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Continous Random Variables


Date: 03/15/2001 at 00:37:10
From: Brian
Subject: Continous random variables

Let c be a constant and consider the density function

    f(y) = (1/c)e^(-y/2)  if Y > or = 0 and
           (1/c)e^(y/2)  if Y < 0

1. find the value of c
2. find the cumulative distribution function F(y)
3. compute F(1)
4. compute P(y>.5)


Date: 03/18/2001 at 22:54:50
From: Doctor Jordi
Subject: Re: Continous random variables

Dear Brian,

The exercise that you present above should help you understand 
continuous random variables and their distributions. I will try to 
give you a few hints that should help you to solve it:

1) The value of c should be such that the total probability integrates 
   to 1.  

My suggestion is to find the integral of the density function from 
minus infinity to plus infinity without paying any attention to the 
value of c for the moment. You will get some number as an answer.  
Choose c such that this number times (1/c) equals 1.

2) The cumulative distribution function F(y) has the same value as 
   P(Y < y); it is the probability that the random variable Y takes 
   on a value less than y. 
 
Given the density function, f(y), F(y) is defined to be

     Integral[-infinity, y] (f(s)ds)
     ^
     | 
    (the definite integral of f(s)ds from negative infinity to y.  

I will write this from now on as Int[-inf, y] (f(s)ds) ) where s is 
just any dummy variable. In this particular problem, you will have to 
split up your integration into two intervals, since the form of the 
density function changes at y = 0.  You will get something like this:

             /  Int[-inf, y](f(s)ds)                     for y =< 0
            |
     F(y) = {
            |   Int[-inf, 0](f(s)ds) + Int[0, y](f(s)ds) for y > 0
             \

If you want to check your work, here are some properties that F(y) 
should have in the end. These properties follow because F(y) is a 
probability function (or a distribution).

         F(-inf) = 0     
       
         F(inf)  = 1
       
         0 < F(y) < 1   for all real y.

         F(y) is a strictly increasing function 
              (meaning F'(y) is always nonnegative)

         F'(y) = f(y)

3) This one should be easy; since once you've found F(y), all you have 
   to do is evaluate it a point.

4) You only have to recall an important law about probability.  
   P(Y > y) = 1 - P(Y < y).  Further, recall that P(Y < y) = F(y).

I hope this is enough to get you started on working this problem.  
Please write back if you have more questions, or if you would like to 
talk more about this.
         
- Doctor Jordi, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Functions

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