Simple Binominal Distribution Question

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,

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

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

Am I on the right track...?

Re: Simple Binominal Distribution Question

Quote:

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,

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

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

In the first summation do not us x twice.

$\displaystyle P(X\leq x)=\sum\limits_{y = 0}^x {\binom{n}{y}\theta^y(1-\theta)^{n-y}} $

Now note that $\displaystyle \sum\limits_{y = 0}^n {\binom{n}{y}\theta^y(1-\theta)^{n-y}}=1 $

Re: Simple Binominal Distribution Question

Plato: thanks.

I understand that $\displaystyle \sum^{n}_{x=0}p(x)=1$.

I also get that the sum of the two sequences must equal 1 because, due to their respective distributions, the two events, $\displaystyle P(X\leq x)$ and $\displaystyle P(Y\geq n-x)$ partition the sample space, so that $\displaystyle P(X\leq x)+P(Y\geq n-x)=1$.

Since $\displaystyle X$ is the probability of success and $\displaystyle Y$ is the probability of failure of the same experiment, the probability that there are $\displaystyle x$ successes in $\displaystyle n$ Bernouli trials must equal the probability that there are $\displaystyle n-x$ failures in the same $\displaystyle n$ Bernouli trials.

Probably I'm being a bit dense, but I don't understanding how to *prove* it.

[I am a bit confused by the notation as well. If I change the notation per your suggestion, then aren't I expressing the sum $\displaystyle P(X=0)+P(X=1)+...+P(X=y)$? Does it make sense to write that?]

Re: Simple Binominal Distribution Question

Quote:

Originally Posted by

**TimM** I understand that $\displaystyle \sum^{n}_{x=0}p(x)=1$.

I also get that the sum of the two sequences must equal 1 because, due to their respective distributions, the two events, $\displaystyle P(X\leq x)$ and $\displaystyle P(Y\geq n-x)$ partition the sample space, so that $\displaystyle P(X\leq x)+P(Y\geq n-x)=1$.

I don't think you can show that.

You do want $\displaystyle P(X\leq x)=P(Y\geq n-x)$ do you not

$\displaystyle \sum\limits_{k = 0}^x \binom{n}{k}\theta^k(1-\theta)^{n-k}=\sum\limits_{k = n-x}^n \binom{n}{k}(1-\theta)^k \cdot\theta^{n-k} $

Now there will be some overlapping there.

You know that $\displaystyle \binom{N}{k}=\binom{N}{N-k}$

This is a post script.

You have two distinct cases.

If $\displaystyle x<n-x$ the proof is straight forward.

In the first sum when $\displaystyle k=3$ and $\displaystyle k=N-3$ the second sum then the two corresponding terms are equal.

However, if $\displaystyle N-x\le x$ you must take special care.

You have overlapping terms is each sum.

Re: Simple Binominal Distribution Question

Plato: Thanks again for your help with this. IN work now, but will have a look at this and try to post my answer tonight.