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Chapter 10
Conditioning on a random variable with
a continuous distribution
At this point in the course, I hope you understand the importance of the conditioning formula E
Y | info
i P^
Fi | info
E
Y | Fi , info
for finite or countably infinite collections of disjoint events Fi for which = ∪i Fi. As a particular case, if X is a random variable that takes only a discrete set of values {x 1 , x 2 ,.. .} then E
Y | info
i P{X^ =^ x^ i^ |^ info}E^
Y | X = x (^) i , info
This formula can be simplified by the introduction of the function h(x) = E
Y | X = x, info
For then E
Y | info
i P{X^ =^ x^ i^ |^ info}h(x^ i^ )^ =^ E^
h(X) | info
In this Chapter, I want to persuade you that a similar formula applies when X has a continuous distribution, with density function f (given the info):
(∗) E
Y | info
= E
h(X) | info
−∞
h(x) f (x) dx.
As a special case, when Y equals the indicator function of an event B, the formula reduces to PB =
−∞
P(B | X = x) f (x) dx.
From now on, I will omit explicit mention of the conditioning information “info”, writ- ing h(x) for E(Y | X = x). There are several ways to arrive at formula (∗). The most direct relies on the plausible assertion that E
Y | X ∈ J
≈ h(x) if J is a small interval with x ∈ J. The error of approximation should disappear as J shrinks to the point x. Split R into a union of disjoint, small intervals Ji = [x (^) i , x (^) i+ 1 ), where x (^) i+ 1 = x (^) i + δ, then condition:
EY =
i P{X^ ∈^ Ji^ }E(Y^ |^ X^ ∈^ Ji^ } ≈^
i δ^ f^ (x^ i^ )h(x^ i^ )^ ≈
−∞
h(x) f (x) dx.
The combined errors of all the approximation should disappear in the limit as δ tends to zero.
Alternatively, we could start from a slightly less intuitive assumption that EY should be nonnnegative if E(Y | X = x) ≥ 0 for every x. If we replace Y by Y − h(X) then we have E
Y − h(X) | X = x
= E
Y | X = x
− h(x) = 0 , which gives E(Y −h(X)) ≥ 0. A similar argument applied to h(X)−Y gives E
h(X) − Y
- Equality (∗) follows.
Remark. Notice that formula (∗) also implies that (∗∗), E (Y g(X )) = E
g(X )h(X )
at least for bounded functions g because E
Y g(X ) | X = x
= g(x)h(x). In advanced probability theory, the treatment of conditional expectations becomes most abstract. Formula (∗∗) is used to define the conditional expectation h(x) = E(Y | X = x). One needs to show that there exists a random variable of the form h(X ), which is uniquely determined up to trivial changes on sets of zero probability, for which Eg(X )
Y − h(X )
= 0 for every bounded g. Essentially h(X ) is the best approximation to Y using only information given by X. With this abstract approach, one then needs to show that conditional expecta- tions have the properties that I have taken as axiomatic for Stat 241.
Example <10.1>: The convolution formula for densities derived from (∗).
The Poisson process is often used to model the arrivals of customers in a waiting line, or the arrival of telephone calls at an exchange. The underlying idea is that of a large popu- lation of potential customers, each of whom acts independently of all the others.
Example <10.2>: A queuing problem with a surprising solution (can be skipped)
Examples for Chapter 10
<10.1> Example. Suppose X and Y are independent random variables with continuous distri- butions. If X has density f and Y has density g then (see Chapter 7) the random variable Z = X + Y has density h(z) =
−∞
g(z − x) f (x) dx
The same formula can be derived from the formula
PB =
−∞
P(B | X = x) f (x) dx,
applied with B = {z ≤ Z ≤ z + δ} for a small, positive δ. Note that P(z ≤ Z ≤ z + δ | X = x) = P(z − x ≤ Y ≤ z − x + δ | X = x) = P(z − x ≤ Y ≤ z − x + δ) because X, Y independent ≈ δg(z − x). Invoke the conditioning formula.
P{z ≤ Z ≤ z + δ} ≈
−∞
δg(z − x) f (x) dx,
which leads us back to the convolution formula.
<10.2> Example. Suppose an office receives two different types of inquiry: persons who walk in off the street, and persons who call by telephone. Suppose the two types of arrival are described by independent Poisson processes, with rate λ (^) w for the walk-ins, and rate λc for
You should be able to write out the necessary conditioning argument for (2′).
The two-step mechanism explains the appearance of the geometric distribution in the problem posed at the start of the Example. The classification of each inquiry as either a walk-in or a call is effectively carried out by a sequence of independent coin tosses, with probability p of a head (= a walk-in). The problem asks for the distribution of the number of tails before the first head. The embedding of the inquiries into continuous time is irrele- vant.
