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.
(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.