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Proof of ratio distribution normal and uniform

  1. Jul 21, 2012 #1
    Let X be standard normal and Y uniform [0,1]
    find the distribution of Z=X/Y.

    i think i managed to deal with this problem up to the integral stage:

    P(X/Y< t ) = P(X<Y*t )

    thought of :
    1.
    [itex]F_{z}(t)[/itex] = [itex]\int^{1}_{0}[/itex][itex]\int^{yt}_{0}f(x,y)dxdy[/itex]

    2.

    [itex]F_{z}(t)[/itex] = [itex]\int^{yt}_{0}f(x)dx[/itex]

    maybe is neither one of them :) , but '1' is the first i thought of...
    *saw in Wikipedia that this problem define the slash distribution, but without proof.
     
  2. jcsd
  3. Jul 21, 2012 #2

    Ray Vickson

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    What on Earth is z (you wrote Fz)? There is no 'z' anywhere in your integrals. Anyway, [itex]P(X < tY) = \int_0^1 \int_{-\infty}^{yt} f(x,y) \, dx \: dy[/itex]. But, it is more informative to get the density [itex]g(t) = \frac{d}{dt} P(X < Yt).[/itex]

    RGV
     
  4. Jul 21, 2012 #3
    the assignment is to find the distribution function, so i suppose to integrate the normal density function (X) with this boundaries ?

    X,Y are independent forgot to mention.

    for the Z and t i thought i can write it Z is the R.V and 't' is just for notation it can be also small 'z' more acceptable.
     
  5. Jul 21, 2012 #4

    uart

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    You could have used small z for t, but I understood what you wrote, [itex]F_Z(t)[/itex] is the cumulative density function for random variable "Z" as a function of parameter "t".

    The integral you had in "1." is correct, but as Ray said, what you really want is the density [itex]\frac{d}{dt} F_Z(t) [/itex]. This is important, because at some point in the derivation you should find you can interchange the order of the integral and the derivative, leaving it a very easy integral.

    TBH I hadn't heard of the 'slash' distribution before. I just goggled it however and it is precisely the distribution that I previously obtained. So it's a yes on that one. :smile:
     
  6. Jul 21, 2012 #5
    sorry but my English is limited , didn't understand this line.
    should i take the derivative of f(x,y) ? and then ?
     
  7. Jul 21, 2012 #6

    uart

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    Maybe the term derivation was confusing there. Read it as,

    This is important, because at some point in the simplification you should find you can interchange the order of the integral and the derivative, leaving it a very easy integral.

    You want to find,
    [tex]\frac{d}{dt} \left\{ \int^{1}_{0} \int^{yt}_{-\infty}f(x,y)dxdy \right\}[/tex]
     
    Last edited: Jul 21, 2012
  8. Jul 21, 2012 #7
    EDIT:
    changing the order of integration in the double integral, you should get:
    [tex]
    \int_{0}^{t}{dx \, \int_{\frac{x}{t}}^{1}{f(x, y) \, dy}}
    [/tex]
    This is a parametric integral with a variable upper bound:
    [tex]
    I(t) = \int_{0}^{t}{dx \, G(x; t)}, \ G(x; t) = \int_{\frac{x}{t}}^{1}{f(x, y) \, dy}
    [/tex]
    To find the total derivative w.r.t. t, we apply the formula:
    [tex]
    I'(t) = G(x = t; t) \cdot 1 + \int_{0}^{1}{dx \, \frac{\partial G}{\partial t}(x, t)}
    [/tex]
    But,
    [tex]
    G(x = t; t) = \int_{1}^{1}{f(t, y) \, dy} = 0
    [/tex]
    and
    [tex]
    \frac{\partial G(x; t)}{\partial t} = f(x, y = \frac{x}{t}) \cdot (-1) \cdot \left(-\frac{x}{t^2}\right) = \frac{x}{t^2} \, f(x; \frac{x}{t})
    [/tex]
    Then, the derivative is:
    [tex]
    I'(t) = \frac{1}{t^2} \, \int_{0}^{1}{dx \, f(x, \frac{x}{t})}
    [/tex]
     
    Last edited: Jul 21, 2012
  9. Jul 21, 2012 #8
    sorry for being infant , but

    [itex]\frac{d}{dt}[/itex][itex]\int^{1}_{0}\Phi(yt)-\Phi(0)dy[/itex]

    is that close to what you meant ?
     
  10. Jul 21, 2012 #9

    Ray Vickson

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    No, no, no. The x integral goes from -infinity to yt, because X ranges over the whole real line.

    RGV
     
  11. Jul 21, 2012 #10
    did wrote to me ?

    so :

    [itex]\frac{d}{dt}[/itex][itex]\int^{1}_{0}\Phi(yt)dy[/itex] ?

    ahhhh
     
  12. Jul 21, 2012 #11

    uart

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    Yes of course, that is what I had Ray. But then I just cut and pasted the OP to save time and missed their error :redface:.

    Indiscriminate cut and pasted corrected. :)
     
  13. Jul 21, 2012 #12

    uart

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    Reverse the order of the derivative and outer integral on the double integral.

    I wrote f(x,y) = f(x) f(y) (using the independence property). Personally I like to write the inner integral explicitly using a notation like "normal_cdf(yt)", and then interchange the order of the outer integral and the derivative, but as Dickfore points out, you don't really have to do it that way.
     
    Last edited: Jul 21, 2012
  14. Jul 21, 2012 #13
    didn't quiet understood what Dickfore did , so tried changing variables in the attached file
    seems to me pretty close but something is missing(besides the boundaries of the p.d.f) or i didn't calculate right ,
     

    Attached Files:

  15. Jul 21, 2012 #14

    uart

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    Yeah, I did it a slightly easier way too.

    Starting with the expression we had before (the one with the double integral):
    [tex]\frac{d}{dt} \left\{ \int^{1}_{0} \int^{yt}_{-\infty}f(x) f(y) dxdy \right\}[/tex]

    Substitute f(y) = 1, since it's the uniform distribution. Then just reverse the order of the outer integral and the derivative so that you can apply the fundamental theorem of calculus directly to the inner derivative-integral combination.

    [tex] \int^{1}_{0} \, \frac{d}{dt} \left\{ \int^{yt}_{-\infty}f(x) dx \right\}\,\, dy[/tex]
    Hint: [itex]y[/itex] is a constant with respect to the derivative, so you can replace [itex]\frac{d}{dt}[/itex] with [itex]y \, \frac{d}{d(yt)}[/itex].

    You end up with a very easy integral.
     
    Last edited: Jul 21, 2012
  16. Jul 22, 2012 #15
    think i managed to do it, in two ways , could you confirm please
     

    Attached Files:

    • Q_4.pdf
      Q_4.pdf
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    Last edited: Jul 22, 2012
  17. Jul 22, 2012 #16

    chiro

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    Hey wuid.

    Just as a check, you should get something like R and simulate say a hundred thousands samples from a normal and a uniform and see if you get roughly the same PDF that you yourself calculated: this is one of the better ways of verifying analytic results when it comes to distributions and probabilistic results.
     
  18. Jul 22, 2012 #17

    uart

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    Yes the second way is how I did it. You made a mistake in the last line. When you substituted [itex]z=y^2/2[/itex] you should have ended up with an exponential term of [itex]e^{-t^2 z}[/itex] instead of the [itex]e^{-t^2 z/2}[/itex] which you have written.
     
  19. Jul 22, 2012 #18
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  20. Jul 22, 2012 #19
    what you mean get something like R ?

    by simulating what you suggest (using Matlab)

    1. generating number for X using x=randn(10^6,1);
    2. generating number for Y using y=rand(10^6,1);
    3. dividing X by Y
    4. what should i do next , where to save the the results ?
     
  21. Jul 22, 2012 #20
    o.k! yes i know i did some silly mistakes was doing this h.w in work...

    what do you think on the first page , the method of changing variables ?
     
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