Probability Density Functions
Probability Density Functions
Definition
Let X be a continuous rv. Then a probability distribution or
probability density function (pdf) of X is a function f (x) such
that for any two numbers a and b with a ≤ b,
Z
P(a ≤ X ≤ b) =
b
f (x)dx
a
That is, the probability that X takes on a value in the interval
[a, b] is the area above this interval and under the graph of the
density function. The graph of f (x) is often referred to as the
density curve.
Probability Density Functions
Probability Density Functions
Figure: P(60 ≤ X ≤ 70)
Probability Density Functions
Probability Density Functions
Definition
A continuous rv X is said to have a uniform distribution on the
interval [A, B], if the pdf of X is
(
1
A≤x ≤B
f (x; A, B) = B−A
0
otherwise
Probability Density Functions
Definition
A continuous rv X is said to have a uniform distribution on the
interval [A, B], if the pdf of X is
(
1
A≤x ≤B
f (x; A, B) = B−A
0
otherwise
The graph of any uniform pdf looks like the graph in the previous
example:
Cumulative Distribution Functions
Cumulative Distribution Functions
Definition
The cumulative distribution function F (x) for a continuous rv
X is defined for every number x by
Z x
F (x) = P(X ≤ x) =
f (y )dy
−∞
For each x, F (x) is the area under the density curve to the left of
x.
Cumulative Distribution Functions
Cumulative Distribution Functions
Proposition
Let X be a continuous rv with pdf f (x) and cdf F (x). Then for
any number a,
P(X > a) = 1 − F (a)
and for any two numbers a and b with a < b,
P(a ≤ X ≤ b) = F (b) − F (a).
Cumulative Distribution Functions
Proposition
Let X be a continuous rv with pdf f (x) and cdf F (x). Then for
any number a,
P(X > a) = 1 − F (a)
and for any two numbers a and b with a < b,
P(a ≤ X ≤ b) = F (b) − F (a).
Cumulative Distribution Functions
Cumulative Distribution Functions
Proposition
If X is a continuous rv with pdf f (x) and cdf F (x), then at every x
at which the derivative F 0 (x) exists, F 0 (x) = f (x).
Cumulative Distribution Functions
Cumulative Distribution Functions
Definition
The expected value or mean valued of a continuous rv X with
pdf f (x) is
Z ∞
µX = E (X ) =
x · f (x)dx
−∞
Cumulative Distribution Functions
Definition
The expected value or mean valued of a continuous rv X with
pdf f (x) is
Z ∞
µX = E (X ) =
x · f (x)dx
−∞
Definition
The variance of a continuous random variable X with pdf f (x)
and mean value µ is
Z ∞
2
σX = V (X ) =
(x − µ)2 · f (x)dx = E [(X − µ)2 ]
−∞
The standard deviation (SD) of X is σX =
p
V (X ).
Cumulative Distribution Functions
Cumulative Distribution Functions
Definition
Let p be a number between 0 and 1. The (100p)th percentile of
the distribution of a continuous rv X , denoted by η(p), is defined
by
Z η(p)
p = F (η(p)) =
f (y )dy
−∞
Normal Distribution
Normal Distribution
Definition
A continuous rv X is said to have a normal distribution with
parameter µ and σ (µ and σ 2 ), where −∞ < µ < ∞ and σ > 0, if
the pdf of X is
f (x; µ, σ) = √
1
2
2
e −(x−µ) /(2σ )
2πσ
We use the notation X ∼ N(µ, σ 2 ) to denote that X is rormally
distributed with parameters µ and σ 2 .
Normal Distribution
Definition
A continuous rv X is said to have a normal distribution with
parameter µ and σ (µ and σ 2 ), where −∞ < µ < ∞ and σ > 0, if
the pdf of X is
f (x; µ, σ) = √
1
2
2
e −(x−µ) /(2σ )
2πσ
We use the notation X ∼ N(µ, σ 2 ) to denote that X is rormally
distributed with parameters µ and σ 2 .
Remark:
1. Obviously, f (x) ≥ 0 for
R ∞all x;1 −(y −µ)2 /(2σ2 )
dy = 1.
2. It is guaranteed that −∞ √2πσ e
Normal Distribution
Normal Distribution
Proposition
For X ∼ N(µ, σ 2 ), we have
E (X ) = µ and V (X ) = σ 2
Normal Distribution
Proposition
For X ∼ N(µ, σ 2 ), we have
E (X ) = µ and V (X ) = σ 2
σ=1
σ=2
σ = 0.5
Normal Distribution
Normal Distribution
The cdf of a normal random variable X is
Z x
F (x) = P(X ≤ x) =
f (y ; µ, σ)dy
−∞
Z x
1
2
2
√
e −(y −µ) /(2σ ) dy
=
2πσ
−∞ Z
x−µ
1
2
2
=√
e −(z) /(2σ ) dz
change of variable:z = y − µ
2πσ −∞
Z x−µ
σ
1
z
2
=√
e −(w ) /2 · σdw
change of variable:w =
σ
2πσ −∞
Z x−µ
σ
1
2
√ e −(w ) /2 dw
=
2π
−∞
Normal Distribution
Normal Distribution
Definition
The normal distribution with parameter values µ = 0 and σ = 1 is
called the standard normal distribution. A random variable
having a standard normal distribution is called a standard normal
random variable and will be denoted by Z . The pdf of Z is
1
2
f (z; 0, 1) = √ e −z /2
2π
−∞<z <∞
The graph of f (z; 0, 1) is called
R z the standard normal (or z) curve.
The cdf of Z is P(Z ≤ z) = −∞ f (y ; 0, 1)dy , which we will
denote by Φ(z).
Normal Distribution
Normal Distribution
Shaded area = Φ(0.5)
Normal Distribution
Normal Distribution
Table A.3
z
···
-1.2
-1.1
···
1.6
1.7
···
Standard Normal Curve Areas
.00
···
0.1151
0.1357
···
0.9452
0.9554
···
.01
···
0.1131
0.1335
···
0.9463
0.9564
···
.02
···
0.1112
0.1314
···
0.9474
0.9573
···
.03
···
0.1094
0.1292
···
0.9484
0.9582
···
.04
···
0.1075
0.1271
···
0.9495
0.9591
···
···
···
···
···
···
···
···
···
.09
···
0.0985
0.1170
···
0.9545
0.9633
···
Normal Distribution
Normal Distribution
Z ∼ N(0, 1), calculate (a)P(Z ≤ 1.61); (b)P(Z > −1.12); and
(c)P(−1.12 < Z ≤ 1.61).
Normal Distribution
Z ∼ N(0, 1), calculate (a)P(Z ≤ 1.61); (b)P(Z > −1.12); and
(c)P(−1.12 < Z ≤ 1.61).
z
···
-1.2
-1.1
···
1.6
1.7
···
.00
···
0.1151
0.1357
···
0.9452
0.9554
···
.01
···
0.1131
0.1335
···
0.9463
0.9564
···
.02
···
0.1112
0.1314
···
0.9474
0.9573
···
.03
···
0.1094
0.1292
···
0.9484
0.9582
···
.04
···
0.1075
0.1271
···
0.9495
0.9591
···
···
···
···
···
···
···
···
···
.09
···
0.0985
0.1170
···
0.9545
0.9633
···
Normal Distribution
Z ∼ N(0, 1), calculate (a)P(Z ≤ 1.61); (b)P(Z > −1.12); and
(c)P(−1.12 < Z ≤ 1.61).
z
···
-1.2
-1.1
···
1.6
1.7
···
.00
···
0.1151
0.1357
···
0.9452
0.9554
···
.01
···
0.1131
0.1335
···
0.9463
0.9564
···
P(Z ≤ 1.61) = 0.9463;
.02
···
0.1112
0.1314
···
0.9474
0.9573
···
.03
···
0.1094
0.1292
···
0.9484
0.9582
···
.04
···
0.1075
0.1271
···
0.9495
0.9591
···
···
···
···
···
···
···
···
···
.09
···
0.0985
0.1170
···
0.9545
0.9633
···
Normal Distribution
Z ∼ N(0, 1), calculate (a)P(Z ≤ 1.61); (b)P(Z > −1.12); and
(c)P(−1.12 < Z ≤ 1.61).
z
···
-1.2
-1.1
···
1.6
1.7
···
.00
···
0.1151
0.1357
···
0.9452
0.9554
···
.01
···
0.1131
0.1335
···
0.9463
0.9564
···
.02
···
0.1112
0.1314
···
0.9474
0.9573
···
.03
···
0.1094
0.1292
···
0.9484
0.9582
···
.04
···
0.1075
0.1271
···
0.9495
0.9591
···
···
···
···
···
···
···
···
···
.09
···
0.0985
0.1170
···
0.9545
0.9633
···
P(Z ≤ 1.61) = 0.9463;
P(Z > −1.12) = 1 − P(Z ≤ −1.12) = 1 − 0.1314 = 0.8686;
Normal Distribution
Z ∼ N(0, 1), calculate (a)P(Z ≤ 1.61); (b)P(Z > −1.12); and
(c)P(−1.12 < Z ≤ 1.61).
z
···
-1.2
-1.1
···
1.6
1.7
···
.00
···
0.1151
0.1357
···
0.9452
0.9554
···
.01
···
0.1131
0.1335
···
0.9463
0.9564
···
.02
···
0.1112
0.1314
···
0.9474
0.9573
···
.03
···
0.1094
0.1292
···
0.9484
0.9582
···
.04
···
0.1075
0.1271
···
0.9495
0.9591
···
···
···
···
···
···
···
···
···
.09
···
0.0985
0.1170
···
0.9545
0.9633
···
P(Z ≤ 1.61) = 0.9463;
P(Z > −1.12) = 1 − P(Z ≤ −1.12) = 1 − 0.1314 = 0.8686;
P(−1.12 < Z ≤ 1.61) = P(Z ≤ 1.61) − P(Z ≤ −1.12) =
0.9463 − 0.1314 = 0.8149.
Normal Distribution
Normal Distribution
Many tables for the normal distribution contain only the
nonnegative part.
Normal Distribution
Many tables for the normal distribution contain only the
nonnegative part.
z
···
1.6
1.7
···
.00
···
0.9452
0.9554
···
.01
···
0.9463
0.9564
···
What is P(Z < −1.63)?
.02
···
0.9474
0.9573
···
.03
···
0.9484
0.9582
···
.04
···
0.9495
0.9591
···
···
···
···
···
···
.09
···
0.9545
0.9633
···
Normal Distribution
Many tables for the normal distribution contain only the
nonnegative part.
z
···
1.6
1.7
···
.00
···
0.9452
0.9554
···
.01
···
0.9463
0.9564
···
.02
···
0.9474
0.9573
···
.03
···
0.9484
0.9582
···
.04
···
0.9495
0.9591
···
···
···
···
···
···
.09
···
0.9545
0.9633
···
What is P(Z < −1.63)?
By symmetry of the pdf of Z , we know that P(Z < −1.63) =
P(Z > 1.63) = 1 − P(Z ≤ 1.63) = 1 − 0.9484 = 0.0516
Normal Distribution
Many tables for the normal distribution contain only the
nonnegative part.
z
···
1.6
1.7
···
.00
···
0.9452
0.9554
···
.01
···
0.9463
0.9564
···
.02
···
0.9474
0.9573
···
.03
···
0.9484
0.9582
···
.04
···
0.9495
0.9591
···
···
···
···
···
···
.09
···
0.9545
0.9633
···
What is P(Z < −1.63)?
By symmetry of the pdf of Z , we know that P(Z < −1.63) =
P(Z > 1.63) = 1 − P(Z ≤ 1.63) = 1 − 0.9484 = 0.0516
Normal Distribution
Normal Distribution
Recall: The (100p)th percentile of the distribution of a continuous
rv X , η(p), is defined by
Z
η(p)
p = F (η(p)) =
f (y )dy
−∞
Normal Distribution
Recall: The (100p)th percentile of the distribution of a continuous
rv X , η(p), is defined by
Z
η(p)
p = F (η(p)) =
f (y )dy
−∞
Similarly, the (100p)th percentile of the standard normal rv Z is
defined by
Z η(p)
1
2
√ e −y /2 dy
p = F (η(p)) =
2π
−∞
Normal Distribution
Recall: The (100p)th percentile of the distribution of a continuous
rv X , η(p), is defined by
Z
η(p)
p = F (η(p)) =
f (y )dy
−∞
Similarly, the (100p)th percentile of the standard normal rv Z is
defined by
Z η(p)
1
2
√ e −y /2 dy
p = F (η(p)) =
2π
−∞
We need to use the table for normal distribution to find (100p)th
percentile.
Normal Distribution
Normal Distribution
e.g. Find the 95th percentile for the standard normal rv Z
Normal Distribution
e.g. Find the 95th percentile for the standard normal rv Z
z
···
1.6
1.7
···
.00
···
0.9452
0.9554
···
.01
···
0.9463
0.9564
···
.02
···
0.9474
0.9573
···
.03
···
0.9484
0.9582
···
.04
···
0.9495
0.9591
···
0.5
···
0.9505
0.9599
···
···
···
···
···
···
.09
···
0.9545
0.9633
···
Normal Distribution
e.g. Find the 95th percentile for the standard normal rv Z
z
···
1.6
1.7
···
.00
···
0.9452
0.9554
···
.01
···
0.9463
0.9564
···
.02
···
0.9474
0.9573
···
.03
···
0.9484
0.9582
···
.04
···
0.9495
0.9591
···
0.5
···
0.9505
0.9599
···
η(95) = 1.645, a linear interpolation of 1.64 and 1.65.
···
···
···
···
···
.09
···
0.9545
0.9633
···
Normal Distribution
e.g. Find the 95th percentile for the standard normal rv Z
z
···
1.6
1.7
···
.00
···
0.9452
0.9554
···
.01
···
0.9463
0.9564
···
.02
···
0.9474
0.9573
···
.03
···
0.9484
0.9582
···
.04
···
0.9495
0.9591
···
0.5
···
0.9505
0.9599
···
···
···
···
···
···
.09
···
0.9545
0.9633
···
η(95) = 1.645, a linear interpolation of 1.64 and 1.65.
Remark: If p does not appear in the table, we can either use the
number closest to it, or use the linear interpolation of the closest
two.
Normal Distribution
Normal Distribution
In statistical inference, the percentiles corresponding to right small
tails are heavily used.
Notation
zα will denote the value on the z axis for which α of the area
under the z curve lies to the right of zα .
Normal Distribution
In statistical inference, the percentiles corresponding to right small
tails are heavily used.
Notation
zα will denote the value on the z axis for which α of the area
under the z curve lies to the right of zα .
zα
Normal Distribution
Normal Distribution
Remark:
1. zα is the 100(1 − α)th percentile of the standard normal
distribution.
Normal Distribution
Remark:
1. zα is the 100(1 − α)th percentile of the standard normal
distribution.
2. By symmetry the area under the standard normal curve to the
left of −zα is also α.
Normal Distribution
Remark:
1. zα is the 100(1 − α)th percentile of the standard normal
distribution.
2. By symmetry the area under the standard normal curve to the
left of −zα is also α.
3. The zα s are usually referred to as z critical values.
Percentile
α (tail area)
zα
90
0.1
1.28
95
0.05
1.645
97.5
0.025
1.96
···
···
···
99.95
0.0005
3.27
Normal Distribution
Normal Distribution
Proposition
If X has a normal distribution with mean µ and stadard deviation
σ, then
X −µ
Z=
σ
has a standard normal distribution. Thus
a−µ
b−µ
≤Z ≤
)
σ
σ
b−µ
a−µ
= Φ(
) − Φ(
)
σ
σ
P(a ≤ X ≤ b) = P(
P(X ≤ a) = Φ(
a−µ
)
σ
P(X ≥ b) = 1 − Φ(
b−µ
)
σ
Normal Distribution
Normal Distribution
Example (Problem 38):
There are two machines available for cutting corks intended for use
in wine bottles. The first produces corks with diameters that are
normally distributed with mean 3cm and standard deviation 0.1cm.
The second produces corks with diameters that have a normal
distribution with mean 3.04cm and standard deviation 0.02cm.
Acceptable corks have diameters between 2.9cm and 3.1cm.
Which machine is more likely to produce an acceptable cork?
Normal Distribution
Example (Problem 38):
There are two machines available for cutting corks intended for use
in wine bottles. The first produces corks with diameters that are
normally distributed with mean 3cm and standard deviation 0.1cm.
The second produces corks with diameters that have a normal
distribution with mean 3.04cm and standard deviation 0.02cm.
Acceptable corks have diameters between 2.9cm and 3.1cm.
Which machine is more likely to produce an acceptable cork?
2.9 − 3
3.1 − 3
≤Z ≤
)
0.1
0.1
= P(−1 ≤ Z ≤ 1) = 0.8413 − 0.1587 = 0.6826
2.9 − 3.04
3.1 − 3.04
P(2.9 ≤ X2 ≤ 3.1) = P(
≤Z ≤
)
0.02
0.02
= P(−7 ≤ Z ≤ 3) = 0.9987 − 0 = 0.9987
P(2.9 ≤ X1 ≤ 3.1) = P(
Normal Distribution
Normal Distribution
Example (Problem 44):
If bolt thread length is normally distributed, what is the probability
that the thread length of a randomly selected bolt is (a)within 1.5
SDs of its mean value? (b)between 1 and 2 SDs from its mean
value?
Normal Distribution
Example (Problem 44):
If bolt thread length is normally distributed, what is the probability
that the thread length of a randomly selected bolt is (a)within 1.5
SDs of its mean value? (b)between 1 and 2 SDs from its mean
value?
µ − 1.5σ − µ
µ + 1.5σ − µ
≤Z ≤
)
σ
σ
= P(−1.5 ≤ Z ≤ 1.5)
P(µ − 1.5σ ≤ X1 ≤ µ + 1.5σ) = P(
= 0.9332 − 0.0668 = 0.8664
Normal Distribution
Example (Problem 44):
If bolt thread length is normally distributed, what is the probability
that the thread length of a randomly selected bolt is (a)within 1.5
SDs of its mean value? (b)between 1 and 2 SDs from its mean
value?
µ − 1.5σ − µ
µ + 1.5σ − µ
≤Z ≤
)
σ
σ
= P(−1.5 ≤ Z ≤ 1.5)
P(µ − 1.5σ ≤ X1 ≤ µ + 1.5σ) = P(
= 0.9332 − 0.0668 = 0.8664
µ+σ−µ
µ + 2σ − µ
≤Z ≤
)
σ
σ
= 2P(1 ≤ Z ≤ 2)
2 · P(µ + σ ≤ X1 ≤ µ + 2σ) = 2P(
= 2(0.9772 − 0.8413) = 0.0.2718
Normal Distribution
Normal Distribution
Proposition
{(100p)th percentile for N(µ, σ 2 )} =
µ + {(100p)th percentile for N(0, 1)} · σ
Normal Distribution
Proposition
{(100p)th percentile for N(µ, σ 2 )} =
µ + {(100p)th percentile for N(0, 1)} · σ
Example (Problem 39)
The width of a line etched on an integrated circuit chip is normally
distributed with mean 3.000 µm and standard deviation 0.140.
What width value separates the widest 10% of all such lines from
the other 90%?
Normal Distribution
Proposition
{(100p)th percentile for N(µ, σ 2 )} =
µ + {(100p)th percentile for N(0, 1)} · σ
Example (Problem 39)
The width of a line etched on an integrated circuit chip is normally
distributed with mean 3.000 µm and standard deviation 0.140.
What width value separates the widest 10% of all such lines from
the other 90%?
ηN(3,0.1402 ) (90) = 3.0+0.140·ηN(0,1) (90) = 3.0+0.140·1.28 = 3.1792
Normal Distribution
Normal Distribution
Proposition
Let X be a binomial rv based on n trials with success probability p.
Then if the binomial probability histogram is not too skewed, X has
√
approximately a normal distribution with µ = np and σ = npq,
where q = 1 − p. In particular, for x = a posible value of X ,
area under the normal curve
P(X ≤ x) = B(x; n, p) ≈
to the left of x+0.5
x+0.5 − np
= Φ( √
)
npq
In practice, the approximation is adequate provided that both
np ≥ 10 and nq ≥ 10, since there is then enough symmetry in the
underlying binomial distribution.
Normal Distribution
Normal Distribution
A graphical explanation for
P(X ≤ x) = B(x; n, p) ≈
= Φ(
area under the normal curve
to the left of x+0.5
x+0.5 − np
)
√
npq
Normal Distribution
A graphical explanation for
P(X ≤ x) = B(x; n, p) ≈
= Φ(
area under the normal curve
to the left of x+0.5
x+0.5 − np
)
√
npq
Normal Distribution
Normal Distribution
Example (Problem 54)
Suppose that 10% of all steel shafts produced by a certain process
are nonconforming but can be reworked (rather than having to be
scrapped). Consider a random sample of 200 shafts, and let X
denote the number among these that are nonconforming and can
be reworked. What is the (approximate) probability that X is
between 15 and 25 (inclusive)?
Normal Distribution
Example (Problem 54)
Suppose that 10% of all steel shafts produced by a certain process
are nonconforming but can be reworked (rather than having to be
scrapped). Consider a random sample of 200 shafts, and let X
denote the number among these that are nonconforming and can
be reworked. What is the (approximate) probability that X is
between 15 and 25 (inclusive)?
In this problem n = 200, p = 0.1 and q = 1 − p = 0.9. Thus
np = 20 > 10 and nq = 180 > 10
Normal Distribution
Example (Problem 54)
Suppose that 10% of all steel shafts produced by a certain process
are nonconforming but can be reworked (rather than having to be
scrapped). Consider a random sample of 200 shafts, and let X
denote the number among these that are nonconforming and can
be reworked. What is the (approximate) probability that X is
between 15 and 25 (inclusive)?
In this problem n = 200, p = 0.1 and q = 1 − p = 0.9. Thus
np = 20 > 10 and nq = 180 > 10
P(15 ≤ X ≤ 25) = Bin(25; 200, 0.1) − Bin(14; 200, 0.1)
25 + 0.5 − 20
15 + 0.5 − 20
≈ Φ( √
) − Φ( √
)
200 · 0.1 · 0.9
200 · 0.1 · 0.9
= Φ(0.3056) − Φ(−0.2500)
= 0.6217 − 0.4013
= 0.2204
Exponential Distribution
Exponential Distribution
Definition
X is said to have an exponential distribution with parameter
λ(λ > 0) if the pdf of X is
(
λe −λx x ≥ 0
f (x; λ) =
0
otherwise
Exponential Distribution
Definition
X is said to have an exponential distribution with parameter
λ(λ > 0) if the pdf of X is
(
λe −λx x ≥ 0
f (x; λ) =
0
otherwise
Remark:
1. Usually we use X ∼ EXP(λ) to denote that the random variable
X has an exponential distribution with parameter λ.
Exponential Distribution
Definition
X is said to have an exponential distribution with parameter
λ(λ > 0) if the pdf of X is
(
λe −λx x ≥ 0
f (x; λ) =
0
otherwise
Remark:
1. Usually we use X ∼ EXP(λ) to denote that the random variable
X has an exponential distribution with parameter λ.
2. In some sources, the pdf of exponential distribution is given by
(
1 − θx
x ≥0
e
f (x; θ) = θ
0
otherwise
The difference is that λ → 1θ .
Exponential Distribution
Exponential Distribution
Exponential Distribution
Exponential Distribution
Proposition
If X ∼ EXP(λ), then
E (X ) =
1
λ
and
V (X ) =
1
λ2
And the cdf for X is
(
1 − e −λx
F (x; λ) =
0
x ≥0
x <0
Exponential Distribution
Exponential Distribution
Proof:
Z
E (X ) =
0
=
=
=
=
=
∞
xλe −λx dx
Z
1 ∞
(λx)e −λx d(λx)
λ 0
Z
1 ∞ −y
ye dy
y = λx
λ 0
Z ∞
1
−y ∞
[−ye |0 +
e −y dy ] integration by parts: u = y , v = −e −y
λ
0
1
[0 + (−e −y |∞
0 )]
λ
1
λ
Exponential Distribution
Exponential Distribution
Proof (continued):
Z ∞
E (X 2 ) =
x 2 λe −λx dx
0
Z ∞
1
= 2
(λx)2 e −λx d(λx)
λ 0
Z ∞
1
= 2
y 2 e −y dy
y = λx
λ 0
Z ∞
1
= 2 [−y 2 e −y |∞
2ye −y dy ]
integration by parts
0 +
λ
0
Z ∞
1
= 2 [0 + 2(−ye −y |∞
+
e −y dy )] integration by parts
0
λ
0
1
−y ∞
= 2 2[0 + (−ye |0 )]
λ
2
= 2
λ
Exponential Distribution
Exponential Distribution
Proof (continued):
2
1
1
V (X ) = E (X 2 ) − [E (X )]2 = 2 − ( )2 = 2
λ
λ
λ
Z x
−λy
λe
dy
F (x) =
0
Z x
=
e −λy d(λy )
0
Z λx
=
e −z dz
0
= −e −z |λx
0
= 1 − e −λx
z = λy
Exponential Distribution
Exponential Distribution
Example (Problem 108)
The article “Determination of the MTF of Positive Photoresists
Using the Monte Carlo method” (Photographic Sci. and
Engr., 1983: 254-260) proposes the exponential distribution
with parameter λ = 0.93 as a model for the distribution of a
photon’s free path length (µm) under certain circumstances.
Suppose this is the correct model.
Exponential Distribution
Example (Problem 108)
The article “Determination of the MTF of Positive Photoresists
Using the Monte Carlo method” (Photographic Sci. and
Engr., 1983: 254-260) proposes the exponential distribution
with parameter λ = 0.93 as a model for the distribution of a
photon’s free path length (µm) under certain circumstances.
Suppose this is the correct model.
a. What is the expected path length, and what is the standard
deviation of path length?
Exponential Distribution
Example (Problem 108)
The article “Determination of the MTF of Positive Photoresists
Using the Monte Carlo method” (Photographic Sci. and
Engr., 1983: 254-260) proposes the exponential distribution
with parameter λ = 0.93 as a model for the distribution of a
photon’s free path length (µm) under certain circumstances.
Suppose this is the correct model.
a. What is the expected path length, and what is the standard
deviation of path length?
b. What is the probability that path length exceeds 3.0?
Exponential Distribution
Example (Problem 108)
The article “Determination of the MTF of Positive Photoresists
Using the Monte Carlo method” (Photographic Sci. and
Engr., 1983: 254-260) proposes the exponential distribution
with parameter λ = 0.93 as a model for the distribution of a
photon’s free path length (µm) under certain circumstances.
Suppose this is the correct model.
a. What is the expected path length, and what is the standard
deviation of path length?
b. What is the probability that path length exceeds 3.0?
c. What value is exceeded by only 10% of all path lengths?
Exponential Distribution
Exponential Distribution
Proposition
Suppose that the number of events occurring in any time interval
of length t has a Poisson distribution with parameter αt (where α,
the rate of the event process, is the expected number of events
occurring in 1 unit of time) and that numbers of occurrences in
nonoverlappong intervals are independent of one another. Then
the distribution of elapsed time between the occurrence of two
successive events is exponential with parameter λ = α.
Exponential Distribution
Proposition
Suppose that the number of events occurring in any time interval
of length t has a Poisson distribution with parameter αt (where α,
the rate of the event process, is the expected number of events
occurring in 1 unit of time) and that numbers of occurrences in
nonoverlappong intervals are independent of one another. Then
the distribution of elapsed time between the occurrence of two
successive events is exponential with parameter λ = α.
e.g.
the number of customers visiting Costco in each hour =⇒
Poisson distribution;
Exponential Distribution
Proposition
Suppose that the number of events occurring in any time interval
of length t has a Poisson distribution with parameter αt (where α,
the rate of the event process, is the expected number of events
occurring in 1 unit of time) and that numbers of occurrences in
nonoverlappong intervals are independent of one another. Then
the distribution of elapsed time between the occurrence of two
successive events is exponential with parameter λ = α.
e.g.
the number of customers visiting Costco in each hour =⇒
Poisson distribution;
the time between every two successive customers visiting Costco
=⇒ Exponential distribution.
Exponential Distribution
Exponential Distribution
Example (Example 4.22)
Suppose that calls are received at a 24-hour hotline according to a
Poisson process with rate α = 0.5 call per day.
Exponential Distribution
Example (Example 4.22)
Suppose that calls are received at a 24-hour hotline according to a
Poisson process with rate α = 0.5 call per day.
Then the number of days X between successive calls has an
exponential distribution with parameter value 0.5.
Exponential Distribution
Example (Example 4.22)
Suppose that calls are received at a 24-hour hotline according to a
Poisson process with rate α = 0.5 call per day.
Then the number of days X between successive calls has an
exponential distribution with parameter value 0.5.
The probability that more than 3 days elapse between calls is
P(X > 3) = 1 − P(X ≤ 3) = 1 − F (3; 0.5) = e −0.5·3 = 0.223.
Exponential Distribution
Example (Example 4.22)
Suppose that calls are received at a 24-hour hotline according to a
Poisson process with rate α = 0.5 call per day.
Then the number of days X between successive calls has an
exponential distribution with parameter value 0.5.
The probability that more than 3 days elapse between calls is
P(X > 3) = 1 − P(X ≤ 3) = 1 − F (3; 0.5) = e −0.5·3 = 0.223.
The expected time between successive calls is 1/0.5 = 2 days.
Exponential Distribution
Exponential Distribution
“Memoryless” Property
Let X = the time certain component lasts (in hours)
and we assume the component lifetime is exponentially distributed
with parameter λ. Then what is the probability that the
component can last at least an additional t hours after working for
t0 hours, i.e. what is P(X ≥ t + t0 | X ≥ t0 )?
Exponential Distribution
“Memoryless” Property
Let X = the time certain component lasts (in hours)
and we assume the component lifetime is exponentially distributed
with parameter λ. Then what is the probability that the
component can last at least an additional t hours after working for
t0 hours, i.e. what is P(X ≥ t + t0 | X ≥ t0 )?
P({X ≥ t + t0 } ∩ {X ≥ t0 })
P(X ≥ t0 )
P(X ≥ t + t0 )
=
P(X ≥ t0 )
1 − F (t + t0 ; λ)
=
F (t0 ; λ)
P(X ≥ t + t0 | X ≥ t0 ) =
= e −λt
Exponential Distribution
Exponential Distribution
“Memoryless” Property
However, we have
P(X ≥ t) = 1 − F (t; λ) = e −λt
Exponential Distribution
“Memoryless” Property
However, we have
P(X ≥ t) = 1 − F (t; λ) = e −λt
Therefore, we have
P(X ≥ t) = P(X ≥ t + t0 | X ≥ t0 )
for any positive t and t0 .
Exponential Distribution
“Memoryless” Property
However, we have
P(X ≥ t) = 1 − F (t; λ) = e −λt
Therefore, we have
P(X ≥ t) = P(X ≥ t + t0 | X ≥ t0 )
for any positive t and t0 .
In words, the distribution of additional lifetime is exactly the same
as the original distribution of lifetime, so at each point in time the
component shows no effect of wear. In other words, the
distribution of remaining lifetime is independent of current age.
Gamma Distribution
Gamma Distribution
Definition
For α > 0, the gamma function Γ(α) is defined by
Z ∞
Γ(α) =
x α−1 e −x dx
0
Gamma Distribution
Definition
For α > 0, the gamma function Γ(α) is defined by
Z ∞
Γ(α) =
x α−1 e −x dx
0
Properties for gamma function:
1. For any α > 1, Γ(α) = (α − 1) · Γ(α − 1) [via integration by
parts];
Gamma Distribution
Definition
For α > 0, the gamma function Γ(α) is defined by
Z ∞
Γ(α) =
x α−1 e −x dx
0
Properties for gamma function:
1. For any α > 1, Γ(α) = (α − 1) · Γ(α − 1) [via integration by
parts];
2. For any positive integer, n, Γ(n) = (n − 1)!;
Gamma Distribution
Definition
For α > 0, the gamma function Γ(α) is defined by
Z ∞
Γ(α) =
x α−1 e −x dx
0
Properties for gamma function:
1. For any α > 1, Γ(α) = (α − 1) · Γ(α − 1) [via integration by
parts];
2. For any positive integer, n, Γ(n) = (n − 1)!;
√
3. Γ( 12 ) = π.
Gamma Distribution
Definition
For α > 0, the gamma function Γ(α) is defined by
Z ∞
Γ(α) =
x α−1 e −x dx
0
Properties for gamma function:
1. For any α > 1, Γ(α) = (α − 1) · Γ(α − 1) [via integration by
parts];
2. For any positive integer, n, Γ(n) = (n − 1)!;
√
3. Γ( 12 ) = π.
e.g. Γ(4) = (4 − 1)! = 6 and Γ( 25 ) =
3
2
· Γ( 32 ) = 32 [ 12 · Γ( 21 )] =
3√
4 π
Gamma Distribution
Gamma Distribution
Definition
A continuous random variable X is said to have a gamma
distribution if the pdf of X is
(
1
x α−1 e −x/β x ≥ 0
α
f (x; α, β) = β Γ(α)
0
otherwise
where the parameters α and β satisfy α > 0, β > 0. The standard
gamma distribution has β = 1, so the pdf of a standard gamma
rv is
(
1
x α−1 e −x x ≥ 0
f (x; α) = Γ(α)
0
otherwise
Gamma Distribution
Gamma Distribution
Remark:
1. We use X ∼ GAM(α, β) to denote that the rv X has a gamma
distribution with parameter α and β.
Gamma Distribution
Remark:
1. We use X ∼ GAM(α, β) to denote that the rv X has a gamma
distribution with parameter α and β.
2. If we let α = 1 and β = 1/λ, then we get the exponential
distribution:
( 1
1
x 1−1 e −x/ λ = λe −λx x ≥ 0
1
1
Γ(1)
f (x; 1, ) = λ
λ
0
otherwise
Gamma Distribution
Remark:
1. We use X ∼ GAM(α, β) to denote that the rv X has a gamma
distribution with parameter α and β.
2. If we let α = 1 and β = 1/λ, then we get the exponential
distribution:
( 1
1
x 1−1 e −x/ λ = λe −λx x ≥ 0
1
1
Γ(1)
f (x; 1, ) = λ
λ
0
otherwise
3. When X is a standard gamma rv (β = 1), the cdf of X ,
Z
F (x; α) =
0
x
y α−1 e −y
dy
Γ(α)
is called the incomplete gamma function.
Gamma Distribution
Remark:
1. We use X ∼ GAM(α, β) to denote that the rv X has a gamma
distribution with parameter α and β.
2. If we let α = 1 and β = 1/λ, then we get the exponential
distribution:
( 1
1
x 1−1 e −x/ λ = λe −λx x ≥ 0
1
1
Γ(1)
f (x; 1, ) = λ
λ
0
otherwise
3. When X is a standard gamma rv (β = 1), the cdf of X ,
Z
F (x; α) =
0
x
y α−1 e −y
dy
Γ(α)
is called the incomplete gamma function.
There are extensive tables of F (x; α) available (Appendix Table
A.4).
Gamma Distribution
Gamma Distribution
Gamma Distribution
Gamma Distribution
Proposition
If X ∼ GAM(α, β), then
E (X ) = αβ
and
V (X ) = αβ 2
Furthermore, for any x > 0, the cdf of X is given by
x
P(X ≤ x) = F (x; α, β) = F
;α
β
where F (•; α) is the incomplete gamma function.
Gamma Distribution
Gamma Distribution
Example:
The survival time (in days) of a white rat that was subjected to a
certain level of X-ray radiation is a random variable
X ∼ GAM(5, 4). Then what is
a. the probability that the survival time is at most 16 days;
Gamma Distribution
Example:
The survival time (in days) of a white rat that was subjected to a
certain level of X-ray radiation is a random variable
X ∼ GAM(5, 4). Then what is
a. the probability that the survival time is at most 16 days;
b. the probability that the survival time is between 16 days and
20 days (not inclusive);
Gamma Distribution
Example:
The survival time (in days) of a white rat that was subjected to a
certain level of X-ray radiation is a random variable
X ∼ GAM(5, 4). Then what is
a. the probability that the survival time is at most 16 days;
b. the probability that the survival time is between 16 days and
20 days (not inclusive);
c. the expected survival time.
Gamma Distribution
Example:
The survival time (in days) of a white rat that was subjected to a
certain level of X-ray radiation is a random variable
X ∼ GAM(5, 4). Then what is
a. the probability that the survival time is at most 16 days;
b. the probability that the survival time is between 16 days and
20 days (not inclusive);
c. the expected survival time.
Chi-Squared Distribution
Chi-Squared Distribution
Definition
Let ν be a positive integer. Then a random variable X is said to
have a chi-squared distribution with parameter ν if the pdf of X
is the gamma density with α = ν/2 and β = 2. The pdf of a
chi-squared rv is thus
(
1
x (ν/2)−1 e −x/2 x ≥ 0
ν/2
f (x; ν) = 2 Γ(ν/2)
0
x <0
The parameter ν is called the number of degrees of freedom
(df) of X . The symbol χ2 is often used in place of “chi-squared”.
Chi-Squared Distribution
Chi-Squared Distribution
Remark:
1. Usually, we use X ∼ χ2 (ν) to denote that X is a chi-squared rv
with parameter ν;
Chi-Squared Distribution
Remark:
1. Usually, we use X ∼ χ2 (ν) to denote that X is a chi-squared rv
with parameter ν;
2. If X1 , X2 , . . . , Xn is n independent standard normal rv’s, then
X12 + X22 + · · · + Xn2 has the same distribution as χ2 (n).
Chi-Squared Distribution
Chi-Squared Distribution