Robert Bartoszynski - Probability and Statistical Inference

(3.23)

To estimate , we can use the principle of maximum likelihood (to be explored in detail in Chapter 11). At present, it suffices to say that this principle suggests using as , an estimator of , the value of that maximizes 3.23. Let us call this value . Let us call this value . It depends on the observed value and hence is itself random. Thus, is defined by the condition

and our objective is to find the maximizer of . Since is a discrete variable, we cannot use methods of finding a maximum based on derivatives. Instead, the method that works in this case is based on the observation that if the function has a maximum (possibly local) at , then and . If two neighboring probabilities have equal values, the ratio equals 1. Consequently, we should study the ratio and find all arguments at which this ratio crosses the threshold 1. After some reduction, we have

The above ratio always exceeds 1 if , so in this case the maximum is not attained. Assume now that . The inequality

is equivalent to

(3.24)

with the equality occurring if and only if . Thus, the maximum is attained at 1

and also at if the latter value is an integer. Let us observe that the result above is consistent with common intuition: The proportion of labeled fish in the whole lake is , and it should be close to the proportion of labeled fish in the second catch. This gives the approximate equation , with the solution .

To supplement their revenues, many states are sponsoring number games or lotteries. The details vary from state to state, but generally, a player who buys a lottery ticket chooses several numbers from a specified set of numbers. We will carry the calculations for the choice of 6 out of 50 numbers which is quite typical. After the sales of tickets close, six winning numbers are chosen at random from the set . All those (if any) who chose six winning numbers share the Big Prize; if there are no such winners, the Big Prize is added to the next week's Big Prize. Those who have five winning numbers share a smaller prize, and so on. Let be the probability that a player has exactly winning numbers. We will compute for and 3. The calculations would be the same if the winning numbers were chosen in advance, but remained secret to the players. We can now represent the situation in a familiar scheme of an urn with 6 winning numbers and 44 losing numbers, and the choice of 6 numbers from the urn (without replacement). This is the same problem as that of labeled fish. The total number of choices that can be made is , while is the number of choices with exactly winning numbers. Thus,