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Proof of Normal approximation to Poisson.

  1. Dec 5, 2009 #1
    I have been looking for a proof of the fact that for a large parameter lambda, the Poisson distribution tends to a Normal distribution. I know the classic proof using the Central Limit Theorem, but I need a simpler one using just limits and the corresponding probability density functions. I was told this was really easy using Stirling's approximation:

    n! ~ sqrt(2*pi*n) * (n/e)^n

    but I just don't see it. Anyone knows this proof?
     
  2. jcsd
  3. May 4, 2011 #2
    First you take the natural logarithm to the Poisson distribution and then apply Stirlings approximation. Then define a new variable

    [tex]y=x-\mu[/tex]

    and assume that y is much smaller than [tex]\mu[/tex]
    By doing this you will end up with a term

    [tex]\ln\left(1+\frac{y}{\mu}\right) [/tex]

    which can be approximated by looking at the Maclaurin series

    [tex]\ln\left(1+\frac{y}{\mu}\right) \approx \frac{y}{\mu} - \frac{y^{2}}{2\mu^{2}}. [/tex]

    Now any term with a power of [tex]\mu[/tex] greater than 2 in the denominator may be approximated as zero due to the assumption that y is much smaller than [tex]\mu[/tex]. When the algebra is done you just takt the exponential function on both sides and you end up with a normal distribution with mean and variance [tex]\mu[/tex].
     
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