# derived probability density function

• September 27th 2007, 09:41 PM
rbenito
derived probability density function
I have a problem about derive a pdf from two marginal pdfs:
random variables x and y are distributed according to the joint probability density function (pdf):
fxy(x,y) = 3x/2 , if 1 <= x <= y <= 2
0 , otherwhise

Random variable z is defined by z = y - x
Determine the probablity density function fz(z)

thanks.
• September 28th 2007, 05:52 PM
BruceBronson
it's been a while since i've done this but i think you need to do a doubble integral over the domain specified on the function you have been given
• September 28th 2007, 09:32 PM
CaptainBlack
Quote:

Originally Posted by rbenito
I have a problem about derive a pdf from two marginal pdfs:
random variables x and y are distributed according to the joint probability density function (pdf):
fxy(x,y) = 3x/2 , if 1 <= x <= y <= 2
0 , otherwhise

Random variable z is defined by z = y - x
Determine the probablity density function fz(z)

thanks.

For any value of $z$ the function $z = y-x$ defines a curve through $xy$ space. The pdf of $p(z)$ of $z$ is the integral of the joint density over the area between the curve for $z$ and $z+\delta z$ divided by $\delta z$ (rather the limit as $\delta z$ goes to zero).

Write the curves in the form: $y=x+z$, then we have:

$
A= \delta z~ \int f(x,x+z) dx
$

So:

$
p(z) = \int f(x,x+z) dx
$

If you can't complete this from here say so and I will finish it.

RonL
• September 30th 2007, 08:14 AM
rbenito
can you complete the integral?
what are the limits?
• September 30th 2007, 10:38 AM
CaptainBlack
Quote:

Originally Posted by rbenito
what are the limits?

$\mathbb{R}$,

but since the support of $x$ is $[1,2]$, that will do, but you have to modify this because the function is non-zero only when $1

RonL
• September 30th 2007, 10:48 AM
rbenito
can you finish the integral?
of the problem to obtain the derived pdf?
• September 30th 2007, 07:16 PM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
For any value of $z$ the function $z = y-x$ defines a curve through $xy$ space. The pdf of $p(z)$ of $z$ is the integral of the joint density over the area between the curve for $z$ and $z+\delta z$ divided by $\delta z$ (rather the limit as $\delta z$ goes to zero).

Write the curves in the form: $y=x+z$, then we have:

$
A= \delta z~ \int f(x,x+z) dx
$

So:

$
p(z) = \int f(x,x+z) dx
$

If you can't complete this from here say so and I will finish it.

RonL

If I've done this right:

Now as $z=y-x,\ p(z)$ is zero for $z \not \in [0,1]$, and for $z \in [0,1]$:

$
p(z) = \int_1^{2-z} f(x,x+z) dx = p(z) = \int_1^{2-z} \frac{3x}{2} dx = \frac{3}{4}~\left[(z-2)^2-1\right]
$

RonL
• October 1st 2007, 08:28 AM
rbenito
what about the integral related to "y"?
I've noticed that you integrated respect to "x", but as far as I know, if you want to derive a pdf from two random variables, the limits must be according to the marginal "x" and "y" pdfs.
• October 1st 2007, 02:26 PM
CaptainBlack
Quote:

Originally Posted by rbenito
I've noticed that you integrated respect to "x", but as far as I know, if you want to derive a pdf from two random variables, the limits must be according to the marginal "x" and "y" pdfs.

It is only a coincidence that it is x, what I actually did was integrate along the
curve g(x,y)=z, it just so happened that x was a convenient parameterisation
of the curve.

RonL