Multinomial distribution

[iocal]

Let \(\displaystyle \displaystyle p(x_1,x_2)=\binom{x_1}{x_2}(\frac{1}{2})^{x_1}\ (\frac{x_1}{15})\ for\ x_2=0,1,.....,x_1\ and\ x_1=1,2,3,4,5\)

(a) Show that \(\displaystyle \displaystyle \sum_{x_1=1}^{5}\sum_{x_2=0}^{x_1}p(x_1,x_2)=1\)
(b) Then show that \(\displaystyle \displaystyle E(x_2)=\frac{11}{6}\)

I am having a little trouble computing the sums. Is it some form of the binomial theorem that I have to use here?
Also I know in part (b) the expectation is given by \(\displaystyle \displaystyle E(x_2)=\sum_{x_1=1}^{5}\sum_{x_2=0}^{x_1}{x_2}\ \binom{x_1}{x_2}(\frac{1}{2})^{x_1}(\frac{x_1}{15})\)
The problem is to evaluate the sums, since from the looks of it the binomial theorem does not apply directly, right?

I've been stuck here for a while so any help is greatly appreciated, thanks.

 

[DrPhil]

Let \(\displaystyle \displaystyle p(x_1,x_2)=\binom{x_1}{x_2}(\frac{1}{2})^{x_1}\ (\frac{x_1}{15})\ for\ x_2=0,1,.....,x_1\ and\ x_1=1,2,3,4,5\)

(a) Show that \(\displaystyle \displaystyle \sum_{x_1=1}^{5}\sum_{x_2=0}^{x_1}p(x_1,x_2)=1\)
(b) Then show that \(\displaystyle \displaystyle E(x_2)=\frac{11}{6}\)

I am having a little trouble computing the sums. Is it some form of the binomial theorem that I have to use here?
Also I know in part (b) the expectation is given by \(\displaystyle \displaystyle E(x_2)=\sum_{x_1=1}^{5}\sum_{x_2=0}^{x_1}{x_2}\ \binom{x_1}{x_2}(\frac{1}{2})^{x_1}(\frac{x_1}{15})\)
The problem is to evaluate the sums, since from the looks of it the binomial theorem does not apply directly, right?

I've been stuck here for a while so any help is greatly appreciated, thanks.

I don't instantly see how to do this, so I will muse a bit.

There are only five values of x_1, so how about evaluating them separately?

Let \(\displaystyle \displaystyle P(x_1) = \sum_{x_2=0}^{x_1}p(x_1,x_2) = \left( \frac{1}{2}\right)^{x_1}\left(\frac{x_1}{15}\right)\sum_{x_2=0}^{x_1} \binom{x_1}{x_2}\)

.....\(\displaystyle \displaystyle P(1) = \dfrac{1}{30}\ \left(1 + 1\right)\)

.....\(\displaystyle \displaystyle P(2) = \dfrac{1}{30}\ \left(1 + 2 + 1\right)\)

.....\(\displaystyle \displaystyle P(3) = \dfrac{1}{40}\ \left(1 + 3 + 3 + 1\right)\)

.....\(\displaystyle \displaystyle P(4) = \dfrac{1}{60}\ \left(1 + 4 + 6 + 4 + 1\right)\)

.....\(\displaystyle \displaystyle P(5) = \dfrac{1}{96}\ \left(1 + 5 + 10 + 10 + 5 + 1\right)\)

Does that help?

 

[iocal]

I don't instantly see how to do this, so I will muse a bit.

There are only five values of x_1, so how about evaluating them separately?

Let \(\displaystyle \displaystyle P(x_1) = \sum_{x_2=0}^{x_1}p(x_1,x_2) = \left( \frac{1}{2}\right)^{x_1}\left(\frac{x_1}{15}\right)\sum_{x_2=0}^{x_1} \binom{x_1}{x_2}\)

.....\(\displaystyle \displaystyle P(1) = \dfrac{1}{30}\ \left(1 + 1\right)\)

.....\(\displaystyle \displaystyle P(2) = \dfrac{1}{30}\ \left(1 + 2 + 1\right)\)

.....\(\displaystyle \displaystyle P(3) = \dfrac{1}{40}\ \left(1 + 3 + 3 + 1\right)\)

.....\(\displaystyle \displaystyle P(4) = \dfrac{1}{60}\ \left(1 + 4 + 6 + 4 + 1\right)\)

.....\(\displaystyle \displaystyle P(5) = \dfrac{1}{96}\ \left(1 + 5 + 10 + 10 + 5 + 1\right)\)

Does that help?

It helps a lot thank you. By the way, it was cool to see that you get 5 rows of Pascal's triangle that way, exluding the first as 0 is not in the support.
Continuing in the same fashion it should be easy to evaluate the expectation as well.
Thanks again.

EDIT: I did part b) and the result is in agreement with what I had to show. Once again I am grateful.