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MTH/STA 561 UNIFORM PROBABILITY DISTRIBUTION Perhaps the simplest possible continuous probability distribution is the uniform distribution. We say that a continuous random variable Y is uniformly distributed over the interval ( ; ) if the probability that the observed value for Y falls in any subinterval of length y, in ( ; ), is proportional to y. The probability that the observed value for Y is smaller than or greater than is zero. Thus, the density for Y is positive only for values between and , and if the probability that the observed value for Y lies in an interval is proportional to the length of that interval, the density function f (y) must be constant for < y < . That is f (y) = c for < y < . Since 1= it must be true that c = 1= ( Z f (y) dy = Z cdy = c ( ); ). Thus, we have established the following distribution. De…nition 1. A random variable Y is said to have a continuous uniform probability distribution on the interval ( ; ) if the probability density function of Y is given by f (y; ; ) = 1= ( 0 ) for < y < elsewhere. Any continuous random variable that follows a uniform distribution is referred to as the uniform random variable. The constants, such as and , that determine the speci…c form of a probability density function are called parameters of the probability density function. Here we employ the symbol f (y; ; ) to emphasize the fact that the uniform distribution depends upon parameters and . Some continuous random variables in the physical, management, and biological sciences have approximately uniform probability distributions. For example, suppose that we are counting events that have a Poisson distribution, such as telephone calls coming into a switchboard. If it is known that exactly one such event has occurred in a given interval, 1 say (0; t), then the actual time of occurrence is distributed uniformly over this interval. Example 1. Suppose that arrivals of customers at a certain checkout counter follow a Poisson distribution. It is known that, during a given 30-minute period, one customer arrived at the counter. What is the probability that the customer arrived during the last 5 minutes of the 30-minute period? Solution. As mentioned above, the actual time T of arrival will follow a uniform distribution over the interval (0; 30); that is, f (t; 0; 30) = 1 30 for 0 t 30 elsewhere. 0 Then the desired probability is P f25 T 30g = Z30 1 30 25 5 1 dt = = = : 30 30 30 6 25 Theorem 1. with parameters Proof. For The cumulative distribution function for the uniform random variable Y and is given by 8 for y < 0 y for < y < F (y; ; ) = : 1 for y <y< , F (y; ; ) = Zy 1 dt = 0 + t t=y = y : t= The graph of the distribution function for uniform random variable look like 2 Example 2. The distribution function of the uniform random variable Y as de…ned in Example 1 is given by 8 for y 0 < 0 y for 0 < y < 30 F (y; 0; 30) = : 30 1 for y 30 Theorem 2. and are The mean and variance of the uniform random variable Y with parameters + ( )2 2 and = V ar (Y ) = ; 2 12 respectively. The moment-generating function is = E (Y ) = et t( mY (t) = et for t 6= 0 ) Proof. By de…nition, E (Y ) = Z ( = 1 y dy = )( + ) = 2( ) y= y2 2 1 2 = y= 2 2( ) + 2 and E Y 2 Z = y 1 2 ) 2+ + 3( ) ( = dy = y= y3 3 1 = y= 2 2 3 = + 3 3( + ) 2 : 3 Thus, 2 V ar (Y ) = E Y = = 2 4 2 2 [E (Y )] = + + + 12 )2 ( 12 2 + 2 3 3 ( + )2 2 2 + 2 2 + 12 = 2 : It is also easy to see that tY mY (t) = E e = Z ty e 1 dy = 3 1 ety t y= = y= et t( et ) : Example 3. Example 1 are The mean and variance of the uniform random variable Y as de…ned in (30 0)2 0 + 30 2 = = 15 and = = 75; 2 12 respectively, whereas the moment-generating function is mY (t) = e30t 1 30t for t 6= 0 Di¤erentiating mY (t) with respect to t, we get d 30te30t e30t + 1 mY (t) = dt 30t2 and d2 t2 [30 (30te30t + e30t ) 30e30t ] 2t (30te30t m (t) = Y dt2 30t4 3 30t 2 30t 900t e 60t e + 2te30t 2t = 30t4 2 30t 900t e 60te30t + 2e30t 2 = 30t3 e30t + 1) By L’Hôpital’s rule, we obtain d 30te30t e30t + 1 mY (t) = lim t!0 dt 30t2 t=0 30 (30te30t + e30t ) 30e30t = lim t!0 60t 900te30t 30e30t 30 = lim = lim = = 15 t!0 t!0 60t 2 2 E (Y ) = and E Y2 d2 900t2 e30t 60te30t + 2e30t 2 m (t) = lim Y t!0 dt2 30t3 t=0 900 (30t2 e30t + 2te30t ) 60 (30te30t + e30t ) + 60e30t = lim t!0 90t2 2 30t 27000t e = lim = lim 300te30t = 300 t!0 t!0 90t2 = Hence, V ar (Y ) = E Y 2 [E (Y )]2 = 300 4 152 = 75: