A message of \(10^{k}\) binary digits is sent along a fibre optic cable with high probabilities \(p_{0}\) and \(p_{1}\) that the digits 0 and 1, respectively, are received correctly. If the probability of a digit in the original message being a 1 is \(\alpha,\) find the probability that the entire message is received correctly. Find the probability \(\beta\) that a randomly chosen digit in the message is received as a 1 and show that \(\beta=\alpha\) if, and only if \[ \alpha=\frac{q_{0}}{q_{1}+q_{0}}, \] where \(q_{0}=1-p_{0}\) and \(q_{1}=1-p_{1}.\) If this condition is satisfied and the received message consists entirely of zeros, what is the probability that it is correct? If now \(q_{0}=q_{1}=q\) and \(\alpha=\frac{1}{2},\) find the approximate value of \(q\) which will ensure that a message of one million binary digits has a fifty-fifty chance of being received entirely correctly. The probability of error \(q\) is proportional to the square of the length of the cable. Initially the length is such that the probability of a message of one million binary bits, among which 0 and 1 are equally likely, being received correctly is \(\frac{1}{2}.\) What would this probability become if a booster station were installed at its mid-point, assuming that the booster station re-transmits the received version of the message, and assuming that terms of order \(q^{2}\) may be ignored?
A candidate finishes examination questions in time \(T\), where \(T\) has probability density function \[ \mathrm{f}(t)=t\mathrm{e}^{-t}\qquad t\geqslant0, \] the probabilities for the various questions being independent. Find the moment generating function of \(T\) and hence find the moment generating function for the total time \(U\) taken to finish two such questions. Show that the probability density function for \(U\) is \[ \mathrm{g}(u)=\frac{1}{6}u^{3}\mathrm{e}^{-u}\qquad u\geqslant0. \] Find the probability density function for the total time taken to answer \(n\) such questions.
Solution: \begin{align*} && M_T(x) &= \mathbb{E}[e^{xT}] \\ &&&= \int_0^{\infty} e^{xt}te^{-t} \d t \\ &&&= \int_0^{\infty}te^{(x-1)t} \d t \\ &&&= \left [ \frac{t}{x-1} e^{(x-1)t} \right]_0^{\infty} - \int_0^\infty \frac{e^{(x-1)t}}{x-1} \d t \\ &&&= \left [ \frac{e^{(x-1)t}}{(x-1)^2} \right]_0^{\infty} \\ &&&= \frac{1}{(x-1)^2} \\ \\ && M_U(x) &= M_{T_1+T_2}(x) \\ &&&= \frac1{(x-1)^4} \\ \\ && I_n &= \int_0^{\infty} t^ne^{(x-1)t} \d t \\ &&&= \left[ \frac{1}{(x-1)}t^ne^{(x-1)t} \right]_0^{\infty} - \frac{n}{(x-1)} \int_0^{\infty}t^{n-1}e^{(x-1)t} \d t \\ &&&= -\frac{n}{(x-1)}I_{n-1} \\ \Rightarrow && I_n &= \frac{n!}{(1-x)^{n+1}} \\ \\ \Rightarrow && \int_0^{\infty} e^{xt} \frac16u^3e^{-u} \d u &= \int_0^{\infty} \frac16u^3e^{(x-1)u} \d u \\ &&&= \frac{1}{(1-x)^4} \\ \Rightarrow && f_U(u) &= \frac16u^3e^{-u} \\ \\ && M_{X_1+\cdots+X_n}(x) &= \frac{1}{(x-1)^{2n}} \\ \Rightarrow && f_{X_1+\cdots+X_n}(t) &= \frac1{(2n-1)!} t^{2n-1}e^{-t} \end{align*} (NB: This is the gamma distribution)