Slide set 12
Stat 330 (Spring 2015)
Last update: January 28, 2015
Stat 330 (Spring 2015): slide set 12
Continuous Random Variables
All properties of discrete RVs have direct counterparts for coninuous RVs.
One basic difference: summations used in the case of discrete RVs are
replaced by integrals.
Summing over (uncountable) infinite many values corresponds to an integral.
For e.g.,we define a cumulative distribution function (cdf) as follows:
Definition: CDF of a X is a continuous random variable:
The function FX (t) := P (X ≤ t) is called the cumulative distribution
function of X.
The only difference to the discrete case is that the cdf of a continuous
variable is not a stairstep function.
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Stat 330 (Spring 2015): slide set 12
Properties of FX
The following properties hold for the cumulative distribution function FX
for random variable X.
•
0 ≤ FX (t) ≤ 1 for all t ∈ R
•
FX is monotone increasing, (i.e. if x1 ≤ x2 then FX (x1) ≤ FX (x2).)
•
limt→−∞ FX (t) = 0 and limt→∞ FX (t) = 1.
However, there is slight difference from the discrete case:
Definition: Probability Density Function For a continuous variable X with
cumulative distribution function FX the density function of X is defined
as:
0
fX (x) := FX
(x).
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Stat 330 (Spring 2015): slide set 12
Properties of density function f (x)
A function fX is a density function of a random variable X, if
(i) fX (x) ≥ 0 for all x,
R∞
(ii) −∞ f (x)dx = 1.
Relationship between fX and FX
Since the density function fX is defined as the derivative of the cumulative
distribution function, we can obtain the cumulative distribution function
from the density by integrating:
Rt
• FX (t) = P (X ≤ t) = −∞ f (x)dx
Rb
• P (a ≤ X ≤ b) = a f (x)dx
It follows that
P (X = a) = P (a ≤ X ≤ a) =
Ra
a
f (x)dx = 0.
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Stat 330 (Spring 2015): slide set 12
Example: pdf
Let Y be the time until the first major failure of a new disk drive.
A possible density function for Y is
f (y) =
e−y
0
y>0
otherwise
First, we need to check, that f (y) is actually a density function. Obviously,
f (y) is a non-negative function on whole of <.
The second condition, f must fulfill to be a density of Y is
Z
∞
Z
f (y)dy =
−∞
∞
e−y dy = −e−y |∞
0 = 0 − (−1) = 1
0
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Stat 330 (Spring 2015): slide set 12
Continuing the disk drive example...
What is the probability that the first major disk drive failure occurs within
the first year?
Z
P (Y ≤ 1) =
1
e−y dy = −e−y |10 = 1 − e−1 ≈ 0.63.
0
What is the cumulative distribution function of Y ?
Z
t
FY (t) =
Z
f (y)dy =
∞
t
e−y dy = 1 − e−t for all t ≥ 0.
0
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Stat 330 (Spring 2015): slide set 12
Continuing the disk drive example...
Density and Distribution functions of the random variable Y .
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Stat 330 (Spring 2015): slide set 12
Compare discrete and continuous RVs
discrete random variable
image Im(X) finite or countable
infinite
cumulative distribution function:
P
FX (t) = P (X ≤ t) = k≤t pX (k)
probability mass function:
pX (x) = P (X = x)
expected value:
P
E[h(X)] = x h(x) · pX (x)
P
E[X] = x x · pX (x)
variance:
2
V
ar[X]
=
E[(X
−
E[X])
]
P
2
x (x − E[X]) pX (x)
continuous random variable
image Im(X) uncountable
Rt
FX (t) = P (X ≤ t) = ∞ f (x)dx
probability density function:
0
fX (x) = FX (x)
R
E[h(X)] = x h(x) · fX (x)
R∞
E[X] = −∞ x · fX (x)dx
=
2
V
ar[X]
=
E[(X
−
E[X])
] =
R∞
(x − E[X])2fX (x)dx
−∞
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Stat 330 (Spring 2015): slide set 12
Some special continuous density functions
Uniform Density
One of the most basic continuous density is the uniform density.
The pdf is:
1
b−a
if a < x < b
0
otherwise
We say that X ∼ U (a, b) i.e., the random variable X is distributed as the
Uniform distribution with parameters a and b
f (x) =
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Stat 330 (Spring 2015): slide set 12
Properties of the Uniform distribution
The cumulative distribution function FX is

 0
x−a
Ua,b(x) := FX (x) =
 b−a
1
if x ≤ a
if a < x < b
if x ≥ b.
We now compute the expected value and variance of a a uniform distribution
on (a, b).
Z
b
E[X] =
x
a
2
1 1 2b
1
dx =
x |a =
b−a
b − a2
b − a2
1
=
= (a + b).
2(b − a) 2
Z b
a+b 2 1
(b − a)2
V ar[X] =
(x −
)
dx = . . . =
.
2
b−a
12
a
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Stat 330 (Spring 2015): slide set 12
Uniform distribution: Example
The(pseudo) random number generator on my calculator is supposed to
create realizations of U (0, 1) random variables.
Define U as the next random number the calculator produces.
What is the probability, that the next number is larger than 0.85?
To answer that, we will compute P (U ≥ 0.85).
We know the density function of U : fU (u) =
Therefore
Z
1
1−0
= 1.
1
P (U ≥ 0.85) =
1du = 1 − 0.85 = 0.15.
0.85
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