Conditional Probability and Conditional Expectation
Abstract
One of the most useful concepts in probability theory is that of conditional probability and conditional expectation. The reason is twofold. First, in practice, we are often interested in calculating probabilities and expectations when some partial information is available; hence, the desired probabilities and expectations are conditional ones. Secondly, in calculating a desired probability or expectation it is often extremely useful to first “condition” on some appropriate random variable.
Keywords
Conditional Probability; Conditional Expectation; Computing Expectations by Conditioning; Computing Probabilities by Conditioning; Conditional Variance Formula
3.1 Introduction
One of the most useful concepts in probability theory is that of conditional probability and conditional expectation. The reason is twofold. First, in practice, we are often interested in calculating probabilities and expectations when some partial information is available; hence, the desired probabilities and expectations are conditional ones. Secondly, in calculating a desired probability or expectation it is often extremely useful to first “condition” on some appropriate random variable.
3.2 The Discrete Case
Recall that for any two events 
and 
, the conditional probability of 
given 
is defined, as long as 
, by
Hence, if 
and 
are discrete random variables, then it is natural to define the conditional probability mass function of 
given that 
, by
for all values of 
such that 
. Similarly, the conditional probability distribution function of 
given that 
is defined, for all 
such that 
, by
Finally, the conditional expectation of 
given that 
is defined by
In other words, the definitions are exactly as before with the exception that everything is now conditional on the event that 
. If 
is independent of 
, then the conditional mass function, distribution, and expectation are the same as the unconditional ones. This follows, since if 
is independent of 
, then
Example 3.1
Suppose that 
, the joint probability mass function of 
and 
, is given by
Calculate the conditional probability mass function of 
given that 
.
Example 3.2
If 
and 
are independent binomial random variables with respective parameters 
and 
, calculate the conditional probability mass function of 
given that 
.
where we have used that 
is a binomial random variable with parameters 
(see Example 2.44). Thus, the conditional probability mass function of 
, given that 
, is
(3.1)
The distribution given by Equation (3.1), first seen in Example 2.35, is known as the hypergeometric distribution. It is the distribution of the number of blue balls that are chosen when a sample of 
balls is randomly chosen from an urn that contains 
blue and 
red balls. (To intuitively see why the conditional distribution is hypergeometric, consider 
independent trials that each result in a success with probability 
; let 
represent the number of successes in the first 
trials and let 
represent the number of successes in the final 
trials. Because all trials have the same probability of being a success, each of the 
subsets of 
trials is equally likely to be the success trials; thus, the number of the 
success trials that are among the first 
trials is a hypergeometric random variable.) ■
Example 3.3
If 
and 
are independent Poisson random variables with respective means 
and 
, calculate the conditional expected value of 
given that 
.
Solution: Let us first calculate the conditional probability mass function of 
given that 
. We obtain
where the last equality follows from the assumed independence of 
and 
. Recalling (see Example 2.37) that 
has a Poisson distribution with mean 
, the preceding equation equals
In other words, the conditional distribution of 
given that 
is the binomial distribution with parameters 
and 
. Hence,
Conditional expectations possess all of the properties of ordinary expectations. For example such identities such as
remain valid.
Example 3.4
There are 
components. On a rainy day, component 
will function with probability 
; on a nonrainy day, component 
will function with probability 
, for 
. It will rain tomorrow with probability 
. Calculate the conditional expected number of components that function tomorrow, given that it rains.
3.3 The Continuous Case
If 
and 
have a joint probability density function 
, then the conditional probability density function of 
, given that 
, is defined for all values of 
such that 
, by
To motivate this definition, multiply the left side by 
and the right side by 
to get
In other words, for small values 
and 
, 
is approximately the conditional probability that 
is between 
and 
given that 
is between 
and 
.
The conditional expectation of 
, given that 
, is defined for all values of 
such that 
, by
Example 3.5
Suppose the joint density of 
and 
is given by
Compute the conditional expectation of 
given that 
, where 
.
Example 3.6
The t-Distribution
If 
and 
are independent, with 
having a standard normal distribution and 
having a chi-squared distribution with 
degrees of freedom, then the random variable 
defined by
is said to be a t-random variable with 
degrees of freedom. To compute its density function, we first derive the conditional density of 
given that 
. Because 
and 
are independent, the conditional distribution of 
given that 
is the distribution of 
, which is normal with mean 
and variance 
. Hence, the conditional density function of 
given that 
is
Using the preceding, along with the following formula for the chi-squared density derived in Exercise 
of Chapter 2,
we obtain the density function of 
:
With
this yields
Example 3.7
The joint density of 
and 
is given by
Solution: The conditional density of 
, given that 
, is given by
Hence, by Proposition 2.1,
Example 3.8
Let 
and 
be independent exponential random variables with rates 
and 
. Find the conditional density of 
given that 
.
Solution: To begin, let us first note that if 
is the joint density of 
, then the joint density of 
and 
is
which is easily seen by noting that the Jacobian of the transformation
is equal to 
.
Applying the preceding to our example yields
where
Now, if 
, then
yielding that 
and that 
given 
is uniformly distributed on 
. On the other hand, if 
, then we use
to obtain
thus yielding the result:
An interesting byproduct of our analysis is that
3.4 Computing Expectations by Conditioning
Let us denote by 
that function of the random variable 
whose value at 
is 
. Note that 
is itself a random variable. An extremely important property of conditional expectation is that for all random variables 
and 
(3.2)If 
is a discrete random variable, then Equation (3.2) states that
(3.2a)
while if 
is continuous with density 
, then Equation (3.2) says that
(3.2b)
We now give a proof of Equation (3.2) in the case where 
and 
are both discrete random variables.
Proof of Equation (3.2) When X and Y Are Discrete. We must show that
(3.3)
Now, the right side of the preceding can be written
and the result is obtained. ■
One way to understand Equation (3.3) is to interpret it as follows. It states that to calculate 
we may take a weighted average of the conditional expected value of 
given that 
, each of the terms 
being weighted by the probability of the event on which it is conditioned.
The following examples will indicate the usefulness of Equation (3.2).
Example 3.9
Sam will read either one chapter of his probability book or one chapter of his history book. If the number of misprints in a chapter of his probability book is Poisson distributed with mean 2 and if the number of misprints in his history chapter is Poisson distributed with mean 5, then assuming Sam is equally likely to choose either book, what is the expected number of misprints that Sam will come across?
Example 3.10
The Expectation of the Sum of a Random Number of Random Variables
Suppose that the expected number of accidents per week at an industrial plant is four. Suppose also that the numbers of workers injured in each accident are independent random variables with a common mean of 2. Assume also that the number of workers injured in each accident is independent of the number of accidents that occur. What is the expected number of injuries during a week?
Solution: Letting 
denote the number of accidents and 
the number injured in the 
th accident, 
, then the total number of injuries can be expressed as 
. Now,
But
which yields
and thus
Therefore, in our example, the expected number of injuries during a week equals 
. ■
The random variable 
, equal to the sum of a random number 
of independent and identically distributed random variables that are also independent of 
, is called a compound random variable. As just shown in Example 3.10, the expected value of a compound random variable is 
. Its variance will be derived in Example 3.19.
Example 3.11
The Mean of a Geometric Distribution
A coin, having probability 
of coming up heads, is to be successively flipped until the first head appears. What is the expected number of flips required?
Solution: Let 
be the number of flips required, and let
Now,
(3.4)
However,
(3.5)
To see why Equation (3.5) is true, consider 
. Since 
, we know that the first flip resulted in heads and so, clearly, the expected number of flips required is 1. On the other hand if 
, then the first flip resulted in tails. However, since the successive flips are assumed independent, it follows that, after the first tail, the expected additional number of flips until the first head is just 
. Hence 
. Substituting Equation (3.5) into Equation (3.4) yields
or
Because the random variable 
is a geometric random variable with probability mass function 
, its expectation could easily have been computed from 
without recourse to conditional expectation. However, if you attempt to obtain the solution to our next example without using conditional expectation, you will quickly learn what a useful technique “conditioning” can be.
Example 3.12
A miner is trapped in a mine containing three doors. The first door leads to a tunnel that takes him to safety after two hours of travel. The second door leads to a tunnel that returns him to the mine after three hours of travel. The third door leads to a tunnel that returns him to his mine after five hours. Assuming that the miner is at all times equally likely to choose any one of the doors, what is the expected length of time until the miner reaches safety?
Solution: Let 
denote the time until the miner reaches safety, and let 
denote the door he initially chooses. Now,
However,
(3.6)
To understand why this is correct consider, for instance, 
, and reason as follows. If the miner chooses the second door, then he spends three hours in the tunnel and then returns to the mine. But once he returns to the mine the problem is as before, and hence his expected additional time until safety is just 
. Hence 
. The argument behind the other equalities in Equation (3.6) is similar. Hence,
Example 3.13
Multinomial Covariances
Consider 
independent trials, each of which results in one of the outcomes 
, with respective probabilities 
. If we let 
denote the number of trials that result in outcome 
, then 
is said to have a multinomial distribution. For 
, let us compute
Because each trial independently results in outcome 
with probability 
, it follows that 
is binomial with parameters 
, giving that 
. To compute 
, condition on 
to obtain
Now, given that 
of the 
trials result in outcome 
, each of the other 
trials independently results in outcome 
with probability
thus showing that the conditional distribution of 
, given that 
, is binomial with parameters 
. Using this yields
Because 
is binomial with parameters 
Hence,
which yields the result
Example 3.14
The Matching Rounds Problem
Suppose in Example 2.31 that those choosing their own hats depart, while the others (those without a match) put their selected hats in the center of the room, mix them up, and then reselect. Also, suppose that this process continues until each individual has his own hat.
(a) Find 
where 
is the number of rounds that are necessary when 
individuals are initially present.
(b) Find 
where 
is the total number of selections made by the 
individuals, 
.
(c) Find the expected number of false selections made by one of the 
people, 
.
Solution: (a) It follows from the results of Example 2.31 that no matter how many people remain there will, on average, be one match per round. Hence, one might suggest that 
. This turns out to be true, and an induction proof will now be given. Because it is obvious that 
, assume that 
for 
. To compute 
, start by conditioning on 
, the number of matches that occur in the first round. This gives
Now, given a total of 
matches in the initial round, the number of rounds needed will equal 1 plus the number of rounds that are required when 
persons are to be matched with their hats. Therefore,
where the final equality used the result, established in Example 2.31, that 
. Since the preceding equation implies that 
, the result is proven. (b) For 
, conditioning on 
, the number of matches in round 1, gives
where 
. To solve the preceding equation, rewrite it as
Now, if there were 
one match in each round, then it would take a total of 
selections. Thus, let us try a solution of the form 
. For the preceding equation to be satisfied by a solution of this type, for 
, we need
or, equivalently,
Now, using the results of Example 2.31 and Exercise 72 of Chapter 2 that 
, the preceding will be satisfied if
and this will be valid provided that 
. That is,
satisfies the recursive equation for 
.
The formal proof that 
, is obtained by induction on 
. It is true when 
(since, in this case, the number of selections is twice the number of rounds and the number of rounds is a geometric random variable with parameter 
). Now, the recursion gives
Hence, upon assuming that 
, for 
and using that 
, we see that