# Stats Problem using Stirling's Approx.

1. Nov 24, 2014

### wintermute++

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. Nov 24, 2014

### haruspex

No, you've lost some terms in the substitution. Please post your working to reach the above expression.

3. Nov 24, 2014

### wintermute++

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

4. Nov 24, 2014

### haruspex

It doesn't say whether k is small compared with n. If it were you could simplify it further.

5. Nov 24, 2014

### wintermute++

If that were the case, it would be approximately 1, correct?

6. Nov 24, 2014

### haruspex

No, I get $n^k e^{k/2n - k^2/n}$

7. Nov 24, 2014

### wintermute++

Would you mind explaining how you got that answer? I get ${n^k \over e^k}$

8. Nov 24, 2014

### haruspex

${n^{n+1/2} \over (n-k)^{n-k+1/2}e^{k}}$
${n^kn^{n-k+1/2} \over (n-k)^{n-k+1/2}e^{k}}$
${n^k \over (1-k/n)^{n-k+1/2}e^{k}}$
${n^k \over e^{-(k/n)(n-k+1/2)}e^{k}}$
${n^k \over e^{-(k/n)(-k+1/2)}}$
${n^k e^{(k/n)(-k+1/2)}}$
${n^k e^{-k^2/n+k/2n}}$
I guess we can drop the k/2n.

9. Nov 24, 2014

### wintermute++

I see, very nice. Many thanks for your help earlier too.