# Evaluating a couple of limits

• Aug 2nd 2012, 03:39 AM
Mathsdog
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?

• Aug 2nd 2012, 04:02 AM
emakarov
Re: Evaluating a couple of limits
Quote:

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

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 $n\to\infty$.
• Aug 2nd 2012, 04:03 AM
Prove It
Re: Evaluating a couple of limits
Quote:

Originally Posted by Mathsdog
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?

\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*}