Probability of Continuous Random Variables
Date: 04/23/2003 at 00:53:47 From: Beryl Subject: Probability of continuous random variables My book says that for any continuous random variable X, P (X=x)=0 By calculus, I can figure out the probability is 0, but by intuition, I can't see why. For example, Let X = the time that a customer spends at a supermarket. Then X is a continuous random variable. Why is the probability of a customer who spends 10 minutes at the supermarket always 0? That is possible for a customer spending exactly 10 minutes there, right? Then why is the probability 0?
Date: 04/23/2003 at 20:53:37 From: Doctor Samus Subject: Re: Probability of continuous random variables Hi Beryl, To calculate the probability of an event, you divide the number of outcomes that produce the event by the total number of outcomes. When the total number of outcomes is finite, then any event that can occur will have a non-zero probability, since the number of outcomes that produce the event isn't zero, and you divide that number by a finite number of total outcomes. Looking at your example, if the customer could only spend 1, 2, 3, ... 10 minutes in the store, then there is a finite number of outcomes, and so the probability of spending 10 minutes in the store would be 1/10, since only one outcome (spending 10 minutes) produces the event, and there are 10 total outcomes. With a continuous random variable, we no longer have the luxury of a finite number of total outcomes. Since there is an infinite number of real numbers in any interval of the real line, any continuous random variable has an infinite number of outcomes. Therefore, for a particular event that is produced by only 1 outcome, the probability is essentially: 1 lim - = 0 x->oo x Even if there is more than one outcome that produces the event, as long as the number of successful outcomes is finite, the probability is still zero, since the probability is essentially: c lim - = 0, where c is the number of successful outcomes. x->oo x Does this make sense? Thanks for writing in, and please write back if you need any more help. - Doctor Samus, The Math Forum http://mathforum.org/dr.math/
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