How Does the Distribution of Heads in Coin Flipping Change with Large N?

[chingcx]

Homework Statement

Flip N fair coins. The distribution for different numbers of heads and tails should be peaked at N/2. When N is very large, the peak will be very high. Let x = N(head)-N/2, required to find an expression for this distribution near the peak, i.e. x<<N.

Homework Equations

Strling Approx
ln(1+x)=x for small x

The Attempt at a Solution

clearly incorrect because square of (N/2)! must be smaller than the product of (N/2+x)! and (N/2-x)!, but I can't find where goes wrong.

 

[weejee]

1. Going over to the 5th line, you omitted -x \ln{(\frac{N}{2}+x)} and x \ln{(\frac{N}{2}-x)}. These give rise to additional corrections.
-x \ln{(\frac{N}{2}+x)}+x\ln{(\frac{N}{2}-x)}=-x\ln{\left(\frac{1+2x/N}{1-2x/N}\right)}\approx -x\ln{(1+4x/N)}\approx -4x^2 /N

2. In the 7th line, you need to be more careful when expanding to the 2nd order. You should proceed as below.
\ln \left(\frac{N/2}{N/2+x}\right) = \ln \left(\frac{1}{1+2x/N}\right) \approx \ln \left( 1-\frac{2x}{N}+\frac{4x^2}{N^2} \right) \approx -\frac{2x}{N}+\frac{4x^2}{N^2}-\frac{1}{2} {\left(-\frac{2x}{N}+\frac{4x^2}{N^2} \right)}^{2}

\approx -\frac{2x}{N}+\frac{4x^2}{N^2}-\frac{1}{2} \frac{4x^2}{N^2} = -\frac{2x}{N}+\frac{2x^2}{N^2}

and similarly

\ln \left(\frac{N/2}{N/2-x}\right) \approx \frac{2x}{N}+\frac{2x^2}{N^2}


Then, the final answer becomes -2x^2 /N rather than 4x^2 /N.

 

[chingcx]

1. Going over to the 5th line, you omitted -x \ln{(\frac{N}{2}+x)} and x \ln{(\frac{N}{2}-x)}. These give rise to additional corrections.
-x \ln{(\frac{N}{2}+x)}+x\ln{(\frac{N}{2}-x)}=-x\ln{\left(\frac{1+2x/N}{1-2x/N}\right)}\approx -x\ln{(1+4x/N)}\approx -4x^2 /N

2. In the 7th line, you need to be more careful when expanding to the 2nd order. You should proceed as below.
\ln \left(\frac{N/2}{N/2+x}\right) = \ln \left(\frac{1}{1+2x/N}\right) \approx \ln \left( 1-\frac{2x}{N}+\frac{4x^2}{N^2} \right) \approx -\frac{2x}{N}+\frac{4x^2}{N^2}-\frac{1}{2} {\left(-\frac{2x}{N}+\frac{4x^2}{N^2} \right)}^{2}

\approx -\frac{2x}{N}+\frac{4x^2}{N^2}-\frac{1}{2} \frac{4x^2}{N^2} = -\frac{2x}{N}+\frac{2x^2}{N^2}

and similarly

\ln \left(\frac{N/2}{N/2-x}\right) \approx \frac{2x}{N}+\frac{2x^2}{N^2}


Then, the final answer becomes -2x^2 /N rather than 4x^2 /N.

Thanks, but it doesn't make sense... when N is very large, the peak should be very narrow, but from the equation, N becomes larger then the exponential decreases much slower.

 

[weejee]

You are right. The width of the peak actually increases, as sqrt(N).
Yet, if you consider (# of heads)/(# of coins), the width decreases as 1/sqrt(N).