# Thread: derived probability density function

1. ## 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.

2. 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

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

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

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

So:

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

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

RonL

4. ## can you complete the integral?

what are the limits?

5. Originally Posted by rbenito
what are the limits?
$\displaystyle \mathbb{R}$,

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

RonL

6. ## can you finish the integral?

of the problem to obtain the derived pdf?

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

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

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

So:

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

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

RonL
If I've done this right:

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

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

RonL

8. ## 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.

9. 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