Joint PDF problem

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#1
Hey folks,
I've got a problem with a joint probability density function question.

Given f(x, y) = 6(x^2)y for 0 <= x <= y, x + y <= 2 and 0 elsewhere, I need to verify that f is a valid joint PDF.

The fact that f(x, y) >= 0 for all x, y is obvious and not difficult to prove. The part I'm having trouble with is showing that:

∫ ∫ f(x,y)*dx*dy = 1

...where both integrals are from -∞ to ∞.

Here's my thinking: based on the bounds given in the function definition, x must be bounded from 0 to 1, since if x is greater than 1 then x + y is greater than 2 and f goes to 0. y must be bounded from 0 to 2.

However, simply integrating with those bounds doesn't work; the answer I get is 4. The problem is that the integral using those bounds includes some area which should not be included according to the definition of the PDF since it would include, for example, x = 1 and y = 2.

However, I don't know what to do to get around this. I think the bounds need to be changed, but I don't know what they should be. I also toyed with the idea of subtracting a different integral, but I think that would invalidate the proof anyway.

Thanks for any help.

Edit: So I'm able to get the correct answer (1) if I subtract the integral ∫ ∫ f(x,y)*dx*dy bounded from x = 0 to 1 and y = 1 to 2, since this is equal to the area that's currently in my integral but shouldn't be. This is the answer I'll go with for now, but can anyone tell me if doing this subtraction is a valid way of showing that the density function is equal to 1 when integrated from -∞ to ∞, or if I need to go about this another way?
 
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Martingale

TS Contributor
#2
Hey folks,
I've got a problem with a joint probability density function question.

Given f(x, y) = 6(x^2)y for 0 <= x <= y, x + y <= 2 and 0 elsewhere, I need to verify that f is a valid joint PDF.

The fact that f(x, y) >= 0 for all x, y is obvious and not difficult to prove. The part I'm having trouble with is showing that:

∫ ∫ f(x,y)*dx*dy = 1

...where both integrals are from -∞ to ∞.

Here's my thinking: based on the bounds given in the function definition, x must be bounded from 0 to 1, since if x is greater than 1 then x + y is greater than 2 and f goes to 0. y must be bounded from 0 to 2.

However, simply integrating with those bounds doesn't work; the answer I get is 4. The problem is that the integral using those bounds includes some area which should not be included according to the definition of the PDF since it would include, for example, x = 1 and y = 2.

However, I don't know what to do to get around this. I think the bounds need to be changed, but I don't know what they should be. I also toyed with the idea of subtracting a different integral, but I think that would invalidate the proof anyway.

Thanks for any help.

Edit: So I'm able to get the correct answer (1) if I subtract the integral ∫ ∫ f(x,y)*dx*dy bounded from x = 0 to 1 and y = 1 to 2, since this is equal to the area that's currently in my integral but shouldn't be. This is the answer I'll go with for now, but can anyone tell me if doing this subtraction is a valid way of showing that the density function is equal to 1 when integrated from -∞ to ∞, or if I need to go about this another way?


x+y<2
so y<2-x

and x<y

so x<y<2-x

=> 0<x<1

so integrate from 0 to 1 with x and x to 2-x with y
 
#3
Thank you for your help. That does indeed work. Unfortunately, I've got another question now... given the same density function, I'm trying to find P(x + y <= 1).

I tried rearranging the equation to get my bounds as you did, and I got...

x + y <= 1
:. y <= 1 - x

and x <= y so...

x <= y <= 1 - x

and also x <= 0.5, since otherwise x + y > 1.

However, integrating using these bounds gives me a probability of 0.03125, which seems like it can't possibly be correct.
 
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#4
Verify

If you are unsure of your answer, you can always find 1 - P(x + y <= 1) = P(x + y > 1).

The boundary equations for this region are {x = y, x + y = 1, x + y = 2, x = 0}. with the region R as { (x, y) : x < y and x + y > 1 and x + y < 2 and x > 0 }.

To integrate over this region, you'll have to account for the change in boundary functions at the point x0, where x0 is the solution to 1 - x = x. You will then consider all y, such that x < y < 2 - x for x0 < x < x1, and 1 - x < y < 2 - x for 0 < x < x0, where x1 is the solution to x = 2 - x. That is x0 = 1/2 and x1 = 1.

So P(x + y > 1) is a double integral over R that you will consider as the sum of one double integral over R1 = { (x, y): 0 < x < 1/2 and 1 - x < y < 2 - x }, and another double integral over R2 = { (x, y): 1/2 < x < 1 and x < y < 2 - x }.

This yields 18/64 + 44/64 = 62/64 = 31/32 = P(x + y > 1). Comparing this to your previous result: P(x + y <= 1) = 1/32 = 1 - P(x + y > 1) = 1 - 31/32, so that P(x + y <= 1) + P(x + y > 1) = P({(x, y): x and y are in sample space}) = 1/31 + 31/32 = 32/32 = 1, which agrees with the probability axiom P(S) = 1, for sample space S.

You may not always have the time or resources to compute the probability of the complement event, but when you do it can't hurt.

:)
 
#10
I already gave the correct answer...I'm just curious what you are thinking
I've just told you what I am thinking. If we integrate over x first and then over y, then we will get the volume equal to 1.

If integrate over y first, and then over x, using the bounds you recommended, then the volume will not be equal to 1. We need to have volume to be 1 over double integration in; this is the property of a joint distribution function.
 

Martingale

TS Contributor
#11
x+y<2
so y<2-x

and x<y

so x<y<2-x

=> 0<x<1

so integrate from 0 to 1 with x and x to 2-x with y
I've just told you what I am thinking: if integrate over x first and then over y, then we will get the volume equal to 1.

If integrate over y first, and then over x, using the bounds you recommended, then the volume will not be equal to 1. We need to have volume to be one; this is the property of a joint distribution function.


\(\int\limits_{0}^{1}\int\limits_{x}^{2-x}6x^2ydydx=1\)

http://www.wolframalpha.com/input/?i=integrate+6%28x^2%29y+from++x%3D0+to+1+and+y%3Dx+to+2-x
 
#12
This however is not equal to 1 but it should be

http://www.wolframalpha.com/input/?i=int+6yx^2+dx+dy%2C+from+y%3D0+to+1%2C+from+x%3Dx+to+2-x
 

Martingale

TS Contributor
#14
This however is not equal to 1 but it should be

http://www.wolframalpha.com/input/?i=int+6yx^2+dx+dy%2C+from+y%3D0+to+1%2C+from+x%3Dx+to+2-x
So????


Why should it be???

Edit: let me clarify since you didn't understand...Why should it be with those particular limits
 
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Martingale

TS Contributor
#16
because it is a property of a joint distribution function. And either way you integrate it should be equal to 1. You are still not sure?:)
you obviously don't know what you are talking about.

Just to end this quickly...if you want to integrate x then y you can do this...


\(\int\limits_{0}^{1}\int\limits_{0}^{y}6x^2ydxdy+\int\limits_{1}^{2}\int\limits_{0}^{2-y}6x^2ydxdy=1\)


though I already showed that it integrated to 1 so this is just a waste of time.
 

Dason

Ambassador to the humans
#17
What he is saying is that you can integrate it in either order but the limits of your integration might need to change to reflect that. If you have variables in your integration limits you can't just switch the order without changing things around.
 
#18
S

Edit: let me clarify since you didn't understand...Why should it be with those particular limits
I don't understand? I think you didn't understand. I stated in my very first post, post number 6.

http://talkstats.com/showpost.php?p=38964&postcount=6

"Different bounds are needed. "

It took you a while to realise this :)
Why was I stating that this will not work with the bounds you provided when we integrate in different order? Because your solution is incomplete. Once again, remember the properties of a valid joint pdf. :)
 
#19
you obviously don't know what you are talking about.
Yes, I do

\(\int\limits_{0}^{1}\int\limits_{0}^{y}6x^2ydxdy+\int\limits_{1}^{2}\int\limits_{0}^{2-y}6x^2ydxdy=1\)
That's better! Much better:)


though I already showed that it integrated to 1 so this is just a waste of time.
It's not. The solution was not complete. But you now you have shown it in different order... So, problem solved. :)
 
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