Probability



7. Continuous probability.

So far we have been considering examples where the outcomes form a finite or countably infinite set. In many situations it is more natural to model a situation where the outcomes could be any real number or vectors of real numbers. In these situations we usually model probabilities by means of integrals.

Example 1. An bank is doing a study of the amount of time, T, between arrivals of successive customers. Assume we are measuring time in minutes. They are interested in the probability that T will have various values. If we imagine that T can be any non-negative real number, then the sample space is the set of all non-negative real numbers, i.e. S = {t: t ( 0}. This is an uncountable set. In situations such as this, the probability that T assumes any particular value may be 0, so we are interested in the probability that the outcome lies in various intervals. For example, what is the probability that T will be less than 2 and 3 minutes.

A common way to describe probabilities in situations such as this is by means of an integral. We try to find a function f(t) such that the probability that T lies in any interval a ( t ( b is equal to , i.e.

(1) Pr{ a ( T ( b } = .

A function f(t) with this property is called a probability density function for the outcomes of experiment. We can regard T as a random variable and f(t) is also called the probability density function for the random variable T.

Since Pr{ T = a } = 0 and Pr{ T = b } = 0, one has Pr{ a  b. Divide the interval a ( t ( b into m equal sized subintervals by points t0, t1, …, tm where tk = a + k((t) with (t = (b – a)/m. Suppose we have a sequence T1, T1, …, Tn, … of repeated independent trials where each Tn has density function f(t). Suppose s1, s1, …, sn, … are the values we actually observe for the random variables T1, T1, …, Tn, … In our computation of = let's approximate all the values of sj that are between tk-1 and tk by tk and then group all the approximate values that equal tk together. Then we have

= (

= =

where gj is the number of times that sj is between tk-1 and tk. As n ( ( one has So as n gets large it is not unreasonable to suppose that . If we let m ( ( then all the above approximations get better and ( which leads to (7).

For an exponential random variable we have

(8) μ = = - te--(t | + = - 0 + 0 - =

We used integration by parts with u = t and dv = (e--(t dt so that du = dt and v = -e--(t.

In the example above where the time between arrivals of bank customers was an exponential random variable with ( = ½, the average time between arrivals is 1/( = 2 min.

Problem 1. The lifetime of a certain type of light bulb has density function f(t) given by f(t) = 0 for t < 100 days and f(t) = 20000/t3 for t > 100 days.

a. Find the probability that the lifetime is more than 110 days.

b. Find the average lifetime.

Problem 2. Taxicabs pass by at an average rate of 20 per hour. Assume the time between taxi cabs is an exponential random variable. What is the probability that a taxicab will pass by in the next minute?

In section 5 we considered machine replacement problems using discrete probability. One can also use continuous probability for these problems.

Example 3 (Machine replacement). Consider the hard drive on my office computer. It costs c1 = $300 to replace if it is replaced before it fails. If it fails before it is replaced, it costs an additional c2 = $1000 in terms of down time for my computer. This is in addition to the $300 replacement cost. Suppose that after it has been installed it is equally likely to fail anytime in the next five years. Suppose every hard drive fails by the end of the 5th year. Let

T = the time the hard drive fails

q = time at which you replace it if it hasn't already failed

C = Cq = the cost of a replacement

Tq = replacement time if it is replaced at time q if it has not already failed.

a. Find the probability density function f(t) and cumulative distribution function F(t) for T.

b. Find the expected cost E(C) of a replacement.

c. Find the expected time E(Tq) of a replacement.

d. Find the long run average cost z(q) = E(C)/E(Tq) of a replacement.

e. When should the hard drive be replaced so as to minimize the long run replacement cost.

Since it is equally likely to be replaced at any time in the next five years, f(t) should be constant for t between 0 and 5 and f(t) should be 0 for t less than 0 and greater than 5. Since the integral of f(t) over all t should be 1, we must have f(t) = 1/5 for 0 ( t ( 5 and f(t) = 0 for t < 0 and t > 5. This is another example of a uniform probability distribution which was discussed earlier. It is constant in an interval and zero elsewhere.

The cumulative distribution function is just the integral of the density function from - ( to t. So F(t) = 0 for t < 0 and F(t) = t/5 for 0 ( t ( 5 and F(t) = 1 for t > 5.

E(C) = c1 + c2F(q) = 300 + 1000q/5 = 300 + 200q.

Tq is an example of a random variable is a mixture of continuous and discrete. It is continuous for t ( q and it is discrete for t = q. It's density function is the same as that of T for t < q, i.e. f(t) = 1/5 for 0 ( t  ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download