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Uniform and Exponential Distribution of Random Variables

Date: 01/16/2006 at 16:49:00
From: troy
Subject: Relationship between uniform distrib and exponential distrib

I've seen statements claiming if you take the natural log of a 
uniformly distributed random variable, it becomes a exponentially 
distributed random variable.  I know it's a true statement, but I
wonder if you would provide a proof?



Date: 01/17/2006 at 09:28:45
From: Doctor George
Subject: Re: Relationship between uniform distrib and exponential distrib

Hi Troy,

Thanks for writing to Doctor Math.

The standard form of the problem includes a minus sign on the
logarithm, so I will include it.

Let X ~ U[0,1] and Y = -ln(X).

   F(y) = P(Y < y)
    Y
        = P[-ln(X) < y]

        = P[X > e(-y)]

        = 1 - P[X < e(-y)]

        = 1 - F[e(-y)]
               X

Now differentiate both sides with respect to y.

   f(y) = f[e(-y)]e(-y) = e(-y)
    Y      X

Does that make sense?  Write again if you need more help.

- Doctor George, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 01/17/2006 at 12:26:42
From: troy
Subject: Thank you (Relationship between uniform distrib and
exponential distrib)

Thank you, Doctor George, for the succinct proof.  I'd like to
continue the discussion below.  Although it's not a complex
derivation, it's definitely non-trivial.  I wondered why most articles
skipped the proof and only made the statement.  Could I ask you to
point me to a reference where I could find similar derivations?

Also on the last equation, f(y) = f[e(-y)]e(-y) = e(-y), 
                            Y      X

it implies that f[e(-y)] = 1. 
                 X

What is the reason for this?  Is it because by definition that Y =
-ln(X), therefore  x = e(-y), so the probability of x = e(-y) is always 1?

Thank you again for the beautiful derivation.

Troy



Date: 01/17/2006 at 13:58:33
From: Doctor George
Subject: Re: Thank you (Relationship between uniform distrib and
exponential distrib)

Hi Troy,

You are on the right track.  Since X is uniform,

   f(x) = 1
    X

now substituting x = e(-y) we get

   f[e(-y)] = 1
    X

As for a reference, most any college level book on Mathematical
Statistics will contain examples similar to this.  Look for a section
on transformation of variables.  Sometimes a book will skip a short
proof like this and have the reader work it out as a problem.

- Doctor George, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Probability

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