统计代写|ST308 Bayesian Analysis

In an experiment on extra-sensory perception (ESP) a person, A, sits in a sealed room and points at one of four cards, each of which shows a different picture. In another sealed room a second person, B, attempts to select, from an identical set of four cards, the card at which A is pointing. This experiment is repeated ten times and the correct card is selected four times.
Suppose that we consider three possible states of nature, as follows.
State 1 : There is no ESP and, whichever card A chooses, B is equally likely to select any one of the four cards. That is, subject B has a probability of 0.25 of selecting the correct card.
Before the experiment we give this state a probability of 0.7 .
State 2 : Subject B has a probability of 0.50 of selecting the correct card.
Before the experiment we give this state a probability of 0.2 .
State 3 : Subject B has a probability of 0.75 of selecting the correct card.
Before the experiment we give this state a probability of 0.1 .
Assume that, given the true state of nature, the ten trials can be considered to be independent.
Find our probabilities after the experiment for the three possible states of nature.
Can you think of a reason, apart from ESP, why the probability of selecting the correct card might be greater than 0.25 ?

In a certain small town there are $n$ taxis which are clearly numbered $1,2, \ldots, n$. Before we visit the town we do not know the value of $n$ but our probabilities for the possible values of $n$ are as follows.
\begin{tabular}{|cccccc|}
\hline$n$ & 0 & 1 & 2 & 3 & 4 \
Probability & 0.00 & 0.11 & 0.12 & 0.13 & 0.14 \
\hline$n$ & 5 & 6 & 7 & 8 & $\geq 9$ \
Probability & 0.14 & 0.13 & 0.12 & 0.11 & 0.00 \
\hline
\end{tabular}
On a visit to the town we take a taxi which we assume would be equally likely to be any of taxis $1,2, \ldots, n$. It is taxi number 5 . Find our new probabilities for the value of $n$.

In a forest area of Northern Europe there may be wild lynx. At a particular time the number $X$ of lynx can be between 0 and 5 with
$$
\operatorname{Pr}(X=x)=\left(\begin{array}{c}
5 \
x
\end{array}\right) 0.6^x 0.4^{5-x} \quad(x=0, \ldots, 5) .
$$
A survey is made but the lynx is difficult to spot and, given that the number present is $x$, the number $Y$ observed has a probability distribution with
$$
\operatorname{Pr}(Y=y \mid X=x)=\left{\begin{array}{ll}
\left(\begin{array}{l}
x \
y
\end{array}\right) 0.3^y 0.7^{x-y} & (0 \leq y \leq x) \
0 & (x<y)
\end{array} .\right.
$$
Find the conditional probability distribution of $X$ given that $Y=2$. (That is, find $\operatorname{Pr}(X=0 \mid Y=2), \ldots, \operatorname{Pr}(X=5 \mid Y=2)$ ).

A particular species of fish makes an annual migration up a river. On a particular day there is a probability of 0.4 that the migration will start. If it does then an observer will have to wait $T$ minutes before seeing a fish, where $T$ has an exponential distribution with mean 20 (i.e. an exponential(0.05) distribution). If the migration has not started then no fish will be seen.
(a) Find the conditional probability that the migration has not started given that no fish has been seen after one hour.
(b) How long does the observer have to wait without seeing a fish to be $90 \%$ sure that the migration has not started?