Download 3.3 Cumulative Distribution Functions

STAT 421 Lecture Notes
3.3
40
Cumulative Distribution Functions
Definition The cumulative distribution function (c.d.f.) of a random variable X is the
function
F (x) = Pr(X ≤ x) ∀x ∈ R.
The c.d.f. exists and is defined in the same manner for all random variables (discrete, continuous, and mixed).
Note that F : R → [0, 1].
Example: Suppose that X ∼ Bin(2, p). Then,


0,




(1 − p)2 ,
F (x) =


(1 − p)2 + 2p(1 − p),




1
x<0
0≤x<1
1≤x<2
x ≤ 2.
0.0
0.2
0.4
F(x)
0.6
0.8
1.0
Numerical values for this c.d.f. can be computed using the R command pbinom(0:n,n,p)
(replace n and p with numerical values). The figure below is a plot of the c.d.f. of a binomial
random variable parameters n = 12 and p = .3. The R code for plotting the graph is in the
script entitled DistributionPlotter.R.
0
2
4
6
8
10
12
x
In this instance (because X is discrete), Pr(X = k) can be recovered from the c.d.f.
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because
Pr(X = k) = Pr(X ≤ k) − Pr(X < k)
= Pr(X ≤ k) − Pr(X ≤ k − 1)
= F (k) − F (k − 1).
Properties of the cumulative distribution function
1. F is nondecreasing. In other words, for all x1 < x2 , F (x1 ) ≤ F (x2 ). This property can
be verified by setting up the events
E1 = {X ≤ x1 } and E2 = {X ≤ x2 }.
Then, E1 ⊂ E2 ⇒ Pr(E1 ) ≤ Pr(E2 ) ⇒ F (x1 ) ≤ F (x2 ).
2. limx→−∞ F (x) = 0 and limx→∞ F (x) = 1. The proof is straightforward if the result of
exercise 12, p. 51 is used.
3. A c.d.f. need not be continuous, as the figure above shows, but instead may be discontinuous at countably many points. On the other hand, the third property states that
every c.d.f. is continuous from the right (from above). Specifically,
lim F (y) = F (x).
y→x+
The notation limy→x+ describes a limiting process where y → x and y > x.
A cumulative distribution function need not be continuous from the left (from below). Instead, it may be that limy→x− F (y) < F (x), in which case, we say that there is a discontinuity
at x, or that there is a jump in F at x. The binomial random variable c.d.f. above illustrates
this phenomenon as there is a jump at every point in the support of X.
If there is a jump in F at x, then the height of the jump is Pr(X = x). On the graph
of a c.d.f., the height of each jump is the probability that the random variable takes on the
value x at the horizontal coordinate of the jump.
If the limit limy→x− F (y) = F (x− ) is not equal to F (x) = Pr(X ≤ x), then limy→x− F (y) =
Pr(X < x). Consequently,
F (x) − F (x− ) = Pr(X ≤ x) − Pr(X < x)
= Pr(X = x).
Computing probabilities from cumulative distribution functions
The following results are useful for computing probabilities involving the random variable
X:
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1. Pr(X > x) = 1 − F (x).
2. For x1 < x2 ,
F (x2 ) − F (x1 ) = Pr(X ≤ x2 ) − Pr(X ≤ x1 )
= Pr(x1 < X ≤ x2 ).
3. Pr(X < x) = F (x− ) = limy→x− F (y). If F (x− ) = F (x), then Pr(X < x) = Pr(X ≤ x).
4. Pr(X = x) = F (x) − F (x− ).
Suppose that X is discrete and that a and b belong to the support of X. Then,
1. Pr(a < X < b) = 0 if and only if F (x) is constant over the open interval (a, b).
2. Pr(X = x) > 0 if and only if F is discontinuous at x. This fact follows from F (x) −
F (x− ) = Pr(X = x) > 0.
Suppose that X is continuous. Then,
1. F is continuous everywhere. Because X is continuous,
0 = Pr(X = x) = F (x) − F (x− )
= F (x) − lim− F (y)
y→x
⇒ F (x) = lim− F (y).
y→x
Thus, F is continuous from below, and since F is continuous from above everywhere
(true of every continuous and discrete r.v.), F is continuous everywhere.
2. The relationship between density functions and cumulative distribution functions is
F (x) = Pr(X ≤ x)
∫ x
=
f (t)dt.
−∞
Furthermore, at every x such that f is continuous,
dF (x)
= f (x).
dx
Recall example 3.2.8 (page 105 in the textbook) where

1


, 0<x
2
f (x) = (1 + x)

0,
otherwise.
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From this definition,
lim f (x) = 1 ̸= 0 = f (0).
x→0+
Hence, f is discontinuous at 0 and F ′ = dF/dx does not exist at 0. Thus, f ̸= F ′ , since F ′
does not exist at 0 whereas f does exist at 0 (in fact, f (0) = 0). Unfortunately, the derivative of the cumulative distribution function is not necessarily the probability density function.
Example Suppose that

1 − e−x/3 , x > 0
F (x) =
0,
x ≤ 0.
F is continuous everywhere, so the p.d.f. is f = F ′ and it is

 1 e−x/3 , x > 0
f (x) = 3
0,
x ≤ 0.
Quantile Functions
Definition 3.3.2 Suppose that X is a random variable with c.d.f. F . We define the function
F −1 according to
F −1 (p) = min{F (x) ≥ p}.
x
That is, F
−1
(p) is the minimum value of x such that F (x) ≥ p.
In some cases, F −1 truly is the inverse of F , but often F −1 is not the inverse function
because F is not one-to-one. If F is one-to-one, then there is a unique value of x such that
F (x) = p, and the inverse function exists. The quantile function maps the open interval
(0, 1) to R. F −1 is defined on the open interval because
lim F (x) = 0 and lim F (x) = 1.
x→−∞
x→∞
F −1 (p) is called the p quantile of X. We also say that F −1 (p) is 100p percentile of X. Several
quantiles are named:
1. F −1 (.5) is the median. Suppose that X is continuous. Then Pr[X ≤ F −1 (.5)] = .5 and
the median partitions the distribution of X into two sets with equal probability.
2. F −1 (.25) is the first quartile.
3. F −1 (.75) is the third quartile.
Example Suppose that X ∼ Unif(a, b). Then,

 1 , a≤x≤b
f (x) = b − a
0,
otherwise,
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

0,
x<a


x − a
, a≤x≤b
F (x) =

b−a


1,
b < x.
and
The quantile function is q = F −1 (p). W can determine a convenient expression for F −1 by
solving the equation
q−a
p=
b−a
for q. The solution is q = a + p(b − a).
The graph of the quantile function is a line connecting the pairs (0, a) to (1, b). The median
of X is q = a + 12 (b − a) ⇒ q = a+b
.
2
The quantile function for discrete distributions
1.0
Consider X ∼ Bin(12, .3). The c.d.f. is shown below. The table below and right shows
the pairings between ranges in p and the quantiles. The ranges can be identified in the Figure as the vertical gaps between the line ends and the points vertically above the line ends.
Since F −1 (p) is defined to be the smallest value of x such that p ≤ F (x), the .4 quantile is
.
.
3 because F (2) = .253 and F (3) = .492.
0.0
0.2
0.4
F(x)
0.6
0.8
F −1 (p)
0
1
2
3
4
5
..
.
0
2
4
6
x
8
10
12
p
(0, .0138]
(.014, .085]
(.085, .253]
(.253, .492]
(.492, .724]
(.724, .882]
..
.