How Does Stirling's Approximation Simplify Poisson to Normal Approximation?

• Mathsboi
In summary, the conversation discusses the use of Stirling's approximation in the context of the Poisson distribution, where n is large and r is small. The approximation is used to simplify the equation n(n-1)(n-2)...(n-r+1) and the discussion delves into the use of logarithms in finding the approximation. The conversation ends with a request for clarification on the correctness of the approximation.
Mathsboi
I read this in a book (it was stats and about poisson approx to normal)

$$n(n-1)(n-2) \cdots (n-r+1) = \frac{n!}{(n-r)!} \approx n^r$$
Stating that "Stirling's approximation" had been used.
So I looked the up and found:

$$\ln n! \approx n\ln n - n\$$

In the poisson distribution n is very large and $$r$$ is very small compared to $$n$$ so all the terms in the given equation approximate to $$n$$... This gives me my $$\approx n^r$$

But I just wondered where the Stirling equation comes into it...

$$\ln (\frac{n!}{(n-r)!}) = \ln(n!) - \ln((n-r)!) \$$
$$\approx n\ln n - n - \left[ (n-r)\ln((n-r)) - (n-r) \right]\$$
$$\approx n\ln n - n - (n-r)\ln((n-r)) + n - r \$$
$$\approx n\ln n - (n-r)\ln((n-r)) -r \$$
...
That's as far as I got...

$$\approx \ln (n^n) - \ln((n-r)^{(r-n)}) -r \$$

Unless taking logs, instead of to base e, to base n...

$$\approx Log_n (n^n) - Log_n ((n-r)^{(r-n)}) -r \$$
Then...
$$Log_n (n^n) - Log_n ((n-r)^{(r-n)}) = r\$$
$$n^n - (n-r)^{(n-r)} = n^r\$$

^ not sure if that's correct though

Can anyone help?

This is "number theory", not "algebra and linear algebra" so I am moving it.

$$\ln(n-r) \approx ln(n) - r/n$$

$$n\ln n - (n-r)\ln((n-r)) -r \approx n \ln n - n \ln n + n r/n + r (\ln n - r/n) - r\approx r \ln n - r^2/n$$

1. What is Stirling's approximation proof?

Stirling's approximation proof is a mathematical theorem that approximates the value of a factorial function as the input number grows larger. It was discovered by Scottish mathematician James Stirling in the 18th century.

2. How does Stirling's approximation proof work?

Stirling's approximation proof is based on the central limit theorem, which states that the sum of a large number of independent and identically distributed random variables will be approximately normally distributed. By applying this theorem to the factorial function, Stirling's approximation can be derived.

3. What is the significance of Stirling's approximation proof?

Stirling's approximation proof has many practical applications in mathematics and science. It can be used to simplify calculations involving factorials, which often arise in probability, statistics, and combinatorics. It is also used in the analysis of algorithms to estimate the running time of algorithms that involve factorials.

4. How accurate is Stirling's approximation proof?

Stirling's approximation is asymptotically accurate, meaning that it becomes increasingly accurate as the input number grows larger. For small values, it may not provide a very accurate estimate. However, as the input number approaches infinity, the approximation becomes more and more accurate.

5. Are there any limitations to Stirling's approximation proof?

While Stirling's approximation is a powerful tool for estimating factorials, it is not exact and has certain limitations. For very large values, the approximation may still introduce significant errors. Additionally, the approximation only works for positive real numbers, so it cannot be used for negative or complex numbers.

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