Chapter 10, Ex.6b :
Let \(p\) be the probability distribution:
\(p= \Bigg(\begin{array} \\0 & 1 & 2 \\ 0 & \frac{1}{3} & \frac{2}{3} \\ \end{array}\Bigg)\)
and let \(p_n = p * p * \dots *p\) be the \(\textit{n-fold}\) convolution of \(p\) with itself.
- Find the ordinary generating function \(h(z)\) and \(h_2(z)\) for \(p\) and \(p_2\) and verify that \(h_2(z)=(h(z))^2\)
Answer:
In this case we’ve got a pmf like:
\(P_X(k) = \begin{cases} 0 & 0 \\ \frac{1}{3}, & 1 \\ \frac{2}{3}, & 2 \end{cases}\)
So for \(p\) we have moment functuion:
\[M_P(t)=\mathbb{E}[e^{tX}]=\frac{1}{3}e^{t}\]
and for \(p_2\) we have moment functuion:
\[M_P(t)=\mathbb{E}[e^{tX}]=0+\frac{1}{3}e^{1t}+\frac{2}{3}e^{2t}=\frac{1}{3}e^{t}+\frac{2}{3}e^{2t}\]
Then \(h(z)\) for \(p\) is the first derivative of it’s momemt generating function evaluated at zero.
\(\frac{d}{dt}\frac{1}{3}e^{t}=\)
\(\frac{d^2}{dx^2}\frac{1}{3}e^{t}=\)