I'm doing past papers in order to revise for exams in June and my university irritatingly doesn't provide any mark schemes and I'm very stuck on a question.
The question says:
Let $g(z)=\frac{2}{3}+(14-5z)^{-\frac{1}{2}}$.
Show that exists a nonnegative integer-valued random variable whose probability generating function is $g$.
Attempt:
So, I thought I wanted to show that there exists $X$ such that $g(z)=\mathbb{E}(z^{X})=\sum_{k=0}^{\infty}p_{X}(k)z^{k}$. So, first I checked that $g(1)=1$ which it does.
Then I just differentiated $g$ to find a general formula for the pmf of $X$. So, I got
$g(0)=\frac{2}{3}+\frac{1}{\sqrt{14}} \Rightarrow p_{X}(0)=\frac{2}{3}+\frac{1}{\sqrt{14}}$
Then $g'(z)=\frac{5/2}{(14-5z)^{\frac{3}{2}}} \Rightarrow g''(z)=\frac{75/4}{(14-5z)^{\frac{5}{2}}} \Rightarrow ... \Rightarrow g^{(k)}(z)=\frac{5^{k}(2k)!}{2^{2k}k!(14-5z)^{\frac{2k+1}{2}}}$
Therefore $p_{X}(k)=\frac{5^{k}(2k)!}{2^{2k}(k!)^{2}(14-5z)^{\frac{2k+1}{2}}}$ for $k=1,2,...$
This just seemed too messy, so I thought my method must be incorrect?
Thanks in advance for your help!
It is fairly easy to see, by looking at the power series expansion of our function, that the coefficients are non-negative. So now the only issue is whether the coefficients have sum $1$. To verify this, put $z=1$.