# Thread: Bounds for integration in joint pdf

1. ## Bounds for integration in joint pdf

Hi,

Let f(x,y) = 3/2 if 0<=x<=1 and x^2<=y<=1; otherwise f(x,y)=0

Find

P(X=1/4, 0<y<3)

So I know I'm suppose to do double integration. My bounds for x are from 0 to 1/4. The less than sign on y is confusing me. Should I take the integration of y from 0+x to 3-x?

Thanks.

2. Originally Posted by shogunhd
Hi,

Let f(x,y) = 3/2 if 0<=x<=1 and x^2<=y<=1; otherwise f(x,y)=0

Find

P(X=1/4, 0<y<3)

So I know I'm suppose to do double integration. My bounds for x are from 0 to 1/4. The less than sign on y is confusing me. Should I take the integration of y from 0+x to 3-x?

Thanks.
$\displaystyle \Pr \left( X = \frac{1}{4}, 0 < Y < 3 \right) = \Pr \left( 0 < Y < 3 \bigg{|} X = \frac{1}{4} \right) \Pr \left( X = \frac{1}{4} \right)$.

But since you have a continuous distribution, $\displaystyle \Pr \left( X = \frac{1}{4} \right) = 0$ ......

So, do you mean find $\displaystyle \Pr \left( X \leq \frac{1}{4}, 0 < y < 3 \right) \,$ ?

If so:

First draw the region of integration. Then:

$\displaystyle \Pr \left( X \leq \frac{1}{4}, 0 < y < 3 \right) = \int_{0}^{1/4} \int_{y = x^2}^{y=1} \frac{3}{2} \, dy\, dx = .....$

3. Originally Posted by mr fantastic
$\displaystyle \Pr \left( X = \frac{1}{4}, 0 < Y < 3 \right) = \Pr \left( 0 < Y < 3 \bigg{|} X = \frac{1}{4} \right) \Pr \left( X = \frac{1}{4} \right)$.

But since you have a continuous distribution, $\displaystyle \Pr \left( X = \frac{1}{4} \right) = 0$ ......

So, do you mean find $\displaystyle \Pr \left( X \leq \frac{1}{4}, 0 < y < 3 \right) \,$ ?

If so:

First draw the region of integration. Then:

$\displaystyle \Pr \left( X \leq \frac{1}{4}, 0 < y < 3 \right) = \int_{0}^{1/4} \int_{y = x^2}^{y=1} \frac{3}{2} \, dy\, dx = .....$
Note:

$\displaystyle \Pr \left( X \geq \frac{1}{4}, 0 < y < 3 \right) = \int_{1/4}^{1} \int_{y = x^2}^{y=1} \frac{3}{2} \, dy\, dx = .....$

4. Originally Posted by mr fantastic
Note:

$\displaystyle \Pr \left( X \geq \frac{1}{4}, 0 < y < 3 \right) = \int_{1/4}^{1} \int_{y = x^2}^{y=1} \frac{3}{2} \, dy\, dx = .....$

the question asks for P(x=1/4, 0<y<3). I initially performed the integration using the first double integral you supplied. I was confused about the y bounds because I know that in a continuous distribution "<" sign can mean <=. Why did you change X<=1/4 to x=>1/4?

5. Originally Posted by shogunhd
the question asks for P(x=1/4, 0<y<3). Mr F says: Two things then:

1. Why did you have "My bounds for x are from 0 to 1/4" then?
2. I have shown why P(x=1/4, 0<y<3) = 0.

I initially performed the integration using the first double integral you supplied. I was confused about the y bounds because I know that in a continuous distribution "<" sign can mean <=. Why did you change X<=1/4 to x=>1/4?
Because I thought you had misposted due to the fact that if it wasn't, then the answer is zero and there's no itegrtaion to do at all ......