3 problems found
A group of biologists attempts to estimate the magnitude, \(N\), of an island population of voles ({\it Microtus agrestis}). Accordingly, the biologists capture a random sample of 200 voles, mark them and release them. A second random sample of 200 voles is then taken of which 11 are found to be marked. Show that the probability, \(p_N\), of this occurrence is given by $$ p_N = k{{{\big((N-200)!\big)}^2} \over {N!(N-389)!}}, $$ where \(k\) is independent of \(N\). The biologists then estimate \(N\) by calculating the value of \(N\) for which \(p_N\) is a maximum. Find this estimate. All unmarked voles in the second sample are marked and then the entire sample is released. Subsequently a third random sample of 200 voles is taken. Write down the probability that this sample contains exactly \(j\) marked voles, leaving your answer in terms of binomial coefficients. Deduce that $$ \sum_{j=0}^{200}{389 \choose j}{3247 \choose {200-j}} = {3636 \choose 200}. $$
Solution: There will be \(200\) marked vols out of \(N\), and we are finding \(11\) of them. There are \(\binom{200}{11}\) ways to chose the \(11\) marked voles and \(\binom{N - 200}{200-11}\) ways to choose the unmarked voles. The total number of ways to choose \(200\) voles is \(\binom{N}{200}\). Therefore the probability is \begin{align*} p_N &= \frac{\binom{200}{11} \cdot \binom{N - 200}{200-11}}{\binom{N}{200}} \\ &= \binom{200}{11} \cdot \frac{ \frac{(N-200)!}{(189)!(N - 389)!} }{\frac{N!}{(N-200)!(200)!}} \\ &= \binom{200}{11} \frac{200!}{189!} \frac{\big((N-200)!\big)^2}{N!(N-389)!} \end{align*} As required and \(k = \binom{200}{11} \frac{200!}{189!}\). We want to maximise \(\frac{(N-200)!^2}{N!(N-389)!}\), we will do this by comparing consecutive \(p_N\). \begin{align*} \frac{p_{N+1}}{p_N} &= \frac{\frac{(N+1-200)!^2}{(N+1)!(N+1-389)!}}{\frac{(N-200)!^2}{N!(N-389)!}} \\ &= \frac{(N-199)!^2 \cdot N! \cdot (N-389)!}{(N+1)!(N-388)!(N-200)!^2} \\ &= \frac{(N-199)^2 \cdot 1 \cdot 1}{(N+1) \cdot (N-388)\cdot 1} \\ \end{align*} \begin{align*} && \frac{p_{N+1}}{p_N} &> 1 \\ \Leftrightarrow && \frac{(N-199)^2 \cdot 1 \cdot 1}{(N+1) \cdot (N-388)\cdot 1} & > 1 \\ \Leftrightarrow && (N-199)^2 & > (N+1) \cdot (N-388) \\ \Leftrightarrow && N^2-2\cdot199N+199^2 & > N^2 - 387N -388 \\ \Leftrightarrow && -398N+199^2 & > - 387N -388 \\ \Leftrightarrow && 199^2+388 & > 11N\\ \Leftrightarrow && \frac{199^2+388}{11} & > N\\ \Leftrightarrow && 3635\frac{4}{11} & > N\\ \end{align*} Therefore \(p_N\) is increasing if \(N \leq 3635\), so we should take \(N = 3636\). \[ \P(\text{exactly } j \text{ marked voles}) = \frac{\binom{389}{j} \cdot \binom{3636 - 389}{200-j}}{\binom{3636}{200}}\] Since \begin{align*} && 1 &= \sum_{j=0}^{200} \P(\text{exactly } j \text{ marked voles}) \\ && &= \sum_{j=0}^{200} \frac{\binom{389}{j} \cdot \binom{3247}{200-j}}{\binom{3636}{200}} \\ \Leftrightarrow&& \binom{3636}{200} &= \sum_{j=0}^{200} \binom{389}{j} \cdot \binom{3247}{200-j} \end{align*}
It is known that there are three manufacturers \(A, B, C,\) who can produce micro chip MB666. The probability that a randomly selected MB666 is produced by \(A\) is \(2p\), and the corresponding probabilities for \(B\) and \(C\) are \(p\) and \(1 - 3p\), respectively, where \({{0} \le p \le {1 \over 3}}.\) It is also known that \(70\%\) of MB666 micro chips from \(A\) are sound and that the corresponding percentages for \(B\) and \(C\) are \(80\%\) and \(90\%\), respectively. Find in terms of \(p\), the conditional probability, \(\P(A {\vert} S)\), that if a randomly selected MB666 chip is found to be sound then it came from \(A\), and also the conditional probability, \(\P(C {\vert} S)\), that if it is sound then it came from \(C\). A quality inspector took a random sample of one MB666 micro chip and found it to be sound. She then traced its place of manufacture to be \(A\), and so estimated \(p\) by calculating the value of \(p\) that corresponds to the greatest value of \(\P(A {\vert} S)\). A second quality inspector also a took random sample of one MB666 chip and found it to be sound. Later he traced its place of manufacture to be \(C\) and so estimated \(p\) by applying the procedure of his colleague to \(\P(C {\vert} S)\). Determine the values of the two estimates and comment briefly on the results obtained.
Write down the probability of obtaining \(k\) heads in \(n\) tosses of a fair coin. Now suppose that \(k\) is known but \(n\) is unknown. A maximum likelihood estimator (MLE) of \(n\) is defined to be a value (which must be an integer) of \(n\) which maximizes the probability of \(k\) heads. A friend has thrown a fair coin a number of times. She tells you that she has observed one head. Show that in this case there are two MLEs of the number of tosses she has made. She now tells you that in a repeat of the exercise she has observed \(k\) heads. Find the two MLEs of the number of tosses she has made. She next uses a coin biased with probability \(p\) (known) of showing a head, and again tells you that she has observed \(k\) heads. Find the MLEs of the number of tosses made. What is the condition for the MLE to be unique?
Solution: \begin{align*} && \mathbb{P}(k \text{ heads} | n\text{ tosses}) &= \binom{n}k 2^{-n} \\ && \mathbb{P}(1 \text{ head} | n\text{ tosses}) &= n2^{-n} \\ \Rightarrow && \frac{ \mathbb{P}(1 \text{ head} | n+1\text{ tosses}) }{ \mathbb{P}(1 \text{ head} | n\text{ tosses}) } &= \frac{n+1}{2n} \end{align*} Which is less than \(1\) unless \(n \geq 1\). Therefore the MLE is \(n = 1\) or \(n= 2\). \begin{align*} \frac{ \mathbb{P}(k \text{ head} | n+1\text{ tosses}) }{ \mathbb{P}(k \text{ head} | n\text{ tosses}) } &= \frac{\binom{n+1}{k}}{2 \binom{n}{k}} \\ &= \frac{(n+1)!(n-k)!}{2n!(n+1-k)!} \\ &= \frac{n+1}{2(n+1-k)} \end{align*} This is less than or equal to \(1\) if \(n+1 = 2(n+1-k) \Leftrightarrow n= 2k-1\), therefore the MLEs are \(2k-1\) and \(2k\). If the coin is biased, we have \begin{align*} && \frac{ \mathbb{P}(k \text{ head} | n+1\text{ tosses}) }{ \mathbb{P}(k \text{ head} | n\text{ tosses}) } &= \frac{\binom{n+1}{k}p^kq^{n+1-k}}{\binom{n}{k}p^kq^{n-k}} \\ &&&= \frac{n+1}{(n+1-k)}q \\ \\ && 1 & \geq \frac{n+1}{(n+1-k)}q \\ \Leftrightarrow && (n+1)(1-q) &\geq k \\ \Leftrightarrow && n+1 & \geq \frac{k}{p} \end{align*} Therefore the probability is increasing until \(n+1 \geq \frac{k}{p}\). If \(\frac{k}p\) is an integer the MLEs are \(\frac{k}{p}-1\) and \(\frac{k}p\), otherwise it is \(\lfloor \frac{k}{p} \rfloor\) and the MLE is unique.