Discontinuities of distribution functions
14.6 Discontinuities of distribution functions
An example of a possible distribution function with discontinuities is shown in Figure
14.3. Using the mass analogy we would say that F has a jump discontinuity at each point which carries a positive amount of mass. As the next theorem shows, the jump is equal to the amount of mass concentrated at that particular point.
Calculus of probabilities
Jump =
= a) F(t) =
F IGURE 14.3 A possible distribution function.
F IGURE
14.4 Illustrating a jump dis- continuity of a distribution function.
THEOREM 14.5. Let F be the distribution function of a one-dimensional random variable Then for each real a we have
lim F(t) = F(a)
= F(a)
= a).
Note: The limit relation in (14.2) tells us that Fis continuousfrom the right at each point
a from the right. On the other hand, Equation (14.3) tells us that as
a, because F(t)+
as t
a from the left, F(t) will approach if and only if the probability = a) is zero. When
= a) is nor zero, the graph of F has a jump discontinuity at a of the type shown in Figure 14.4.
Proof. The existence of the limits follows at once from the monotonicity and ness of F. We prove now that the limits have the values indicated. For this purpose we
use part (b) of Theorem 14.2. If t > a we write
F(t) = F(a) + P(a < X
t)
if < a we write (14.5)
F(t) = F(a)
P(t < X
a).
Letting a+ in (14.4) we find
lim F(t) = F(a) + lim
<X
t) ,
whereas if
a- in (14.5) we obtain
lim F(t) = F(a)
Therefore to prove (14.2) and (14.3) we must establish two equations: (14.6)
limP(a < X
519 and
Discontinuities of distribution functions
These can be justified intuitively as follows: When a+ , the half-open interval (a, shrinks to the empty set. That is, the intersection of all half-open intervals (a,
for > a, is empty. On the other hand, when
a- the half-open interval a] shrinks to the point a. (The intersection of all intervals
a] for < a is the set (a}.) Therefore, if probability behaves in a continuous fashion, Equations (14.6) and (14.7) must be valid. To convert this argument into a rigorous proof we proceed as follows:
For each integer n
let
To prove (14.6) it suffices to show
0 as
Let
denote the event
The sets are disjoint and their union
is the event a < X a + 1. By countable additivity we have
On the other hand, Equation (14.8) implies that
so from (14.9) we obtain the relation
The convergence of the series is a consequence of (14.9). But the series on the left of (14.10) is a telescoping series with sum
lim
Therefore (14.10) implies that = 0, and this proves (14.6).
A slight modification of this argument enables us to prove (14.7) as well. Since
a) =
< x < a) +
we need only prove that
< X < a) = 0.
Calculus of probabilities
For this purpose we introduce the numbers
n In this case we consider the events
given by
for n = 1, 2, 3, . . . . These are disjoint and their union is the event a 1 < X < a, so we have
We now note that
and we complete the proof as above.
The most general type of distribution is any real-valued function F that has the following properties :
(a) F is monotonically increasing on the real axis, (b) F is continuous from the right at each point, (c)
F(t) = 0 and
F(t) = 1.
In fact, it can be shown that for each such function F there is a corresponding set function P, defined on the Bore1 sets of the real line, such that P is a probability measure which
assigns the probability F(b) F(a) to each half-open interval (a, b]. For a proof of this statement, see H. Cramer, Mathematical Methods
Statistics, Princeton University Press, Princeton, N.J., 1946.
There are two special types of distributions, known as discrete and continuous, that are of particular importance in practice. In the discrete case the entire mass is concentrated at a finite or countably infinite number of points, whereas in the continuous case the mass is smeared, in uniform or varying thickness, along an interval (finite or infinite). These two types of distributions will be treated in detail in the next few sections.