## 2.5 CONDITIONAL PMF'S AND INDEPENDENCESuppose we are told that an experimental outcome is contained in event A. We then wish to explore the probabilistic behavior of random variables X and Y, given A. Following the definition of conditional probability, we introduce the conditional compound pmf, If event A is stated in terms of the specific experimental value of one of the random variables of the experiment, we introduce another notation: These notations and definitions extend in an obvious way to situations with more than two random variables.
Suppose that a minibus with capacity for five passengers departs from
a commuter station. Observation has shown that the bus never departs
empty (with
no passengers) but that each possible positive number of passengers
is equally likely to be on the bus at departure time. Passengers
are of two types: male and female. Given that the departing bus
contains exactly n passenger (n = 1, 2,..., 5), each possible
combination of male and female passenger has been found to be equally
likely.Example 2: Minibusa. Identify the sample space and joint probability mass function for this experiment. b. Determine the marginal pmf for the number of females on the mini bus. c. Determine the joint conditional pmf for the number of females and the number of males on the minibus, given that the bus departs at full capacity. d. Determine the conditional marginal pmf for the number of females, given that there are at least twice as many females as males on the minibus. Solution:There are two random variables in this experiment: a. A complete listing of their possible (paired) values constitutes the sample space for this experiment. These points occupy the nearly triangular region shown in Figure 2.2. The region is not perfectly triangular since the origin (N _{F} = 0, N_{m} = 0) is excluded because the
minibus never departs empty. The region is bounded above by the line
n_{f} + n_{m} = 5, which expresses the capacity constraint for the minibus.To determine the joint pmf for N _{f} and N_{m}, we must use the conditional
information given in the word statement. We know that any particular
positive total number of passengers, ranging up to 5, is equally likely. LetA _{i} = event that i total passengers are on the busWe know that P{A _{1}} = 1/5, i = 1, 2,. .., 5. Points in an event A_{i} lie on
the line n_{F} + n_{m} = i, as shown for A_{3} in Figure 2.2. Given that an
outcome of the experiment is contained in event A_{i}, we know that
each of the points in A_{i} is equally likely. Since the number of points
in A_{i} is equal to i + 1, we haveThis is the answer to part (a). It says, roughly, that the probability that {N _{f} = n_{f} and N_{m} = n_{M}}
is equal to 1/5 divided by 1 plus the sum
n_{F} + n_{m}. For i = 3, for instance,
The complete joint pmf is shown in Figure 2.3. b. Once we have the joint pmf for N _{F} and N_{m}, we can readily answer
any question about the experiment. The marginal pmf for N_{F} is found
by invoking (2.4), which simply asks us to sum over all values of N_{M}
at each particular fixed value for N_{F}. For instance, to obtain
P{N_{f} = 3) = P_{Nf}(3), we sum the probabilities corresponding to the
(finest-grained) events {N_{f } = 3, N_{m} = 0), {N_{f} = 3, Nm = 1}, and
{N_{f} = 3, N_{m} = 2}, yielding 1/20 + 1/25 + 1/30 = 37/300. The complete pmf
is shown in Figure 2.4.c. If we are given conditional information that the bus departs at full capacity, we know that the experimental outcome is contained in event As (i.e., n _{F} + n_{m} = 5). Thus, invoking (2.5),
This corresponds to a straight line of probability masses, each having mass 1/6, at the integer points on the line n _{F} + n_{m} = 5
(n_{F}, n_{m} 0).d. Let B = event that "there are at least twice as many females as males on the minibus" We want P _{Nf}(n_{f} | B). First we work in the original joint sample
space to determine finest-grained outcomes contained in the event B.
Clearly, these are points n_{F}, n_{m} satisfying the inequality 2n_{m}
n_{F}.
This corresponds to points lying on or below the line n_{m} = I n_{F} (shown
in Figure 2.3). Summing the probabilities of the eight finestgrained
outcomes satisfying this inequality, we find that P{B} = 124/300 = 31/75.
Then, to find the conditional marginal pmf for N_{F}, given B, we
Simply sum the probabilities at a fixed value for n_{F} over all values
of n_{m} contained in B, then scale by 1/31/75. For instance,
The entire conditional marginal pmf is displayed in Figure 2.5. Notice how the conditional information has shifted the pmf for N _{F} toward
greater numbers of females (compare to Figure 2.4).
Just as events can be independent, so, too, can random variables be independent. Intuitively, if X and Y are independent, any information regarding the value of one tells us nothing new about the value of the other. Formally, random variables X and Y are independent if and only if P _{Y|X}(y | x) = p_{y}(y) for all possible values of x and y. Show that the definition
of independence of X and Y implies thatExercise 2.4: Independence of Random Variablespand pGiven an arbitrary number N of random variables, they are said to be mutually independent if their joint pmf factors into the product of the corresponding N marginal pmf's. Sometimes random variables may be independent but conditionally dependent; or, they may be dependent but conditionally independent. The definition of conditional independence is just what we expect: random variables X and Y are said to be conditionally independent given event A if and only if Exercise 2.5: Conditional Independence Show that for two random
variables X and Y that are conditionally independent given event A,Example 2: (continued)In the minibus example, argue that N _{f} and N_{m} are not independent.
Does there exist any nontrivial event A such that, given A, N_{f} and
N_{m} are conditionally independent? |