91影视

Let's start with the function \(f(\sigma) = pe^{\sigma} + qe^{-\sigma}\). We want to find a positive number \(\sigma_{0}\) such that \(f(\sigma_{0}) = 1\) and \(f(\sigma) > 1\) for all \(\sigma > \sigma_{0}\). To find this, first set \(f(\sigma) = 1\): \(1 = pe^{\sigma} + qe^{-\sigma}\) Now, we can apply a substitution, by setting \(x = e^{\sigma}\). Then, \(e^{-\sigma} = \frac{1}{x}\). Now, we have: \(1 = px + \frac{q}{x}\) Solve for x: \(x^{2} - x + q = 0\) Now, we can use the quadratic formula to find x: \(x = \frac{1\pm\sqrt{1-4q}}{2}\). Remember that \(\frac{1}{2} < q < 1\), so both roots are positive. Take smaller root \(x=\frac{1-\sqrt{1-4q}}{2}\). We want to find the corresponding \(\sigma_{0}\) value, so: \(\sigma_{0} = \ln{x} = \ln\left(\frac{1-\sqrt{1-4q}}{2}\right)\) This is the positive number \(\sigma_{0}\) we searched for.

To find the probability \(\mathbb{P}\{\tau_{1}<\infty\}\), we can apply the result which states that this probability equals \( \frac{1}{f(\sigma_{0})} = 1\), since \(f(\sigma_{0}) = 1\). Therefore, the probability \(\mathbb{P}\{\tau_{1}<\infty\}\) is also 1.

To compute \(\mathbb{E}\alpha^{r_{1}}\), we first need to find the moment generating function, \(M_{r_{1}}(t)\), and then evaluate it at \(\ln{\alpha}\). We know that: \(M_{r_{1}}(t) = \sum_{k=0}^{\infty} \mathbb{P}\{r_{1}=k\} e^{kt}\) We also know that \(\mathbb{P}\{r_{1}=k\} = p(1-q)^{k}\) for a geometric distribution with probability of success p. Therefore: \(M_{r_{1}}(t) = p\sum_{k=0}^{\infty}(1-q)^{k}e^{kt}\) Now, we can find \(M_{r_{1}}(\ln{\alpha})\): \(M_{r_{1}}(\ln{\alpha}) = p\sum_{k=0}^{\infty}(1-q)^{k}\alpha^{k}\) This is a convergent geometric series with common ratio \(\alpha(1-q)\), so we can compute its sum: \(M_{r_{1}}(\ln{\alpha}) = \frac{p}{1-\alpha(1-q)}\) Finally, this is the expected value of \(\alpha^{r_{1}}\): \(\mathbb{E}\alpha^{r_{1}} = \frac{p}{1-\alpha(1-q)}\)

Since \(\mathbb{P}\{\tau_{1}=\infty\}>0\), we have \(\mathbb{E} \tau_{1}=\infty\), as given in the problem statement. This means that the expected value of the indicator function, \(\mathbb{E}\left[\mathbb{I}_{\{r_{1}<\infty\}}\right],\) is equivalent to \(\mathbb{P}\{\tau_{1}<\infty\}\). From part (ii), we found that \(\mathbb{P}\{\tau_{1}<\infty\} = 1\). Thus, the expected value of the indicator function \(\mathbb{E}\left[\mathbb{I}_{\{r_{1}<\infty\}}\right]\) is also 1.