# Joint probability function

• Apr 27th 2009, 09:57 AM
notecardheart
Joint probability function
I understand all the concepts of joint PDFs but I just can't get the right answer to this one. The function is:

f(x,y)=3x^(1/2)e^(-2y)

I need to find the probabilty over these two regions:

a) P (0 ≤ x ≤ 1; 3yhttp://www.webassign.net/images/infinity.gif)
b) P({P ({(x, y) | 2/3x ≤ 1; 0 ≤ yhttp://www.webassign.net/images/infinity.gif)
2

Explanations would be much appreciated. Thanks
• Apr 28th 2009, 02:13 AM
Moo
Hello,
Quote:

Originally Posted by notecardheart
I understand all the concepts of joint PDFs but I just can't get the right answer to this one. The function is:

f(x,y)=3x^(1/2)e^(-2y)

I need to find the probabilty over these two regions:

a) P (0 ≤ x ≤ 1; 3yhttp://www.webassign.net/images/infinity.gif)
b) P({P ({(x, y) | 2/3x ≤ 1; 0 ≤ yhttp://www.webassign.net/images/infinity.gif)
2

Explanations would be much appreciated. Thanks

By using wise definitions of the cumulative density functions, you should understand :)

$\displaystyle P(a<X<b)=\int_a^b f_X(x) ~dx$, where $\displaystyle f_X$ is the pdf of X

For a joint distribution, it's the same reasoning :
$\displaystyle P(a<X<b ~,~ c<Y<d)=\int_a^b \int_c^d f(x,y) ~dy dx$

So for the first one for example, it's $\displaystyle \int_0^1 \int_3^\infty f(x,y) ~dy dx$
• Apr 28th 2009, 10:58 PM
matheagle
I would like to know the support of this joint density.
The rvs seem to be independent.