Simple Binominal Distribution Question

Hi all,

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

Question is:

Given , prove that .

Now obviously can easily prove using formula for binomial distribution that , and on up the sequence to . But I don't know how to prove in general that the sums of the two series are such that .

If I write the two sequences as sums I get,

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

, prove that

.

Now obviously can easily prove using formula for binomial distribution that

, and on up the sequence to

. But I don't know how to prove in general that the sums of the two series are such that

.

If I write the two sequences as sums I get,

In the first summation do not us x twice.

Now note that

Re: Simple Binominal Distribution Question

Plato: thanks.

I understand that .

I also get that the sum of the two sequences must equal 1 because, due to their respective distributions, the two events, and partition the sample space, so that .

Since is the probability of success and is the probability of failure of the same experiment, the probability that there are successes in Bernouli trials must equal the probability that there are failures in the same 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 ? Does it make sense to write that?]

Re: Simple Binominal Distribution Question

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.