Pdf of random variables

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Homework Statement


Given:
The joint probability distribution function of X and Y:
f(x,y) =
2xe^(-y), x > 0, y > x^2
0, otherwise

Obtain the pdf of V = (X^2)/Y


The Attempt at a Solution



The interval of V is (0,1) because Y is always greater than X^2.

Fv(v) = P(V <= v)
= P(X^2/Y <= v)
= P(Y <= X^2/v)
= double integral from (0 to ?) and (x^2 to x^2/v) of (2xe^(-y)) dy dx

I can't figure out what the limits of integration would be the first integral, and I'm not sure if the limits for the second integral are correct either. I'm not really sure if I'm doing this right... Any help?
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


Given:
The joint probability distribution function of X and Y:
f(x,y) =
2xe^(-y), x > 0, y > x^2
0, otherwise

Obtain the pdf of V = (X^2)/Y


The Attempt at a Solution



The interval of V is (0,1) because Y is always greater than X^2.

Fv(v) = P(V <= v)
= P(X^2/Y <= v)
= P(Y <= X^2/v)

You mean P(X2≤Yv) = P(Y≥X2/v)

= double integral from (0 to ?) and (x^2 to x^2/v) of (2xe^(-y)) dy dx

I can't figure out what the limits of integration would be the first integral, and I'm not sure if the limits for the second integral are correct either. I'm not really sure if I'm doing this right... Any help?

Your upper limit (?) would be ∞ and everything would be OK except for your inequality reversal above. It's OK for limits of ∞.

I assume you have drawn a picture of y = x2 and y =(1/v)x2 for 0 < v < 1 and, of course, x > 0. So what do you get if you fix your inequality?
 
  • #3
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Your upper limit (?) would be ∞ and everything would be OK except for your inequality reversal above. It's OK for limits of ∞.

I assume you have drawn a picture of y = x2 and y =(1/v)x2 for 0 < v < 1 and, of course, x > 0. So what do you get if you fix your inequality?

The limits of the second integral would be x^2/V to infinity, right?
I haven't drawn the picture, I don't know how to draw the picture given the information. How do I use the joint pdf to draw the picture?
 
  • #4
LCKurtz
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The limits of the second integral would be x^2/V to infinity, right?

Not sure what second integral you refer to. You have to set it up again over the region described by the correct inequality.
I haven't drawn the picture, I don't know how to draw the picture given the information. How do I use the joint pdf to draw the picture?

You will never get such problems figured out if you don't learn to draw the regions.

The pdf itself isn't directly used in drawing the picture. You just need to know the region where it is non-zero. That was given to you as y > x2 with x > 0. You can draw that region can't you? Then you can draw where y > (1/v)x2 for v < 1. Integrate your pdf over the intersection of those two regions.
 
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Not sure what second integral you refer to. You have to set it up again over the region described by the correct inequality.


You will never get such problems figured out if you don't learn to draw the regions.

The pdf itself isn't directly used in drawing the picture. You just need to know the region where it is non-zero. That was given to you as y > x2 with x > 0. You can draw that region can't you? Then you can draw where y > (1/v)x2 for v < 1. Integrate your pdf over the intersection of those two regions.

Ohh I see... I drew a picture now and I got the limits for the x integral to be 0 to infinity and the limits for the y integral to be 0 to x^2/v. Then, when I solve the integral, I get:

ve^(-x^2/v) (evaluated from 0 to infinity) + x^2 (evaluated from 0 to infinity), which gives me an undefined answer b/c of the x^2...
 
  • #6
Ray Vickson
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Ohh I see... I drew a picture now and I got the limits for the x integral to be 0 to infinity and the limits for the y integral to be 0 to x^2/v. Then, when I solve the integral, I get:

ve^(-x^2/v) (evaluated from 0 to infinity) + x^2 (evaluated from 0 to infinity), which gives me an undefined answer b/c of the x^2...

I, personally, find it easier to do such problems using conditional distributions. The marginal densities of X and Y are:
[tex] \displaystyle f_X(x) = \int_{x^2}^{\infty} 2xe^{-y}\, dy = 2x e^{-x^2} [/tex]
and
[tex] \displaystyle f_Y(y) = \int_{0}^{\sqrt{y}} 2xe^{-y}\, dx = y e^{-y} [/tex]

We have [tex] \displaystyle P\{V \leq v \}= \int_{0}^{\infty} f_Y(y) P\{X^2/Y \leq v |Y = y \} = \int_{0}^{\infty} f_Y(y) P\{X^2 <= vy \} \, dy [/tex]
and we can find
[tex] P\{X^2 <= vy \} [/tex] from [tex] f_X[/tex].

RGV
 
  • #7
LCKurtz
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Ohh I see... I drew a picture now and I got the limits for the x integral to be 0 to infinity and the limits for the y integral to be 0 to x^2/v.

How can y go from 0 to x2/v when y is greater than x2/v? Have you made the correction I pointed out in my first response to you?
 
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How can y go from 0 to x2/v when y is greater than x2/v? Have you made the correction I pointed out in my first response to you?

oops, I made a mistake with setting up the integral, the limits of the y integral should be x^2/v to infinity.

One more question: How do you know when setting up the problem, if it's gonna be P(V <= v) or P(V >= v)? How how does that effect how the integrals are set up?
 
  • #9
LCKurtz
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oops, I made a mistake with setting up the integral, the limits of the y integral should be x^2/v to infinity.

One more question: How do you know when setting up the problem, if it's gonna be P(V <= v) or P(V >= v)? How how does that effect how the integrals are set up?

The cumulative distribution for a random variable V is defined:

FV(v) = P(V ≤ v)

so that is the natural thing to calculate. Of course that is always 1 - P(V > v) if that should happen to be easier to calculate.
 

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