# Thread: Help with Joint Probability Density Function

1. ## Help with Joint Probability Density Function

Hi all,

I am stuck with the problem below:

The Joint pdf of random variables X and Y: f(x,y) = k/y , 0<x<y, 0<y<1 and f(x,y) = 0, otherwise.

First of all, the question asked to draw a diagram showing the support of the joint pdf.
Probably this is the part that i am unable to do, which leads to problems with later part of the question.

The question asked to calculate k, I integrate with respect to x over the range 0 to y, and then integrate with respect to y over the range 0 to 1.
The answer I get is k = 1, is it correct? Did i integrate the correct range for x and y?

After that the question asked to find: P(0<=X<=1/4, 0<=Y<=1/3).
For X, the range i used to integrate over function f is 0 to 1/4. However, for Y, i am not so sure about it, I guess it is from 1/4 to 1/3? Or 0 to 1/3 for Y?

Any help regarding the problem would be appreciated.
Thank you.

2. Originally Posted by Watts84
Hi all,

I am stuck with the problem below:

The Joint pdf of random variables X and Y: f(x,y) = k/y , 0<x<y, 0<y<1 and f(x,y) = 0, otherwise.

First of all, the question asked to draw a diagram showing the support of the joint pdf.
Probably this is the part that i am unable to do, which leads to problems with later part of the question.

The question asked to calculate k, I integrate with respect to x over the range 0 to y, and then integrate with respect to y over the range 0 to 1.
The answer I get is k = 1, is it correct? Did i integrate the correct range for x and y?

After that the question asked to find: P(0<=X<=1/4, 0<=Y<=1/3).
For X, the range i used to integrate over function f is 0 to 1/4. However, for Y, i am not so sure about it, I guess it is from 1/4 to 1/3? Or 0 to 1/3 for Y?

Any help regarding the problem would be appreciated.
Thank you.
The support is the triangular region defined by the lines y = 1 and y = x and the y-axis. Is that what you had?

3. Yes, I did manage to get that... Thank you.
In that case, the double integral range would be from 0<x<1, 0<y<1?
If thats the case, it looked like integrating the whole square, instead of the right triangle?

4. no, it's a triangle 0<x<y<1

5. Originally Posted by Watts84
Yes, I did manage to get that... Thank you.
In that case, the double integral range would be from 0<x<1, 0<y<1?
If thats the case, it looked like integrating the whole square, instead of the right triangle?
Have you been taught how to set up and solve double integrals over a region of the xy-plane? If not, then there's little point in attempting questions like this one until you.

The integral you're expected to set up and solve is $\int_{x = 0}^{x = 1} \int_{y = x}^{y = 1} f(x, y) \, dy \, dx$. Then equate the result of this calculation to 1 in order to solve for k.

6. I prfer dxdy, it has 0 as both lower bounds
and the integration is a lot easier.

$\int_0^1\int_0^y f(x,y)dxdy$

7. Originally Posted by matheagle
I prfer dxdy, it has 0 as both lower bounds
and the integration is a lot easier.

$\int_0^1\int_0^y f(x,y)dxdy$

This is just basic calc3
I see now why baldeagles are an endangered species ..... Their lower bounds are zero.

8. Yes, i have been taught how to do double integral over x-y plane. I also prefer the lower bounds to be set to zero too, much easier..Thanks Matheagle.
Been quite confuse in deciding the range for integration.
Thanks a lot of clarified my doubts.
As for the last part: Finding P(0<=X<=1/4, 0<=Y<=1/3).
The integration will look like this: $
\int_{x = 0}^{x = 1/4} \int_{y = x}^{y = 1/3} f(x, y) \, dy \, dx
$
?
If i wish to integrate by dxdy, range of x will be from 0 to 1/4. How about y? $
\int_{y = ?}^{y = ?} \int_{x = 0}^{x = 1/4} f(x, y) \, dx \, dy
$
?

9. dxdy needs two regions, dydx only one, draw it

$\int_0^{1/4}\int_x^{1/3} fdydx$

To do dxdy you need to split y into two parts (0,1/4) and (1/4,1/3)

$\int_0^{1/4}\int_0^y fdxdy+ \int_{1/4}^{1/3}\int_0^{1/4} fdxdy$