Probability theory. quick question regarding conditionalizing the binomial dist

[Applejacks01]

Hello,
so suppose we have B(n,p), where n is discretely uniformly distributed on the integers of the interval (1,5)

Is the expected value 3p, and is the variance 3p-p^2
?

I arrived at those answers by treating n as another variable, so np/5 summed over all n is 3p, and similar logic for E(x^2) yields the variance.

Also when I use the equation E(V(Y/n)) + V(E(Y/n)) I get the same variance.

However this will give a non-zero variance for p =1, which makes NO sense.

Please advise.

Thank you very much

 

[Ray Vickson]

Hello,
so suppose we have B(n,p), where n is discretely uniformly distributed on the integers of the interval (1,5)

Is the expected value 3p, and is the variance 3p-p^2
?

I arrived at those answers by treating n as another variable, so np/5 summed over all n is 3p, and similar logic for E(x^2) yields the variance.

Also when I use the equation E(V(Y/n)) + V(E(Y/n)) I get the same variance.

However this will give a non-zero variance for p =1, which makes NO sense.

Please advise.

Thank you very much

If p = 1, every trial gives "success". Is there variation in the number of trials?

RGV

 

[Applejacks01]

I didn't think about it like that. You're right..there is variation. I suppose that would make sense.

So in general, for the binomial given n, where n is distributed by f(n) we have the variance is:
Sum(Sum f(n)*B(n,p)*x^2 over all x) over all n) - (Sum(Sum f(n)*B(n,p)*x over all x) over all n))^2

Which can be simplified or evaluated much quicker using the law of total variance.

Thanks for your help!

 

[Applejacks01]

Hi, I actually have one more question.
suppose I wanted to calculate the probability that x=#successes =2.
Do I need to conditionalize f(n) = distribution of n(# trials) such that n>=2, or do I just assume that the probabilities when n=0 and 1 are 0??

 

[Ray Vickson]

Hi, I actually have one more question.
suppose I wanted to calculate the probability that x=#successes =2.
Do I need to conditionalize f(n) = distribution of n(# trials) such that n>=2, or do I just assume that the probabilities when n=0 and 1 are 0??

$$P\{X=2\} = \sum_{n=1}^5 f(n) P\{X=5 | n \}.$$
What do you think the values of P{X=5|n} are for n = 1 and 2?

RGV

 

[Applejacks01]

Well actually I think that P(X=2) is Sum C(n,2) *p^2 * (1-p)^(n-2) from n = 1 to 5, and when n equals 1, C(n,2) =0.

But to directly answer your question,I would say the values are 0.

Sorry for not using latex BTW. I'm on my lunch break. Just trying to learn.

 

[Ray Vickson]

Well actually I think that P(X=2) is Sum C(n,2) *p^2 * (1-p)^(n-2) from n = 1 to 5, and when n equals 1 or 2 C(n,2) =0.

But to directly answer your question,I would say the values are 0.

Sorry for not using latex BTW. I'm on my lunch break. Just trying to learn.

Your answer for P(X=2) is correct; the point is that you use the original f(n), and not some conditional version of it, and you let P{X=k|n} take care of the zeros or not.

Don't worry about not using LaTeX; what you wrote is clear enough and not very complicated, so it is perfectly readable without ambiguity.

RGV

 

[Applejacks01]

Your answer for P(X=2) is correct; the point is that you use the original f(n), and not some conditional version of it, and you let P{X=k|n} take care of the zeros or not.

Don't worry about not using LaTeX; what you wrote is clear enough and not very complicated, so it is perfectly readable without ambiguity.

RGV

Thank you so much Ray, you've been very helpful!