# double integral 'prob'

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• Aug 10th 2010, 08:24 AM
bluesblues
double integral 'prob'
I really need help with this question.

the density function is $f(x,y)=\frac{1}{y}exp(-\frac{x}{y})exp(-y)$ for x,y>0

the question is to calculate p(X>Y) and

the hint they give is:

[you will need to calculate a double integral over the area(y>0,x>y)]

$\int_0^\infty) \int_o^y f(x,y) dx dy$ = 0.6321 =p(x>y) ?

$1 - \int_o^\infty \int_o^y f(x,y) dx dy$ = 0.3679 = p(x>y) ?

I am basically confused as to whether integrating from 0 to y for dx and 0 to infinity for dy works and if it follows the hint at all. the answer doing so gives 0.6321. I am also thinking that the answer should be 1 minus this using the principle p(x>y)-1-p(y<x) which I am almost certain doesnt follow the hint. Is my first method corret? as you can clearly see i dont know what integrating using my first method represents. I was under the impression it should equal 1, although this clearly can't be the answer.

any ideas what I should do? thanks
• Aug 10th 2010, 08:47 AM
AllanCuz
Quote:

Originally Posted by bluesblues
I really need help with this question.

the density function is $f(x,y)=\frac{1}{y}exp(-\frac{x}{y})exp(-y)$ for x,y>0

the question is to calculate p(X>Y) and

the hint they give is:

[you will need to calculate a double integral over the area(y>0,x>y)]

I am basically confused as to whether integrating from 0 to y for dx and 0 to infinity for dy works and if it follows the hint at all. the answer doing so gives 0.6321. I am also thinking that the answer should be 1 minus this using the principle p(x>y)-1-p(y<x) which I am almost certain doesnt follow the hint. Is my first method corret? as you can clearly see i dont know what integrating using my first method represents. I was under the impression it should equal 1, although this clearly can't be the answer.

any ideas what I should do? thanks

As long as I'm reading this correctly, we want the density as given by the equation with the bounds of $x,y > 0$ and $x>y$?

Then yes, our y would be bounded from 0 to x and our x would be bounded from 0 to infinity. I don't see why this wouldn't work.
• Aug 10th 2010, 09:05 AM
bluesblues
thanks for the reply, yes those bounds are correct evident in the question but do you not mean x bounded from 0 to y and y bounded from 0 to infinity?

Im afraid what you have said wouldnt work as you'd be integrating -exp(-x/y)exp(-y) w.r.t y which is quite hard...

also putting the end limit after this of x wouldnt give a difinitive number ,,
• Aug 10th 2010, 09:20 AM
AllanCuz
Quote:

Originally Posted by bluesblues
thanks for the reply, yes those bounds are correct evident in the question but do you not mean x bounded from 0 to y and y bounded from 0 to infinity?

Im afraid what you have said wouldnt work as you'd be integrating -exp(-x/y)exp(-y) w.r.t y which is quite hard...

also putting the end limit after this of x wouldnt give a difinitive number ,,

No, I meant the bounds I said. If you want to make a change of bounds that's fine just make sure they agree with the graph. But no...having infinity as a limit does not neccessarily gaurentee an infinite result from the integral (for illustration purposes look at: http://en.wikipedia.org/wiki/Gaussian_integral ). Then again, maybe this equation doesn't extend to infinity in the first quadrent, I dont know because I dont have my graphing calculator on me!

Here's what you do,

Graph the function with your graphing calculator and find the bounds in the first quadrent. If you find that x goes to infinity write them as I have said, and if you find this hard, then switch the bounds of integration. If you find that x does not go to infinity then put whatever bounds you do find into your integral.
• Aug 10th 2010, 09:29 AM
bluesblues
I have to look at my notes regarding changing the order of integration of thats what you mean. im not sure what you mean by changing bounds. The order of integration i would do is the integral wrt x then the integral wrt y,,,,i hope you can see how i disagree that the upper limit as x for the integration wrt y wouldnt give a numbered answer. id prefer not to use a graphing calculator with this question as its from an exam where theyre nt alowed. im not sure i could sketch the function without it,,i could try. to sum up,,,what do you mean by changing the bounds? do you mean the limits? and for what reason would i want to change them,,,

$\int_o^\infty \int_o^y f(x,y) dx dy$ = 0.6321 =p(x>y) ?

$1 - \int_o^\infty \int_o^y f(x,y) dx dy$ = 0.3679 = p(x>y) ?
• Aug 10th 2010, 09:36 AM
AllanCuz
Quote:

Originally Posted by bluesblues
I have to look at my notes regarding changing the order of integration of thats what you mean. im not sure what you mean by changing bounds. The order of integration i would do is the integral wrt x then the integral wrt y,,,,i hope you can see how i disagree that the upper limit as x for the integration wrt y wouldnt give a numbered answer. id prefer not to use a graphing calculator with this question as its from an exam where theyre nt alowed. im not sure i could sketch the function without it,,i could try. to sum up,,,what do you mean by changing the bounds? do you mean the limits? and for what reason would i want to change them,,,

Find change of bounds here: http://www.mathhelpforum.com/math-he...on-146568.html Scroll down to change of bounds.

You cannot switch the order of integration without changing your bounds, and how you do so, is dependent on what the function looks like.
• Aug 10th 2010, 09:41 AM
bluesblues
thanks,,,the functionn i posted at the top,,,i dont think it is sketchable,,,,,furthermore this exam question wouldn't ask you to change to poler cordinates
• Aug 10th 2010, 09:44 AM
AllanCuz
Quote:

Originally Posted by bluesblues
thanks,,,the functionn i posted at the top,,,i dont think it is sketchable,,,,,furthermore this exam question wouldn't ask you to change to poler cordinates

Dude....what are you talking about? Changing the limits of integration has nothing to do with polar co-ordinates nor did I tell you to switch to them. The problem gives you a double integral and defined bounds for x and y and by doing so it gives us a given order of integration. If we want to switch that order of integration we can definately do that!!! But we need to change our bounds (i.e. our x and y limits) as well...we cannot just freely change the order of integration without changing the bounds..

This is a general principle that you should have been taught right when you learned about double integrals....
• Aug 10th 2010, 09:52 AM
bluesblues
ok,,but without knowing the shape of the curve according to your notes theres no way I can find the bounds. your notes dont ring a bell and i've done double integration before.

your obviously trying to show me another way of doing this by what i can only assume is changing the order of integration,,,is there really need for this,,,,could you show me what you mean
• Aug 10th 2010, 10:09 AM
bluesblues
as far as im aware what allancuz was suggesting initially was this: and to change it from here?? using new bounds?!

$\int_o^x \int_0^\infty f(x,y) dx dy = exp(-x) +x -1$ which gives no answer,,,as i can't do bounds,,can't use graphic calc in this type question.
• Aug 10th 2010, 10:20 AM
11rdc11
The correct change of bounds would be $\int^{\infty}_{0} \int^{x}_{0} dy dx$ There is no need for a graphing calc for this problem. The region you have to sketch is just a triangle.
• Aug 10th 2010, 10:29 AM
bluesblues
thankyou,,,just tried doing this presumably i try the y integral first and I cant work it out.

Furthermore in order to aid this question I have stumbled upon this in which the solution is 1/2 which proved my method of working is wrong and your method of change of bounds is wrong it is,,,

the method they used for this would be the same as the problem in this thread
rv's have joind density fn exp(-x-y) for x,y>0

find p(x>y) answer given in 1/2 : O

could someone please show me some kind of solution for both problems and how they obtained it using changed bounds thanks
• Aug 10th 2010, 11:06 AM
bluesblues
a triangle?
• Aug 10th 2010, 11:17 AM
yeKciM
Quote:

Originally Posted by bluesblues
a triangle?

if u have x>y and x,y >= 0

so that's where you are integrating ...
and can go from zero to infinity on x, and from zero to x on y-axis

Edit: didn't see correct, i saw (x,y)>=0 but it's (x,y)>0 (and sorry that it looks like that it isn't y=x line but it is, I just zoom it at one side so it looks like it's less than that :D )
• Aug 10th 2010, 11:24 AM
bluesblues
thanks this was the reasoning i was trying to do in my head without going over old noted it seems to be correct what you have put
although should not the line be y=x nvm,,,but having done this integration a hundred times i still keep getting ... as an answer

$\int_o^\infty \int_0^x \frac{1}{y}exp(-\frac{x}{y})exp(-y)dy dx =,,$
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