4. Continuous Random Variable 连续型随机变量 Continuous random variables appear when we deal will quantities that are measured on a continuous scale. For instance, when we measure the speed of a car, the amount of alcohol in a person's blood, the tensile strength of new alloy. We shall learn how to determine and work with probabilities relating to continuous random variables in this chapter. We shall introduce to the concept of the probability density function. 4.1 Continuous Random Variable 1. Definition Definition 4.1.1 A function f(x) defined on (??, ?) is called a probability density function （概率密度函数）if: (i) f(x)?0 for any x?R; ?(ii) f(x) is intergrable (可积的) on (??, ?)and ???f(x)dx?1. Definition 4.1.2 Let f(x) be a probability density function. If X is a random variable having distribution function xF(x)?P(X?x)????f(t)dt, (4.1.1) then X is called a continuous random variable having density function f(x). In this case, x2P(x1?X?x2)? 2. 几何意义 x1?f(t)dt. (4.1.2) xF(x)?P(X?x)?P((X,Y)|X?x, 0?Y?f(X))? ???f(t)dt 46 x2 P(x1?X?x2)?x1?f(t)dt 3. Note In most applications, f(x) is either continuous or piecewise continuous having at most finitely many discontinuities. Note 1 For a random variable X, we have a distribution function. If X is discrete, it has a probability distribution. If X is continuous, it has a probability density function. Note 2 Let X be a continuous random variable, then for any real number x, P(X?x)?0. 0?P(X?x)??0?P(X?x)?lim??0x??xx??f(x)dx f(x)dx?0 ?xP(a?X?b)?P(a?X?b)?P(a?X?b)?P(a?X?b) 4. Example Example 4.1.2 Find k so that the following can serve as the probability density of a continuous random variable: f(x)?k (???x??) 21?x Solution To satisfy the conditions (4.1.1), k must be nonnegative, and to satisfy the condition (4.1.2) we must have ????f(x)dx?kdx?k??1 2?1+x???so that k?1?. (Cauchy distribution 柯西分布) □ Example 4.1.3 Calculating probabilities from the probability density function If a random variable has the probability density 47 ?3e?3x for x?0f(x)?? ?0 for x?0Find the probability from that it will take on value (a) between 0 and 2; (b) greater than 1. Solution Evaluating the necessary integrals, we get 2(a) P(0?x?2)?3e?3xdx?1?e?6?0.9975 ?0?(b) P(x?1)?3e?3xdx?e?3?0.0498 ?1Example 4.1.4 Determining the distribution function of X, it is known ?3e?3x for x?0 f(x)?? ?0 for x?0Solution Performing the necessary integrations, we get ?0 for x?0?F(x)??x?3t ?3x3edt?1?e for x?0???0 P(x?1)?F(1)?1?e?3?0.9502 □ 5. mean If the integral (4.1.3) does not converges absolutely（绝对收敛）, we say the mean of X does not exist. Definition 4.1.2 Let X be a continuous random variable having probability density function f(x). Then the mean (or expectation) of X is defined by ???E(X)????xf(x)dx, (4.1.3) The mean of continuous random variable has the similar properties as discrete random variable. If g(X) is an integrable function of a continuous random variable X, having density function f(x), mean of g(X) is ?E(g(X))?provided the integral converges absolutely. Example 4.6.4 ???g(x)f(x)dx 48 Let X be a random variable having Cauchy distribution, the probability density function is given by f(x)?(a) Find E(X); (b) Let 1, ???x?? ?(1?x2)?X, 0?X?1 g(X)???0, elsewhereFind E(g(X)). Solution (a) Since the integral ?|x|dx diverges（发散）, E(X) does not exist. 2??(1?x)??1?(b) E(g(X))?6. variance ???g(x)f(x)dx?? xln2dx?. 2?(1?x)2?0Similarly, the variance and standard deviation of a continuous random variable X is defined by ?2?D(X)?E((X??)2), (4.1.4) Where ??E(X)is the mean of X, We easily get ?? is referred to as the standard deviation. ??D(X)?Example 4.1.5 2???x2f(x)dx??2. (4.1.5) Determining the mean and variance using the probability density function ?3e?3x for x?0With reference to the example 4.1.3: f(x)?? ?0 for x?0find the mean and variance of the given probability density. Solution Performing the necessary integrations, we get ????and ????xf(x)dx??x?3e?3xdx?01 3??2??(x??)2f(x)dx??(x?)3e?3xdx?. ??013194.2 Uniform Distribution 均匀分布 49 The uniform distribution, with the parameters a and b, has probability density function ?1 for a?x?b,? f(x)??b?a??0 elsewhere, whose graph is shown in Figure 4.2.1. f(x) To proof 1 b?a0 a b x Figure 4.2.1 The uniform probability density in the interval (a, b) ?????f(x)dx?1. To find the distribution function. The distribution function of the uniform distribution is ?0 for x?a,?x?a?F(x)?? for a?x?b, ?b?a??1 for x?b. Note that all values of x from a and b are “equally likely”, in the sense that the probability that x lies in an interval of width ?xentirely contained in the interval from a to b is equal to ?x/(b?a), regardless of the exact location of the interval. To find the mean and variance of the uniform distribution b???x?a1a?b dx?b?a2And 1a2?ab?b2 EX= ?x? dx?b?a3a2 b2 Thus, a2?ab?b2?a?b?(b?a)2?? ?? ??312?2?22 50