Bivariate problem need help asap

**kermit** · December 22nd 2009, 12:45 AM

The joint Pdf of a bivariate Rv xy

Fxy(X,Y)= {k(xy).............0<y<2x<2
0...................otherwise

K a constant

Here you will first neeb to identify and delineate properly the triangle above which pdf is not zero????

What does this mean???

Also find K
find marginal pdfs of x and y

**mr fantastic** · December 22nd 2009, 01:58 AM

Originally Posted by kermit

The joint Pdf of a bivariate Rv xy

Fxy(X,Y)= {k(xy).............0<y<2x<2
0...................otherwise

K a constant

Here you will first neeb to identify and delineate properly the triangle above which pdf is not zero????

What does this mean???

Also find K
find marginal pdfs of x and y

It means draw the region defined by 0 < y < 2x < 2. The region is the area bounded by the x-axis and the lines y = 2x and x = 1.

This question is essentially the same as the one you asked here: http://www.mathhelpforum.com/math-he...lp-please.html. Please review that thread and learn from it.

**kermit** · December 22nd 2009, 05:04 AM

Can you let me know if this is correct

$ \int_{y=0}^{y=2}\int_{x= \frac{y}{2}}^{x=1}k(xy) \ dx \ dy = 1 $

Thanks

**mr fantastic** · December 23rd 2009, 01:03 AM

Originally Posted by kermit

Can you let me know if this is correct

$ \int_{y=0}^{y=2}\int_{x= \frac{y}{2}}^{x=1}k(xy) \ dx \ dy = 1 $

Thanks

The setup is correct.

**kermit** · December 28th 2009, 02:13 AM

$ k\int_{0}^{2}\left[\left({\dfrac{x^2y}{2}}\right)\right]_{\frac{y}{2}}^{1} \ dy=1 $

$ k\int_{0}^{2} \left( \frac{y}{2} - \frac{(y/2)^2 y}{2}\right) \ dy=1 $

Please sort out the next piece of LaTeX yourself - CB

$ k[({\dfrac{y}{\cfrac{2}{2}})^2 }-({\dfrac{y}{\cfrac{2}{\cfrac{2}{3}}}})^3 {\dfrac{y}{2}}^2]_{0}^{2}=1 $

can anyone tell me if this is integrated correctly. or have i gone wrong somewhere please

**CaptainBlack** · December 28th 2009, 07:52 AM

Originally Posted by kermit

$ k\int_{0}^{2}\left[\left({\dfrac{x^2y}{2}}\right)\right]_{\frac{y}{2}}^{1} \ dy=1 $

$ k\int_{0}^{2} \left( \frac{y}{2} - \frac{(y/2)^2 y}{2}\right) \ dy=1 $

Please sort out the next piece of LaTeX yourself - CB

$ k[({\dfrac{y}{\cfrac{2}{2}})^2 }-({\dfrac{y}{\cfrac{2}{\cfrac{2}{3}}}})^3 {\dfrac{y}{2}}^2]_{0}^{2}=1 $

can anyone tell me if this is integrated correctly. or have i gone wrong somewhere please

Not likely as no one is going to be able to parse your final expressions without working out the real answer, and then they still won't know if that is what you intended.

It would also have been nice if you had said what $f(x,y)$ was earlier.

CB

**mr fantastic** · December 28th 2009, 09:54 PM

Originally Posted by kermit

$ k\int_{0}^{2}\left[\left({\dfrac{x^2y}{2}}\right)\right]_{\frac{y}{2}}^{1} \ dy=1 $

$ k\int_{0}^{2} \left( \frac{y}{2} - \frac{(y/2)^2 y}{2}\right) \ dy=1 $ <---

Please sort out the next piece of LaTeX yourself - CB

$ k[({\dfrac{y}{\cfrac{2}{2}})^2 }-({\dfrac{y}{\cfrac{2}{\cfrac{2}{3}}}})^3 {\dfrac{y}{2}}^2]_{0}^{2}=1 $

can anyone tell me if this is integrated correctly. or have i gone wrong somewhere please

Life would be easier if you simplified your expressions before continuing to integrate.

After simplifying the line I labelled <--- you have $k \int_0^2 \frac{y}{2} - \frac{y^3}{4} \, dy = 1$ and you shouldn't need re-assurance on how to finish this.

Thread: Bivariate problem need help asap

LinkBack

Thread Tools

Display

Bivariate problem need help asap

Similar Math Help Forum Discussions

Bivariate Normal Distribution Problem

COMBINATION PROBLEM - PLEASE HELP! I NEED ASAP

Word problem ... Please help ASAP

Probability Bivariate Random Variable Problem

[SOLVED] statistics problem need help ASAP please.

Search Tags