How to evaluate a truncated binomial series in the infinity limit

Question

For a standard binomial series, if it is truncated in the following way: $$\sum_{k=0}^{n'}{n \choose k}(1-x)^{n-k}x^k$$ with $n'

Numerical calculations seem to suggest that it will give a step function. How to show this analytically?

Answer 1

This answer assumes some knowledge in probability. Let

$$\tag{1}f_{n'}(x):=\sum_{k=0}^{n'}\binom{n}{k}(1-x)^{n-k}x^k.$$

You desire an estimate of the ratio $\dfrac{f_{n/2}(x)}{f_{n}(x)}=f_{n/2}(x)$

(because $f_n(x)=1$ by binomial formula.)

This is the same as looking for probability $P(X

By the Central Limit Theorem (https://en.wikipedia.org/wiki/Central_limit_theorem)), this discrete distribution can be approximated by the normal distribution $N(mean=nx,\sigma^2=nx(1-x)).$

Thus, we have the analytical approximation:

$$\tag{2}\dfrac{1}{\sqrt{2 \pi n x (1-x)}}\int_0^{n/2}\exp\left(-\dfrac{(u-nx)^2}{2nx(1-x)}\right) du$$

(note that in the integral, the variable is $u$)

Instead of dwelving (as I had made at first) into complicated approximations, we are going to use the following simpler, more intuitive, argument.

In fact, it is well known that the "mass" of a $N(m,\sigma^2)$ is almost concentrated on $[m-3\sigma,m+3\sigma] \ $ (precisely at 99.7 %, see (https://www.encyclopediaofmath.org/index.php/Three-sigma_rule)).

Consider the figure below, for $n=100$. Let

• $u=f_1(x):=nx-3\sqrt{nx(1-x)}$ (blue curve),

• $u=f_2(x):=nx+3\sqrt{nx(1-x)}$ (red curve), and

• $u=g(x)=n/2 \ $ (green line).

where the blue and red curves are classical "confidence curves".

These curves help us to situate where is the so-called "mass". For example, for $x=0.2$, all the mass is concentrated on interval $[15,25]$. In this case, the mass is included in the interval of integration $[0,50]$, thus the value of the integral is $\approx 1.$

Depending on the value of $x$, the "mass" of the normal approximation, which is between the blue curve and the red curve, will be, either inside the interval of integration (thus the result is 1), on its fringes (integral between 0 and 1), or outside (nul integral). More precisely in this case ($n=100$) ,

• when $x<0.44$, the mass is inside the integration domain.

• when $0.44green curve, i.e., there is a progressive shrinking of the mass situated into the interval of integration progressively to 0.

• when $x>0.56$, the "mass" of the equivalent gaussian is too "at the right": the integral is 0.

Remark 1 : We have explained the rather steep decreasing of the probability, that "jumps" from 1 to 0 in the rather short interval $[0.44,0.56]$ for $n=100$. With larger values of $n$, the jump takes place in a progressively narrower space.

Remark 2 : What I just write is in the same "spirit" as Ian's answer.

Remark 3 : It would have been possible to approximate the $B(n,x)$ distribution by a Poisson distribution $P(\lambda)$ with $\lambda \approx np$, but it is only valid under the restrictive hypothesis that $\lambda\approx np <30$ (classical hypothesis).

Answer 2

You are writing $P(X_n \leq n')$ where $X_n$ is distributed as Bin(n,x). If $n'=cn$ for some $00$.

But now there exists $\epsilon$ such that $[0,c]$ is disjoint from $(x-\epsilon,x+\epsilon)$ whenever $cx$. Therefore we may use the above WLLN calculation to infer that:

• if $c
• if $c>x$ then $\lim_{n \to \infty} P(P_n \leq c)=1$.

We can additionally apply the central limit theorem to find that if $c=x$ then the limiting probability is $1/2$.

So you are right that in the case you investigated you should see a step function, where the step is at $x=1/2$.

You need better estimates in order to make all of this quantitative (i.e. to say how big $n$ needs to be for these limits to make valid approximations).