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Chapter 4, Sections 12 Continuous Random Variables

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Title: Chapter 4, Sections 12 Continuous Random Variables


1
Chapter 4, Sections 1-2Continuous Random
Variables
  • Definition
  • Probability Distribution

? John J Currano, 03/21/2009
2
A random variable a real-valued function, Y, on
a sample space. Support of Y the set of real
numbers that Y can actually assume. Discrete
random variable finite or countably infinite
support. We calculated probabilities associated
with discrete random variables by assigning
positive probabilities to the sample points, thus
defining the discrete random variables
probability function, and summing to get the
probabilities of other events.
3
  • Now consider random variables which can assume
    any real value, or, at least, any real value in
    some interval. For example
  • a persons height
  • a persons weight
  • the time to recovery from a disease
  • the lifetime of a washing machine, etc.
  • Section 4.1 of the text has a good introduction
    to the topic.
  • To get some insight into how to handle these
    random variables (where probabilities cannot be
    summed as in the discrete case since the support
    is uncountable), we shall define the (cumulative)
    distribution function (cdf) of a random variable
    and investigate it for discrete random variables.

4
Definition. The (cumulative) distribution
function (cdf) of a RV, Y, is the function F (y)
P(Y ? y) its domain is the set of all real
numbers. Note For a discrete random variable, Y,
.
Theorem. Properties of a Distribution Function,
F (y).
  • 0 ? F (y) ? 1 for ?? lt y lt ?

5
Example.
The distribution function (cdf) of Y is
Notice that in this example, F(y) is never 1, yet
6
Definition. The (cumulative) distribution
function (cdf) of a RV, Y, is the function F (y)
P(Y ? y) its domain is the set of all real
numbers.
Theorem. Properties of a Distribution Function,
F (y).
4. F(y) is a nondecreasing function of y. That
is, if y1 and y2 are any values such that
y1 lt y2, then F (y1) ? F (y2).
(??, y1 ? (??, y2, so F(y1) P(Y?(??, y1 ) ?
P(Y?(??, y2 ) F(y2)
5. F (y) is continuous from the right at each y
in (??, ?) that is, for every real number y,
(??,t ? (??,y as t ? y, so F(t) P(Y?(??,t
) ? P(Y?(??,y ) F(y)
6. P (a lt Y ? b) P (Y ? b) ? P (Y ? a)
F (b) ? F (a). This may be different from P (a
? Y ? b) P (Y ? b) ? P (Y lt a).
7
Definition. The (cumulative) distribution
function (cdf) of a RV, Y, is the function F (y)
P(Y ? y) its domain is the set of all real
numbers.
Definition. Let Y be a random variable with
distribution function F(y). Y is said to be a
continuous random variable if the distribution
function F(y) is continuous for ? lt y lt ? and if
the derivative of F(y) exists and is continuous
except for, at most, a finite number of points in
any finite interval.
8
Definition. Y is a continuous random variable if
its distribution function F(y) is continuous for
? lt y lt ? and if F ?(y) exists and is continuous
except for, at most, a finite of points in any
finite interval.
Theorem. If Y is a continuous random variable,
then P(Y y) 0 for every real number y. Thus,
P(Y y) P(Y lt y) for all real numbers y.
Otherwise, F(y) would have a jump discontinuity
at any point y0 for which P(Y y0) gt 0.
Definition. Let Y be a continuous random
variable with distribution function F(y). Then
the function f (y) F ?(y) is called the
(probability) density function (pdf) of Y. Its
domain is the set of all y for which the
derivative exists, but the domain can be (and
usually is) extended to the set of all real
numbers by defining its values at the other real
numbers to be any convenient values.
9
Relations Between F(y) and f (y) and Properties
of f (y)
  • F ?(y) f (y) for ? lt y lt ?. This is
    just the definition of f (y).
  • f (y) ? 0 for ? lt y lt ?, since F
    nondecreasing and f (y) F ?(y)
  • F(y) F(a) by Fundamental
    Theorem of Calculus
  • P(Y ? y) P(Y ? a) by the definition of the
    cdf, F
  • P(a lt Y ? y) by the additivity of P
  • P(a ? Y ? y) since P(Y a) 0

In particular, is the area
bounded above by the graph of f, below by the
horizontal axis, and on the sides by the vertical
lines through a and b.
10
Relations Between F(y) and f (y) and Properties
of f (y)
d) so we have the following relations
between F (y) and f (y)
e)
To check if a function f (y) is a density
function, check that (1) f (y) ? 0 for
? lt y lt ? and (2)
11
  • Example 1. (p. 168 4.16) Let
  • where c is a constant.
  • Find c.
  • Find F (y).
  • Graph f (y) and F (y).
  • Use F (y) in (b) to find P (1 ? Y ? 2).
  • Use the geometric figure for f (y) to calculate P
    (1 ? Y ? 2).

In doing several parts of this problem (and
similar problems), in particular (a), (b), and
(d), it is useful to rewrite the definition of f
(y) as follows to make it clear that it is
defined piecewise in 3 pieces
12
It is often more convenient if the right endpoint
of each interval is in the interval we can do
this provided the distribution is continuous.
12
13
(c)
(a) Find c Method 1. Use the fact that

14
(c)
(a) Find c Method 2. Use the fact that the
area beneath the graph of f (y), a triangle
here, must be 1
15
(c)
(b) Find F (y) Since f (y) is defined piecewise
in 3 pieces
and, if 0 lt y ? 2, then
16
(c)
(b)
(c) Graph F (y)
17
(d) Use F (y) to find
(e) Use the geometric figure for f (y) to
calculate P (1 ? Y ? 2).
18
Definition. Let 0 ?? p ? 1. The pth quantile ?p
of a RV, Y, (also called the 100pth percentile of
Y) is the smallest value such that P(Y ? ?p)
F(?p)?? p. The median of Y is ?0.50, the 50th
percentile. The first and third quartiles are the
25th- and 75th-percentiles, respectively. Notes
1. For a continuous random variable Y, ?p is
the smallest value such that P(Y ? ?p) F(?p)?
p. This is not necessarily true for discrete
RVs. 2. To find ?p for a continuous random
variable Y and a given value of p, try to solve
the equation F(y)? pfor y ?p.
19
In Example 2, we found the formula above for
F(y). Suppose we need to find the 75th percentile
(the 0.75 quantile or the third quartile). Since
F(0) 0 and F(1) 1, ?0.75 lies between y 0
and y 2. So set the formula for F(y) in this
interval equal to 0.75 and solve for y
The solutions to this quadratic equation are y
1 and y 3. Since ?0.75 is in the interval
(0, 2), we obtain ?0.75 1. This agrees with the
result in part (e). Exercise. Find the median of
Y.
20
Example 4.2, p. 162 Given F(y), find f(y).
21
(a) We need to check (1) Nonnegativity (2) f
(y) integrates to 1
22
  • Example 3. Let
  • p. 168 4.14a Graph f (y).
  • p. 168 4.14b Find the distribution function,
    F (y).
  • Find using the cdf, F (y).
  • Find using the pdf, f (y).

b) Geometrically 0 lt y lt 1 1 ? y lt
2
or F (y) 1 ? area of
23
Example 4. The density function of a continuous
random variable Y is
, where k is a constant.
  • Find k.
  • Find P (Y ? 2).

b) P (Y 2)
Complete solutions for Examples 2 4 are in the
document 4.02.Examples.pdf on the faculty web.
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