Simple Binominal Distribution Question

**TimM** · September 7th 2011, 12:10 PM

Hi all,

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

Question is:

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

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

If I write the two sequences as sums I get,

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

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

Am I on the right track...?

**Plato** · September 7th 2011, 12:21 PM

Originally Posted by TimM

Hi all,

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

Question is:

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

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

If I write the two sequences as sums I get,

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

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

In the first summation do not us x twice.
$P(X\leq x)=\sum\limits_{y = 0}^x {\binom{n}{y}\theta^y(1-\theta)^{n-y}}$

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

**TimM** · September 7th 2011, 01:43 PM

Plato: thanks.

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

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

Since $X$ is the probability of success and $Y$ is the probability of failure of the same experiment, the probability that there are $x$ successes in $n$ Bernouli trials must equal the probability that there are $n-x$ failures in the same $n$ 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 $P(X=0)+P(X=1)+...+P(X=y)$ ? Does it make sense to write that?]

**Plato** · September 7th 2011, 02:05 PM

Originally Posted by TimM

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

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

I don't think you can show that.
You do want $P(X\leq x)=P(Y\geq n-x)$ do you not
$\sum\limits_{k = 0}^x \binom{n}{k}\theta^k(1-\theta)^{n-k}=\sum\limits_{k = n-x}^n \binom{n}{k}(1-\theta)^k \cdot\theta^{n-k}$
Now there will be some overlapping there.
You know that $\binom{N}{k}=\binom{N}{N-k}$

This is a post script.
You have two distinct cases.
If $x<n-x$ the proof is straight forward.
In the first sum when $k=3$ and $k=N-3$ the second sum then the two corresponding terms are equal.

However, if $N-x\le x$ you must take special care.
You have overlapping terms is each sum.

**TimM** · September 8th 2011, 01:24 AM

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.

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