Prove that $F(N/2;N,x)+F(N/2;N,1-x)=1$ where $F$ is binomial CDF

Question

I have the following claim:

$$F(N/2;N,p)+F(N/2;N,1-p)=1$$ where $F$ is a binomial CDF with exactly $N/2$ successes in $N$ total trials, and with each trial having success probability $p$.

Is it true?

Answer 1

Note that a bernoulli rv with parameter $1-p=P(X=1)$ can be re-interpreted as $p=(X=0)$ Extending this to the binomial rv with parameter $p:B(p,N)$ (i.e., sum of N i.i.d bernoulli rv's), your claim is equivalent to asking "What is the probability that a N tosses yields at most $\frac{N}{2}$ heads OR at most $\frac{N}{2}$ tails". Since the number of heads and tails adds to $N$ at most one of these statements will always be true, so $P(B(p,N)\leq \frac{N}{2} \cup (N-B(p,N))\leq \frac{N}{2})=P(B(p,N)\leq \frac{N}{2})+P((N-B(p,N))\leq \frac{N}{2})=1$ (since the two events are disjoint.

Now, by noting that $B(1-p,N)=N-B(p,N)$ (i.e., they are mirror images of each other)

$P(B(p,N)\leq \frac{N}{2})+P((N-B(p,N))\leq \frac{N}{2})=P(B(p,N)\leq \frac{N}{2})+P(B(1-p,N))\leq \frac{N}{2})=1\;\;\;\square$