Normal approximation to the binomial

I know that the normal can be used as an approximation to the binomial. I was looking for a derivation of this, and I found it via google in a math forum. Doctor Anthony begins:

Derivation of the Normal distribution from the Binomial distribution --------------------------------------------------------------------- Let a variate take values 0, k, 2k, 3k, ..., nk with probabilities given by successive terms of (q + p)^n.

What's with the k? Well, we're eventually going to want non-integer terms. The expansion of (q + p)ⁿ is familiar:

qn + nqn-1 p + n(n-1)/2 qn-2 p2 + ...

The i^th term of the expansion is C(n,i).

Then the mean m = npk and the variance s^2 = npqk^2

OK. Notice use of the multiplication rule for variance from the other day.

Suppose: y = probability of occurrence of rk = C(n,r) p^r q^(n-r) Also let: y' = probability of occurrence (r+1)k = C(n,r+1)p^(r+1) q^(n-r-1) Then: y' - y = C(n,r+1)p^(r+1) q^(n-r-1) - C(n,r)p^r q^(n-r) n!p^r q^(n-r-1) = ---------------[(n-r)p - (r+1)q] (r+1)! (n-r)!

Hmm... I know that

y = [n! / r! (n-r)!] pr qn-r y' = [n! / (r+1)! (n-r-1)!] pr+1 qn-r-1 y' - y = ?

Lucie, you got some factoring to do. Let's deal with q and p first.

The left term has q^n-r-1 and the right term has q^n-r, so we can factor out
q^n-r-1, leaving a factor of q on the right-hand term in the brackets.

Similarly we can factor out p^r from both sides leaving a factor of p on the left.

The combination expressions expand as shown above. We can factor out n! from both sides. We can factor out 1/(r+1)! from both sides, if we first multiply top and bottom of the right-hand term by (r+1), leaving (r+1) on the top.

Similarly, we can factor out (n-r)! from both sides, if we first multiply top and bottom of the left-hand term by (n-r), leaving (n-r) behind on the top. So everything checks out so far. Next, he wants to divide by y:

And: y' - y 1 1 ------ = ------[np - r(p+q) - q] = ------[np - r - q] y (r+1)q (r+1)q (Equation 1)

Hmm...again. We're dividing the expression we had above by y.

y = [n! / r! (n-r)!] pr qn-r

We have:

n!pr qn-r-1 y' - y = ---------------[(n-r)p - (r+1)q] (r+1)! (n-r)!

So both n! and (n-r)! terms cancel. We also cancel r!, leaving a factor of (r+1) on the bottom. The p^r cancels, and the q^n-r also cancels leaves a factor of q on the bottom. So I get:

y' - y 1 ------ = ------[(n-r)p - (r+1)q] y (r+1)q

Now we have to figure out how to rearrange the term in brackets:

[(n-r)p - (r+1)q]

Expand, and then substitute for p + q = 1:

np - rp - rq - q np - r(p+q) - q np - r - q

It checks out. Doctor Anthony continues:

Let: x = rk - npk, so that x is now the variate measured from the mean. Then: r = x/k + np and r+1 = x/k + np + 1 Thus: k(r+1) = x + k + npk k^2 (r+1)q = (x + k + npk)qk

So far so good.

Multiply top and bottom of the righthand side of Equation 1 by k^2. Then: y' - y [(np-r)k - qk]k ------ = --------------- [note that (np-r)k = -x] y [x + k + npk]qk

Go back to what we had, and then multiply top and bottom by k²:

y' - y [np - r - q] k^2 ------ = ---------------- y (r+1)q k^2

Hmm... The top is fine, but on the bottom we had

(r+1) q k2

We need to get to:

[x + k + npk]qk

He says:

[note that (np-r)k = -x]

OK, so we have:

(r+1) q k2 (rk + k) q k Since: (np - r) k = -x rk = npk + x Substituting: (x + k + npk) q k

Moving on to substitute for (np-r) k = -x on top and multiplying out on the bottom yields:

(-x - kq)k = ---------------- npqk2 + (x+k)qk

Finally, we now let k = dx, so that y' - y = dy and let n ->infinity in such a way that nk^2 is finite. Equation 2 can then be written as: dy (-x - q dx)dx ---- = ---------------- y s^2 + (x+dx)q dx

The only tricky part here was that we've replaced npqk² by s².
Now he says:

As dx -> 0 this becomes: dy -x dx ---- = ------ y s^2

And we're there! If we integrate the left side we get ln(y), and the right side is
-x² / 2s²

y = A exp { -x² / 2s² }