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Simplifying this expression for large values of D

  1. Aug 28, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that:
    p(r'+e) [tex]\cong[/tex] p(r')*exp( -3e2 / 2s2 ) for large D

    given that:
    p(r) = rD-1 * exp( -r2 / 2s2 )
    r' = sqrt(D) * s

    2. Relevant equations
    This arises out of a D-dimensional gaussian distribution and r' is the radius at which the probability is maximized.

    3. The attempt at a solution
    I've gotten to the point:

    p(r'+e) [tex]\propto[/tex] exp( -(r'+e)2 / 2s2 + (D-1)ln(r'+e) )

    And the solution manual says that using the Taylor expansion for ln, will help everything reduce down, but all I get is a mess. What really baffles me is where the dependance on D went.

    Thanks very much for your help
    -Patrick
     
  2. jcsd
  3. Aug 28, 2009 #2
    According to Wolfram Alpha, this is true for large s but not for large D.
     
  4. Aug 28, 2009 #3
    Thanks for helping foxjwill. I'm almost certain that the result is correct though. There's an probabilistic interpretation of it, and it requires that this holds true for large D not s.

    Wolfram Alpha, though, seems like an extremely useful tool. I haven't heard of it before. May I ask exactly what you typed into the query box?

    Thanks
    -Patrick
     
  5. Aug 28, 2009 #4

    gabbagabbahey

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    For large [itex]D[/itex], what can you say about the size of [itex]\frac{e}{r'}[/itex]? Use that to Taylor expand the logarithm:

    [tex]\ln(r'+e)=\ln(r'(1+\frac{e}{r'}))=\ln(r')+\ln(1+\frac{e}{r'})[/tex]
     
  6. Aug 28, 2009 #5
    First, I typed in

    Code (Text):
    (exp(-3epsilon^2/(2s^2)) *r^(D-1)*exp(-r^2/(2s^2)) /  (r+epsilon)^(D-1)*exp(-(r+epsilon)^2/(2s^2)) )
    Then I copied the result given in the Alternate form box:
    Code (Text):
    r^(D-1) (r+epsilon)^(1-D) e^(-(r^2+r epsilon+2 epsilon^2)/s^2)
    Finally, I pasted that into the input box, replaced each instance of r with "(s*sqrt(D))", then prepended "limit" and appended "as D->infinity":
    Code (Text):
     limit (s*sqrt(D))^(D-1) ((s*sqrt(D))+epsilon)^(1-D) e^(-((s*sqrt(D))^2+(s*sqrt(D)) epsilon+2 epsilon^2)/s^2) as D->infinity
    Wolfram Alpha says this equals 0.
     
  7. Aug 28, 2009 #6

    Dick

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    Something in this isn't quite right. If you work out the expansion around sqrt(D)*s you get a linear term in e. That's because the maximum value of your p(r) is at sqrt(D-1)*s NOT sqrt(D)*s, isn't it? Why don't you expand around the maximum? And I also don't get exp(-3*e^2/s^2) for the correction factor several different ways. I get just exp(-e^2/s^2).
     
    Last edited: Aug 28, 2009
  8. Aug 29, 2009 #7
    Thank you very much for your help.
    I haven't gotten then answer actually. It's a textbook question that asks me to prove the statement. (Exercise 1.20 from Pattern Recognition and Machine Learning by Bishop).

    You're right. The maximum is actually at sqrt(D-1)*s, which the solution manual says can be approximated to sqrt(D)*s for large D. I will try expanding around sqrt(D-1)*s and see if I can get it.

    Thank you for helping me.
    -Patrick
     
  9. Aug 29, 2009 #8
    Hi Dick,
    Would you mind showing me how to arrived at exp(-e^2/s^2)?
    I've tried expanding around r = sqrt(D-1)*s
    and by noticing that (D-1)*s^2 = r^2

    I simplified the equation to
    p(r+e) approx. = exp( -(r+e)^2 + r^2*[ r+e-1 - ((r+e-1)^2)/2 ] )

    By the way: is there a general rule for deciding when some approximations are valid and when some aren't?
    eg. for large D: can I say e/r -> 0? or that r+e -> r?
     
  10. Aug 29, 2009 #9

    Dick

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    Where did the '(r+e-1)' stuff come from? Start with your expression in the first post exp(-(r'+e)^2/(2s^2)+(D-1)ln(r'+e)). Expand the (r'+e)^2, expand ln(r'+e) using fabbagabbahey's suggestion. Keep the linear and quadratic terms in the ln series. Collect like powers of e. What do you get?
     
  11. Aug 30, 2009 #10
    Hi Dick,
    Thank you very much for your help and patience.

    Here's my work so far, I have an extra term at the end that I cannot get rid of

    numerator:
    [tex]-(r+e)^2+2(D-1)s^2(ln(r)+ln(1+e/r))[/tex]
    [tex]=-(r+e)^2 + 2r^2(ln(r)+ln(1+e/r))[/tex]
    [tex]=-r^2-2re-e^2+2r^2(r-1-(r-1)^2/2+e/r-(e/r)^2/2)[/tex]
    [tex]=-r^2-2re-e^2+2r^2(r-1)(1-(r-1)/2)+2re-e^2[/tex]
    [tex]=-r^2-2e^2+r^2(r-1)(1-r)[/tex]

    Thanks
    -Patrick
     
  12. Aug 30, 2009 #11

    Dick

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    You do not replace ln(r) with (r-1)-(r-1)^2/2. You can only series expand small quantities. r (or rather you should be writing r') is not necessarily small. Just leave that term as ln(r'). Remember you want f(r')*exp(...). The ln(r') is part of the f(r').
     
  13. Aug 30, 2009 #12
    Of course I can't expand around a large number. I should know that. Thanks again Dick.

    I left ln(r') as it is, and finally I'm left with:

    [tex]-r'^2+2r'^2ln(r') - 2e^2[/tex] in the numerator.

    The first two terms are part of f(r'), and the last term is the correction term. I get a different coefficient (-2) in front of e^2 though. It's different from both your answer and the textbook answer. I'll keep on looking for what I did wrong.

    Thank you very much for your help.
    -Patrick
     
  14. Aug 30, 2009 #13

    Dick

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    You multiplied the quantity you are expanding by 2*s^2. You should divide it back out and get f(r')exp(-e^2/s^2).
     
  15. Aug 30, 2009 #14
    Right. I forgot I did that.
    Thanks again Dick.
     
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