Originally Posted by

**TimM** Hi all,

Simple question that I've been staring at for too long to think about sensibly.

Question is:

Given $\displaystyle X \sim Bi(n, \theta ), Y \sim Bi(n, 1- \theta )$, prove that $\displaystyle P(X\leq x)=P(Y\geq n-x)$.

Now obviously can easily prove using formula for binomial distribution that $\displaystyle P(X=x)=P(Y=n-x)$, and on up the sequence to $\displaystyle P(X=0)=P(Y=n)$. But I don't know how to prove in general that the sums of the two series are such that $\displaystyle P(X\leq x)=P(Y\geq n-x)$.

If I write the two sequences as sums I get,

$\displaystyle P(X\leq x)=\sum ^x_{x=0} \left(\begin{array}{cc}n\\x\end{array}\right) \theta^x (1- \theta)^{n-x}$

$\displaystyle P(Y\geq n-x)=\sum ^n_{y=n-x} \left(\begin{array}{cc}n\\y\end{array}\right) \theta^{n-y} (1- \theta)^{y}$