3  A numerical example of hours probabilities

This section shows how the probability distribution of an individual’s hours of work is generated, using a simple hypothetical numerical example. Suppose for the purposes of this example that takes only discrete values, for . In general, let denote the proportion of values equal to and let denote the proportion less than or equal to The value of (the probability of producing the highest utility) is thus obtained as the addition of terms corresponding to (4):

(5)    

Consider a situation in which there are just four hours levels of work available, so that The values of associated with each hours level, to , are respectively 5, 7.5, 10 and 9. For the purpose of this example for a single individual, it is not necessary to specify either the form of the function, , or the precise discrete hours levels. Clearly, if the s were to represent utility precisely, would always be unambiguously chosen.

As above, suppose that all values for are drawn independently from the same discrete distribution with four possible outcomes, so and let take the values shown in Table 1. In this hypothetical example the arithmetic mean value of is non-zero.

Table 1– Hypothetical discrete distribution of the error term

	1	2	3	4
	-2.0	0.0	2.0	3.5
	0.1	0.3	0.4	0.2
	0.1	0.4	0.8	1.0

The selection of an hours level is, as explained in section 2, associated in this case with the ‘random draws’ from the four distributions, each identical to the one shown in Table 1. For example, a set of random values for to may be say 2, 0, -2 and 2 respectively. These give rise to utilities, , of 7, 7.5, 8, and 11 for the hours levels to respectively. Hence it is clear that is chosen in this case. It can be seen that, given a draw of -2 from the distribution of the option can dominate (that is, ) if takes either of the values 0, 2 or 3.5. The conditional probability of dominating, given this selection from is thus , found by adding the relevant values of in Table 1. The enumeration of all possible combinations of this type is most efficiently carried out following the approach underlying equation (5).

Consider the probability of selecting hours level . The relevant values are shown in Table 2. The second column, headed shows the differences in the values of ; these are all positive, as hours level 3 has, by assumption, the highest value of . The column headed relates to the values and probabilities when is drawn for The first row shows that when , that is when the term must be less than in order to ensure that hours level 3 has a higher value of From the assumed distribution in Table 1, there is a probability of 0.8 that is less than 3. This is shown in the second row of Table 2. Similarly, when , hours level gives higher utility than only if ; this has a probability of 0.1.

Table 2– Conditional probabilities for hours level 3

1	5		30.8	51	71	8.51
2	2.5		0.50.4	2.50.8	4.51	61
4	1		-10.1	10.4	30.8	4.51
Conditional probability that  and  and			0.032	0.320	0.800	1.0

The conditional probability that is chosen, when , is therefore . The final column of Table 2 shows that when so that , hours level 3 always dominates and the conditional probability of it being chosen is 1. The overall probability is thus given by:

(6)    

(7)    

(8)    

Similar calculations show that , and . The resulting probability distribution of hours clearly depends in a complex way on the distribution of the ‘error’ term.

This example has been constructed in order to illustrate the way in which the hours distribution for an individual is derived from the underlying stochastic specification and utility levels. In practice more structure has to be imposed by specifying a precise form for the error distribution . A special case using a continuous distribution is examined in the next section, which is necessarily more technical than the previous discussion.