Important Distributions/Functions of
Random Variables
ECE 313
Probability with Engineering Applications
Lecture 22 - November 10, 1999
Professor Ravi K. Iyer
University of Illinois
Iyer - Lecture 22
ECE 313 - Fall 1999
Normal or Gaussian Distribution
Example 1
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An analog signal received at a detector (measured in microvolts)
may be modeled as a Gaussian random variable N(200, 256) at
a fixed point in time. What is the probability that the signal will
exceed 240 microvolts? What is the probability that the signal is
larger than 240 microvolts, given that it is larger than 210
microvolts?
P( X > 240) = 1 − P( X ≤ 240)
240 − 200 

16
= 1 − FZ (2.5)
= 1 − FZ 

≈ 0.00621
Iyer - Lecture 22
ECE 313 - Fall 1999
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P( X ≥ 240 | X ≥ 210) =
P( X ) ≥ 240)
P( X ≥ 210)
240 − 200 
1 − FZ 


16
=
210 − 200 

16
1 − FZ 

0.00621
=
0.26599
≈ 0.02335
ECE 313 - Fall 1999
Normal or Gaussian Distribution
Example 1 (cont.)
Next:
Where:
 0,

2
−
(
x
−
)
µ
,

f ( x) =  1
exp
2


 2σ

ασ 2
2
∞
−
(
t
−
µ
)
1


exp 
2
 dt
σ 2π
 2σ

0
α=∫
x < 0,
x ≥ 0,
ECE 313 - Fall 1999
It is of interest to define a truncated normal density:
Normal or Gaussian Distribution
(cont.)
Iyer - Lecture 22
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Iyer - Lecture 22
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−∞
∫ f (t )dt = 1
∞
Normal or Gaussian Distribution
(cont.)
The introduction of α insures that
so that f is the density of a nonnegative random variable.
ECE 313 - Fall 1999
For µ > 3σ, the value of α is close to 1, and for most practical
purposes it may be omitted, so that the truncated normal density
reduces to the usual normal density.
Iyer - Lecture 22
Failure Rate of the Normal Distribution
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ECE 313 - Fall 1999
The normal distribution is IFR, which implies that it can be used
to model the behavior of components during the wear-out
phase.
Iyer - Lecture 22
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Normal or Gaussian Distribution
Example 2
Assuming that the life of a given subsystem, in the wear-out
phase, is normally distributed with µ = 10,000 hours and σ =
1,000 hours, determine the reliability for an operating time of
500 hours given that
Ð (a) The age of the component is 9,000 hours,
Ð (b) The age of the component is 11,000.
ECE 313 - Fall 1999
The required quantity under (a) is R9,000(500) and under (b) is
R11,000(500).
R9,000 (500) =
=
R(9, 500)
R(9, 000)
∫ f (t )dt
∞
9,500
∞
∫ f (t )dt
9,000
ECE 313 - Fall 1999
Note that with the usual exponential assumption these two
quantities will be identical.
But in the present case:
Normal or Gaussian Distribution
Example 2 (cont.)
Iyer - Lecture 22
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∞
∫ f (t )dt
µ − 0.5σ
∞
∫ f (t )dt
µ −σ
(from tables)
1 − FX ( µ − 0.5σ ) 1 − FZ ( −0.5) FZ (0.5)
=
=
FZ (1)
1 − FX ( µ − σ )
1 − FZ ( −1)
R 9,000 (500) =
=
0.6915
=
0.8413
= 0.8219
ECE 313 - Fall 1999
Noting that µ − 0.5σ = 9, 500 and µ − σ = 9, 000, we have :
Normal or Gaussian Distribution
Example 2 (cont.)
Iyer - Lecture 22
Normal or Gaussian Distribution
Example 2 (cont.)
Similarly, since µ + 1.5σ = 11, 500 and µ + σ = 11, 000, we have :
1 − FX ( µ + 1.5σ )
1 − FX ( µ + σ )
(from tables)
R11, 000 (500) =
0.0668
=
0.1587
= 0.4209
Thus, unlike the exponential assumption, R11, 000 (500) < R9, 000 (500);
that is, the subsystem has aged.
It can be shown that the normal distribution is a good approximation to
the (discrete) binomial distribution for large n, provided p is not close
ECE 313 - Fall 1999
to 0 or 1. The corresponding parameters are µ = np and σ 2 = np(1 − p).
Iyer - Lecture 22
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The Uniform or Rectangular
Distribution
a < x < b,
otherwise,
ECE 313 - Fall 1999
A continuous random variable X is said to have a uniform
distribution over the interval (a,b) if its density is given by:
1

,
f ( x) =  b − a
0,
x≥b
a ≥ x < b,
x < a,
And the distribution function is given by:
0,
x - a
F( x ) = 
,
b - a
1,
Y = Φ( X )
Knowledge of some characteristic of the system, with
knowledge of the input, allows some estimate of the behavior at
the output.
E. g., the input random variable X and its density f(x) are known
and the input-output behavior is characterized by:
Functions of a Random Variable
Iyer - Lecture 22
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ECE 313 - Fall 1999
To compute the density of the random variable Y.
We assume that Φ is continuous or piecewise continuous, so
Y = Φ( X ) will be a random variable.
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Functions of a Random Variable
Example (Y=X2)
2
ECE 313 - Fall 1999
otherwise.
y > 0,
Let Y = Φ( X ) = X
X could denote the measurement error in a
certain physical experiment, and Y would then be the square of
the error.
Note that FY ( y) = 0 for y ≤ 0. For y > 0 :
= FX ( y − FX ( − y )
= P( − y ≤ X ≤ y )
= P( X 2 ≤ y )
FY ( y) = P(Y ≤ y)
Y
1
[ f ( y ) + f X ( − y )],

f ( y) =  2 y X
 0,
And by differentiation the density of Y is:
Then:
fX ( x) =
x2
1 −2
e ,
2π
y
−∞< x <∞
y ≤ 0.
y > 0,
1 −y/2
 1  1 −y/2
+
,
e
e


fY ( y) =  2 y  2π
2π
 0,
−
 1

e 2,
fY ( y) =  2πy
 0,
ECE 313 - Fall 1999
y ≤ 0,
y > 0,
As a special case of Example 1, let X have the standard normal
distribution [N(0,1)] so that:
Functions of a Random Variable
X=Std. Normal
Iyer - Lecture 22
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We conclude that Y has a gamma distribution with α = 1/2 and
λ = 1/2.
Functions of a Random Variable
X=Std. Normal (cont.)
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ECE 313 - Fall 1999
Since GAM ( 1 2 , n / 2) = Xn2 , it follows that if X is standard normal
then Y = X2 is a chi-square distributed with one degree of
freedom.
Iyer - Lecture 22
Let X be uniformly distributed on (0,1).
Show Y = -λ-1 1n(1-X) has an exponential distribution with
parameter λ > 0.
Example, Simulation Application
X=Uniform
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Y is a nonnegative random variable implying FY(y) = 0 for y ≤ 0.
For y > 0:
FY ( y) = P(Y ≤ y) = P[ − λ−11n(1 − X ) ≤ y]
= P[1n(1 − X ) ≥ − λy]
= P[(1 − X ) ≥ e − λy ]
since e x is an increasing function of x,
= P( X ≤ 1 − e − λy )
= FX (1 − e − λy
ECE 313 - Fall 1999
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Iyer - Lecture 22
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Since X is uniform over (0, 1), FX(x) = x, 0 ≥ x ≥ 1.
Thus, FY ( y) = 1 − e − λy
Y is exponentially distributed with parameter λ.
Example (cont.), Use in Simulation
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This fact can be used in Monte Carlo simulation. In such
simulation programs, it is important to be able to generate
values of variables with known distribution functions (random
deviates).
Most computer systems provide built-in functions to generate
random deviates from the uniform distribution over (0, 1), say u.
To generate a random deviate, y, of an exponentially distributed
random variable Y with the parameter λ, then from this example
it follows that y = − λ−11n(1 − u).
ECE 313 - Fall 1999
Theorem (generalize Ex. 2, 3)
 f [Φ −1 ( y)][ (Φ −1 )′ ( y) ],
fY ( y) =  X
0,
otherwise,
y ∈ Φ( I )
Let X be a continuous random variable with density fX that is
nonzero on a subset I of real numbers (i. e., f X ( x ) > 0, x ∈ I
and f X ( x ) = 0, x ∉1.)
Let Φ be a differentiable monotone function whose domain is I
and whose range is the set of real numbers.
Then Y = Φ(X) is a continuous random variable with the density,
fY(y) given by:
Iyer - Lecture 22
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ECE 313 - Fall 1999
Where Φ-1 is the uniquely defined inverse of Φ and (Φ-1)Õ is the
derivative of the inverse function.
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Proof:
Theorem (cont.)
Ð We prove the theorem assuming that Φ(x) is an increasing function
of x. The proof for the other case follows in a similar way.
since Φ is monotone increasing
FY ( y) = P(Y ≤ y) = P[Φ( X ) ≤ y]
= P[ X ≤ Φ −1 ( y)],
= FX [Φ −1 ( y)]
ECE 313 - Fall 1999
Taking derivatives and using the chain rule, we get the required
result.
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Now let φ be the distribution function, F, of a random variable, X, with
density f.
Example 1, Simulation Issues
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Applying the theorem:
1, 0 < y < 1,
fY ( y) = 
0, otherwise.
Therefore, the random variable Y=F(X) has the density given by:
Y = F( X ) and FY ( y) = FX ( FX−1 ( y)) = y.
FY ( y) = FX (φ −1 ( y))
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ECE 313 - Fall 1999
In other words, if X is a continuous random variable with CDF of F, then
the new random variable Y=F(X) is uniformly distributed over the
interval (0,1).
Iyer - Lecture 22
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Example 1, Simulation Issues (cont.)
ECE 313 - Fall 1999
This idea can be used to generate a random deviate x of X by:
1. Generating a random number u from a uniform distribution of (0,1)
and then
2. Using the relation x=F-1(u) as illustrated in the figure below.
The generation of the (0,1) random number can be done easily.
Question is whether x=F-1(u) can be expressed in a closed
mathematical form. This is possible for distributions such as the
exponential; for distributions such as the normal, we must use other
techniques.
Iyer - Lecture 22