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How to find the bounds

  1. Sep 4, 2006 #1
    given
    [tex] \frac{x + \sqrt{x^{2} - 4y}}{2} \geq 0[/tex]
    [tex] \frac{x - \sqrt{x^{2} - 4y}}{2} \geq 0[/tex]

    how do I find the bound of values of x and y?

    my first answer were [tex]x\geq 0[/tex] and [tex] y\geq 0[/tex], but I discovered flaws in my solution.

    any hints?

    thanks.
     
  2. jcsd
  3. Sep 4, 2006 #2

    StatusX

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    You have x-a>0 and x+a>0, where a>0. Do you want both of these satisfied at the same time? Because then the second one is superfluous. Either way, if you have x-a>0, then x>a, so x^2>a^2 But when you square, you introduce extraneous negative roots, so what you really want to say is (assuming a is positive) x>a iff x^2>a^2 AND x>0. Can you go from here?
     
  4. Sep 4, 2006 #3
    quick question...
    why is the second one superfluous?

    I'm writing this out here...
    since [tex] a = \sqrt{x^{2} - 4y}[/tex]
    then
    [tex] a^{2} = x^{2} - 4y[/tex]

    so assuming a is positive,
    I have
    [tex] x \geq \sqrt{x^{2} - 4y}[/tex]
    iff
    [tex]x^{2} \geq x^{2} - 4y[/tex], which is the same as [tex]y \geq 0[/tex]
    and
    [tex] x \geq 0[/tex]
     
  5. Sep 4, 2006 #4

    StatusX

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    It's superfluous in the sense that if (x,y) satisfy the second of your inequalities, they must also satisfy the first. You also need x^2-4y>0 so the square root is defined.
     
  6. Sep 4, 2006 #5
    my problem lies with the integration...this question came from this bigger problem of mine

    I have already f(x,y) where
    [tex]f(x,y) = \lambda^{2} e^{-\lambda x} \frac{1}{\sqrt{x^{2} - 4y}[/tex]

    with
    [tex] \frac{x + \sqrt{x^{2} - 4y}}{2} \geq 0[/tex]
    and
    [tex] \frac{x - \sqrt{x^{2} - 4y}}{2} \geq 0[/tex]

    I need to get f(x)
    so I have to get
    [tex]\int f(x,y) dy[/tex]


    thanks again for the help.
     
    Last edited: Sep 4, 2006
  7. Sep 4, 2006 #6
    can anyone confirm that the bounds for y is
    [tex] \frac{x^{2}}{4} \geq y \geq 0[/tex]?

    thanks
     
  8. Sep 4, 2006 #7

    StatusX

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    Yes, that looks right.
     
  9. Sep 4, 2006 #8
    thanks for the help, StatusX

    and
    [tex] x \geq 0[/tex], right?

    I have sent you a PM, btw.
     
  10. Sep 4, 2006 #9
    Here's the complete question and my step by step solution:
    For some reason, there seems to be something wrong with my solution,

    I hope you guys can check what I did wrong

    --------------

    The question asks me to find E(Y|X)

    I'm given the following:
    u and v are independent exponential distributions
    thus
    [tex]f(u) = \lambda e^{-\lambda u}, u \geq 0[/tex]
    and
    [tex]f(v) = \lambda e^{-\lambda v}, v \geq 0[/tex]

    I am also given the following facts:
    x = u+v
    and
    y = uv
     
    Last edited: Sep 4, 2006
  11. Sep 4, 2006 #10
    my solution is as follows:

    since u and v are independent,
    I have f(u,v) = f(u)f(v)
    thus
    [tex]f(u,v) = \lambda^{2} e^{-\lambda(u+v)}[/tex]
    for
    [tex]u \geq 0
    v \geq 0[/tex]

    I need f(x,y)
    this is solved through transformation of variables

    thus
    [tex]f(x,y) = \lambda^{2} e^{-\lambda(x)} |J|[/tex]

    where J is the Jacobian
    [tex] |J| = |\frac{\partial u}{\partial x} \frac{\partial v}{\partial y} - \frac{\partial u}{\partial y}\frac{\partial v}{\partial x}|[/tex]

    since x = u+v and y =uv

    I was able to solve for u and v in terms of x and y

    where
    [tex]u = \frac{x + \sqrt{x^{2} - 4y}}{2} [/tex]
    [tex]v = \frac{x - \sqrt{x^{2} - 4y}}{2} [/tex]
    or vice versa
     
  12. Sep 4, 2006 #11
    solving I was able to get
    [tex]\frac{\partial u}{\partial x} = \frac{1}{2} + \frac{x}{2}(x^{2} - 4y)^{-\frac{1}{2}}[/tex]
    [tex]\frac{\partial u}{\partial y} = -(x^{2} - 4y)^{-\frac{1}{2}}[/tex]
    [tex]\frac{\partial v}{\partial x} = \frac{1}{2} - \frac{x}{2}(x^{2} - 4y)^{-\frac{1}{2}}[/tex]
    [tex]\frac{\partial v}{\partial y} = (x^{2} - 4y)^{-\frac{1}{2}}[/tex]

    thus
    [tex] |J| = (x^{2} - 4y)^{-\frac{1}{2}}[/tex]

    and
    [tex]f(x,y) = \lambda^{2} e^{-\lambda x} (x^{2} - 4y)^{-\frac{1}{2}}[/tex]

    with bounds
    [tex]x \geq 0[/tex]
    [tex] \frac{x^{2}}{4} \geq y \geq 0[/tex]
    (as solved from above)
     
  13. Sep 4, 2006 #12
    my BIG BIG problem is when I checked whether f(x,y) is a distribution
    I got:

    [tex]\int_{0}^{\infty}\int_{0}^{\frac{x^{2}}{4}}\lambda^{2} e^{-\lambda x} (x^{2} - 4y)^{-\frac{1}{2}} dy dx = \frac{1}{2} \neq 1[/tex]

    Is there anything I did wrong?

    I checked and triple-checked my solution already, and I can't find anything wrong with it!!!

    help please!
     
  14. Sep 4, 2006 #13
    my solution for the integral:

    [tex]\int_{0}^{\infty}\int_{0}^{\frac{x^{2}}{4}}\lambda ^{2} e^{-\lambda x} (x^{2} - 4y)^{-\frac{1}{2}} dy dx[/tex]


    [tex] = \int_{0}^{\infty}\lambda^{2} e^{-\lambda x} \int_{0}^{\frac{x^{2}}{4}}(x^{2} - 4y)^{-\frac{1}{2}} dy dx[/tex]
    [tex] = \int_{0}^{\infty}\lambda^{2} e^{-\lambda x} [-\frac{1}{2}(x^{2} - 4y)^{\frac{1}{2}} ]_{0}^{\frac{x^{2}}{4}}dx[/tex]
    [tex] = \int_{0}^{\infty}\lambda^{2} e^{-\lambda x} [-\frac{1}{2}(-x)dx[/tex]
    [tex] = \int_{0}^{\infty}\frac{\lambda^{2}x e^{-\lambda x}}{2}dx[/tex]

    using integration by parts
    [tex] = \frac{\lambda^{2}}{2}([\frac{-x e^{-\lambda x}}{\lambda}]_{0}^{\infty} - \int_{0}^{\infty}\frac{e^{-\lambda x}{\lambda} dx)[/tex]
    [tex]=\frac{\lambda^{2}}{2} (0 - \int_{0}^{\infty}\frac{e^{-\lambda x}{\lambda} dx)[/tex]
    [tex]= \frac{\lambda^{2}}{2}(-\frac{1}{\lambda^{2}}[e^{-\lambda x}]_{0}^{\infty})[/tex]
    [tex]= -\frac{1}{2}(0-1)[/tex]
    [tex] = \frac{1}{2} [/tex]
     
  15. Sep 4, 2006 #14

    StatusX

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    So what you want to do is evaluate the integral:

    [tex] \int_0^\infty \int_0^\infty \lambda^2 e^{-\lambda(u+v)} du dv[/tex]

    by looking at the integrand in (x,y) space instead of (u,v) space, where x=u+v and y=uv. Is that correct?

    If so, you're correct in the way you've approached the problem. You want to find the jacobian to transform du dv to dx dy, and then find the region in (x,y) space corresponding to the 1st quadrant in (u,v) space.

    I think the problem is that x and y, as defined, do not uniquely determine a point (u,v). Because the defining equation are symmetric with respect to u and v, if (u,v) corresponds to the point (x,y), so does (v,u). In other words, x and y are bad coordinates.

    That being said, this isn't a huge problem. As long as you restrict to one side of the line u=v (eg, the set of points (u,v) with [itex]0 \leq u \leq v [/itex]), x and y are still good coordinates. Luckily, the integral will be the same on each side of the line, so you can just compute the integral over one side and then double it.

    The key point is to be careful with square roots. For example, [itex]x^2-4y = (u-v)^2[/itex] is true, but [itex]\sqrt{x^2-4y}= u-v[/itex] is not true unless u>v, and if v>u this becomes [itex]\sqrt{x^2-4y}= v-u [/itex]. Once you pick a side, you can pick the sign of the square root. In fact, by implicitly taking the positive root, you already integrated over the region u>v>0, and so your result of 1/2 makes perfect sense: you only integrated over half the first quadrant.
     
    Last edited: Sep 4, 2006
  16. Sep 5, 2006 #15
    statusX,
    thank you very much for the assitance.

    I finally was able to get the correct form:

    it should be:
    [tex]f(x,y) = 2 \lambda^{2} e^{-\lambda x} (x^{2} - 4y)^{-\frac{1}{2}}[/tex]

    correct?
     
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