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Math Help - Evaluating a couple of limits

  1. #1
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    Evaluating a couple of limits

    I wonder if someone could clarify these limits for me. They arise in the derivation of the poisson distribution from the binomial distribution as \lim_{ n \to +\infty} and \lim_{ p \to 0}

    The first is

    [1- (\lambda/n)]^{n} \rightarrow  \mathrm{e}^{-\lambda}

    as n \to +\infty.

    I don't see how this works. If anything I would have thought that

    (1- (\lambda/n))^{n} = \mathrm{e}^{n \cdot \log{(1-\lambda/n)}} \to \mathrm{e}^{0} = 1

    as n \to +\infty.


    The second is that

    \frac{n!}{(n-k)!(n-\lambda)^k = \frac{n(n-1) \ldots (n-k+1)}{(n - \lambda) (n-\lambda) \ldots (n - \lambda)}

    is a "quantity that tends to 1 as n \to +\infty (since \lambda remains constant)."

    What precisely is the role of \lambda? Is the critical condition not that the numerator and denominator grow at the same rate, and if so, how can this be demonstrated?

    Thanks in advance. MD
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  2. #2
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    Re: Evaluating a couple of limits

    Quote Originally Posted by Mathsdog View Post
    If anything I would have thought that

    (1- (\lambda/n))^{n} = \mathrm{e}^{n \cdot \log{(1-\lambda/n)}} \to \mathrm{e}^{0} = 1

    as n \to +\infty.
    \lim_{n\to\infty} n\ln(1-\lambda/n)\ne0. Rather, using the Maclaurin series for ln, n\ln(1-\lambda/n)=n(-\lambda/n+o(1/n))\to-\lambda as n\to\infty.

    Quote Originally Posted by Mathsdog View Post
    \frac{n!}{(n-k)!(n-\lambda)^k} = \frac{n(n-1) \ldots (n-k+1)}{(n - \lambda) (n-\lambda) \ldots (n - \lambda)}

    is a "quantity that tends to 1 as n \to +\infty (since \lambda remains constant)."
    If k is constant, then \lim_{n\to\infty}\frac{n(n-1) \dots (n-k+1)}{(n - \lambda) (n-\lambda) \dots (n - \lambda)} = \lim_{n\to\infty}\frac{n}{n-\lambda} \lim_{n\to\infty}\frac{n-1}{n-\lambda}\dots\lim_{n\to\infty}\frac{n-k+1}{n-\lambda}. Each of these limits is 1. Indeed, \frac{n-i}{n-\lambda}= \frac{n-\lambda+\lambda-i}{n-\lambda}= 1+\frac{\lambda-i}{n-\lambda}\to1 as n\to\infty.
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    Re: Evaluating a couple of limits

    Quote Originally Posted by Mathsdog View Post
    I wonder if someone could clarify these limits for me. They arise in the derivation of the poisson distribution from the binomial distribution as \lim_{ n \to +\infty} and \lim_{ p \to 0}

    The first is

    [1- (\lambda/n)]^{n} \rightarrow  \mathrm{e}^{-\lambda}

    as n \to +\infty.

    I don't see how this works. If anything I would have thought that

    (1- (\lambda/n))^{n} = \mathrm{e}^{n \cdot \log{(1-\lambda/n)}} \to \mathrm{e}^{0} = 1

    as n \to +\infty.


    The second is that

    \frac{n!}{(n-k)!(n-\lambda)^k = \frac{n(n-1) \ldots (n-k+1)}{(n - \lambda) (n-\lambda) \ldots (n - \lambda)}

    is a "quantity that tends to 1 as n \to +\infty (since \lambda remains constant)."

    What precisely is the role of \lambda? Is the critical condition not that the numerator and denominator grow at the same rate, and if so, how can this be demonstrated?

    Thanks in advance. MD
    \displaystyle \begin{align*} \lim_{n \to \infty}\left( 1 - \frac{ \lambda}{n}\right)^n &= \lim_{n \to \infty}e^{\ln{\left[\left(1 - \frac{\lambda}{n}\right)^n\right]}} \\ &= \lim_{n \to \infty}e^{ n\ln{ \left( 1 - \frac{\lambda}{n} \right) } } \\ &= \lim_{n \to \infty} e^{ \frac{ \ln{ \left( 1 - \frac{\lambda}{n} \right) } }{ \frac{1}{n} } } \\ &= e^{\lim_{n \to \infty}\frac{\ln{ \left( 1 - \frac{\lambda}{n} \right) }}{\frac{1}{n}}} \\ &= e^{ \lim_{n \to \infty}\left(\frac{\frac{1}{n - \lambda} - \frac{1}{n}}{-\frac{1}{n^2}}\right) } \textrm{ by L'Hospital's Rule} \\ &= e^{ \lim_{n \to \infty}\left[ \frac{\frac{\lambda}{n(n + \lambda)}}{-\frac{1}{n^2}} \right] } \\ &= e^{ \lim_{n \to \infty}\left(-\frac{n^2\lambda}{n^2 + n\lambda}\right) } \\ &= e^{\lim_{n \to \infty}\left(-\frac{\lambda}{1 + \frac{\lambda}{n}}\right)} \\ &= e^{-\lambda} \end{align*}
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