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Stats Problem using Stirling's Approx.

  1. Nov 24, 2014 #1
    1. The problem statement, all variables and given/known data
    Let n and k be positive integers such that both n and n − k are large. Use Stirling’s formula to write as simple
    an approximation as you can for Pn,k.

    2. Relevant equations
    ## \lim_{n \rightarrow \infty} {(2 \pi)^{1/2}n^{n+1/2}e^{-n} \over n!} = 1 ##

    3. The attempt at a solution
    Since ## n - k ## is also large, I assume you apply Stirling's approximation to both the numerator and denominator of ## {n! \over (n-k)!} ##.

    ## {(2 \pi)^{1/2}n^{n+1/2}e^{-n} \over (2 \pi)^{1/2}(n-k)^{n-k+1/2}e^{-n+k}} = {n^{n+1/2} \over (n-k)^{n-k+1/2}e^{k}} ##

    I'm teaching myself statistics so I don't have anyone to turn to for help. Does this answer look correct?
     
  2. jcsd
  3. Nov 24, 2014 #2

    haruspex

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    No, you've lost some terms in the substitution. Please post your working to reach the above expression.
     
  4. Nov 24, 2014 #3
    Step 1. ## n! = (2 \pi)^{1/2}n^{n+1/2}e^{-n} ## and ## (n-k)! = (2 \pi)^{1/2}(n-k)^{n-k+1/2}e^{-n+k} ##

    Step 2. ## {n! \over (n-k)!} = {(2 \pi)^{1/2}n^{n+1/2}e^{-n} \over (2 \pi)^{1/2}(n-k)^{n-k+1/2}e^{-n+k}} ##
     
  5. Nov 24, 2014 #4

    haruspex

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    Sorry, I misread it as Cn,k. So I agree with your answer now.
    It doesn't say whether k is small compared with n. If it were you could simplify it further.
     
  6. Nov 24, 2014 #5
    If that were the case, it would be approximately 1, correct?
     
  7. Nov 24, 2014 #6

    haruspex

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    No, I get ##n^k e^{k/2n - k^2/n}##
     
  8. Nov 24, 2014 #7
    Would you mind explaining how you got that answer? I get ## {n^k \over e^k} ##
     
  9. Nov 24, 2014 #8

    haruspex

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    ##{n^{n+1/2} \over (n-k)^{n-k+1/2}e^{k}}##
    ##{n^kn^{n-k+1/2} \over (n-k)^{n-k+1/2}e^{k}}##
    ##{n^k \over (1-k/n)^{n-k+1/2}e^{k}}##
    ##{n^k \over e^{-(k/n)(n-k+1/2)}e^{k}}##
    ##{n^k \over e^{-(k/n)(-k+1/2)}}##
    ##{n^k e^{(k/n)(-k+1/2)}}##
    ##{n^k e^{-k^2/n+k/2n}}##
    I guess we can drop the k/2n.
     
  10. Nov 24, 2014 #9
    I see, very nice. Many thanks for your help earlier too.
     
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