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Q: Statistics: joint density ( Answered 4 out of 5 stars,   1 Comment )
Question  
Subject: Statistics: joint density
Category: Science > Math
Asked by: weizai-ga
List Price: $5.00
Posted: 25 Sep 2002 02:33 PDT
Expires: 25 Oct 2002 02:33 PDT
Question ID: 68812
Find the joint density of X+Y and X/Y are independent exponential
random variable with parameter with parameter L, Lamda.  Show that X+Y
adn X/Y are independent.  Show all the work.
Answer  
Subject: Re: Statistics: joint density
Answered By: calebu2-ga on 25 Sep 2002 08:50 PDT
Rated:4 out of 5 stars
 
weizai,

The way we go about solving a problem like this (or at least the way I
go about it) is to use the change of variables theorem.

This tells us how to write the joint density of a set of functions of
other variables. For more on this theorem, see the links I posted
below.

In general: 

suppose we have two variables X and Y with joint density f_xy(x,y)

Suppose also we have two variables U and V defined as :

U = g1(x, y) and V = g2(x, y)

and we want to know what the joint density of U and V is, f_uv(u,v)

We define the jacobian, J

J = det(dg1/dx dg1/dy)
       (dg2/dx dg2/dy)

Then the equation for f_uv(u,v) is :

f_uv(u,v) = f_xy(x,y) / J   (we substitute in functions of u and v for
x and y)

So in your question we have :

f_xy(x,y) = L^2 exp{-L(x+y)}

u = x + y
v = x / y

Then du/dx = 1, du/dy = 1, dv/dx = 1/y, dv/dy = -x/(y*y)

So J = 1/y - (-) x/(y*y) = (x + y)/(y*y)

We can also write :

y = u - x
vy = x

so vu - vx = x

or x = uv/(1+v)
and y = u/(1+v)

Hence f_uv(u,v) = L^2 exp{-L(x+y)} / ((x + y)/(y*y))

So f_uv(u,v) = L^2 exp{-Lu} * (u/((1+v)^2))

------

Now to prove independence it is enough to show that the distribution
function F_uv(u,v) is bilaterally separable. ie.

Integral(u=0, u=inf) of f_uv(u,v) gives f_u(u)
Integral(v=0, v=inf) of f_uv(u,v) gives f_v(v)

f_u(u) = L^2 exp{-Lu} * u * Int((1+v)^-2)

= L^2 exp{-Lu} * u * [-(1+inf)^-2] - [-(1+0)^-2)] = L^2 exp{-Lu} * u

Likewise L^2 exp{-Lu} * u is the equation for a gamma distribution, so

f_v(v) = (1+v)^-2

Hence f_uv(u,v) = f_u(u) * f_v(v), so the two are independent. (see
Mathworld for a formal treatment, I kind of rushed that bit)

Anyway, good luck with your studies and if there's any way we can help
you in the future, be sure to come back and post more questions!

Regards

Calebu2-ga

---------

Useful Links

Mathworld : Exponential Distribution
http://mathworld.wolfram.com/ExponentialDistribution.html

Mathworld : Jacobian
http://mathworld.wolfram.com/Jacobian.html

Mathworld : Joint Distribution Function
http://mathworld.wolfram.com/JointDistributionFunction.html

Mathworld : Change of Variables Theorem
http://mathworld.wolfram.com/ChangeofVariablesTheorem.html

Cornell Lecture notes : Change of Variables Theorem
http://instruct1.cit.cornell.edu/Courses/btry408/node101.html

Google Search term :
jacobian change of variables distribution
weizai-ga rated this answer:4 out of 5 stars

Comments  
Subject: Re: Statistics: joint density
From: math_zip_at_yahoo-ga on 25 Sep 2002 14:53 PDT
 
The question was not correctly stated. The right one should be:

Find the joint density of X+Y and X/Y, which X and Y are independent
exponential
random variable with parameters L and Lamda.  Show that X+Y and X/Y
are independent.

Calebu2-ga's answer is correct for the corrected question.

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