dx = ∫ ∞ ∫ x Interchange the order of integration. This gives E[Y ] = ∫ ∞ ∫ ∞ 0 y 0 0 ∫ 0 ∫ 0 dyf(x) dx− dyf(x) dx. −∞ x (5.6) ∫ 0 ∫ y f(x) dx dy − f(x) dx dy. (5.7) −∞ −∞ This immediately gives the result of the theorem. Note that if X is a continuous random variable and g is a function defined for the values of X and with real values, then Y = g(X) is also a random variable. The following theorem is particularly important and convenient. If a random variable Y = g(X) is expressed in terms of a continuous random variable, then this theorem gives the expectation of Y in terms of probabilities associated to X. Theorem 5.2 The expectation of a function Y = g(X) of a continuous variable X may be expressed in terms of the probability density function f associated with X by Proof: E[g(X)] = E[g(X)] = ∫ ∞ 0 ∫ ∞ −∞ g(x)f X (x) dx. (5.8) ∫ 0 P [g(X) > y] dy − P [g(X) < y] dy. (5.9) −∞ Write this in terms of the density. This gives ∫ ∞ ∫ ∫ 0 ∫ E[g(X)] = f X (x) dx dy − f X (x) dx dy. (5.10) 0 g(x)>y −∞ g(x)
5.1. MEAN 33 Now we can interchange the order of integration. This gives ∫ E[g(X)] = This simplifies to ∫ E[g(X)] = g(x)>0 g(x)>0 ∫ g(x) 0 ∫ ∫ 0 dyf X (x) dx − dyf X (x) dx. (5.11) g(x)
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