Verify and Explain Binomial R.V. Identities

[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.

 

[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..

 

[knowLittle]

What do you mean by opposite order?

 

[knowLittle]

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

 

[mathman]

What do you mean by opposite order?

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)!]}p^k(1-p)^n-k

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

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

 

[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**

Please, help.

 

[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.

 

[knowLittle]

...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.

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?

 

[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).

 

[knowLittle]

I understand:

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

But, I do not understand this part:

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).

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.

 

[chiro]

I understand: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.

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)?

 

[mathman]

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.

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