# Verify and Explain Binomial R.V. Identities

1. Apr 13, 2012

### knowLittle

If X and Y are binomial random variables with respective parameters (n,p) and (n,1-p), verify and explain the following identities:
a.) P{X<=i}= P{Y>=n-i};
b.) P{X=k}= P{Y=n-k}

Relevant Equations:
P{X=i}=nCi *p^(i) *(1-p)^(n-i), where nCi is the combination of "i" picks given "n".

Distribution function
$p\left\{ x\leq i\right\} =\sum _{k=0}^{i}\left( n_{k}\right) p^{k}\left( 1-p\right) ^{n-k}$, where n_k is $(_{k}^{n})$
Solution:
I don't know, how to start the problem. Please, help.

2. Apr 13, 2012

### mathman

If you examine the terms of the binomial you will see that the terms of (n,p) and (n,1-p) are the same except they are in opposite order from each other..

3. Apr 13, 2012

### knowLittle

What do you mean by opposite order?

4. Apr 14, 2012

### knowLittle

I know that * p and 1-p * are complement of each other, but how does this help me?

5. Apr 14, 2012

### mathman

Term of index k for (n,p) (a) = term of index n-k for (n,1-p) (b). As a result the problems you are addressing involve terms from (n,p) being equal to the corresponding (opposite order) terms in (n,1-p).

(a) {n!/[k!(n-k)!]}pk(1-p)n-k

(b) {n!/[(n-k)!k!]}(1-p)n-kpk

Notice that (a) and (b) are exactly the same.

6. Apr 15, 2012

### knowLittle

From what I posted originally:
If X and Y are binomial random variables with respective parameters (n,p) and (n,1-p), verify and explain the following identities:
a.) P{X<=i}= P{Y>=n-i};
b.) P{X=k}= P{Y=n-k}

I have proven part b.) and yes it's the same.
How do I prove part a.)? Particularly, P{Y>=n-i}.

I know that
P{Y>=n-i}=1-P{Y=0}-... up until Y=**A number lesser than (n-i)-1**

7. Apr 15, 2012

### mathman

For part a, the X probability is gotten by adding all the terms from the beginning to k, while the Y probability is gotten by adding all the terms from n-k to the end. Since these terms match term by term, the sums match.

8. Apr 15, 2012

### knowLittle

What do you mean by " the end"?
I know that P{Y>=n-k} has to be described as
1-P{Y=0}-...-P{Y=n-k-1}

Could you develop further on what you said?

9. Apr 16, 2012

### mathman

P(Y≥n-k) = P(Y=n-k) + P(Y=n-k+1) + ... + P(Y=n).

P(Y=n-k) = P(X=k), P(Y=n-k+1) = P(X=k-1), ...., P(Y=n) = P(X=0).

Therefore P(Y≥n-k)=P(X≤k).

Last edited: Apr 16, 2012
10. Apr 17, 2012

### knowLittle

I understand:
But, I do not understand this part:
I know by one of the identities that P(Y=n-k) = P(X=k), but I don't know from where do you get the rest.

11. Apr 17, 2012

### chiro

Use the property of the combination. In other words what are the properties of combinations in pascal's triangle? (What is nCk vs nC[n-k] and similar properties in relation to your question)?

12. Apr 18, 2012

### mathman

Replace k, term by term, by k-1, k-2, etc. until you get to 0.