learning about statistics lecture fourteen

Ghazaleh Parvini Fall 2024
Reasoning Under Uncertainty
Lecture 14

Announcements
• HW6 will be published on Thursday, Oct 24th (and the deadline in Oct 30)
• Discussion on Friday, Oct 25

Uniform Distribution
De
fi
nition
• A random variable has a uniform distribution and it is referred to as a
continuous uniform random variable if and only if its probability density is
given by
The parameters and are real constants with
X
u(x, α, β) =
{
1
β − α
for α < x < β
0 elsewhere
α β α < β

Gamma Distribution
Motivation
• In some problems we saw random variables having probability density of the
form
Where and must be such that the total area under the curve is
equal to 1.
• So we need to evaluate .
f(x) =
{
kxα−1
e− x
β for x > 0
0 elsewhere
α > 0,β > 0 k
k

Gamma Function
•
for
• Gamma Function satis
fi
es the recursion formula
• for , and since
•
we can see that when
Γ(α) =
∫0
yα−1
e−y
dy α > 0
Γ(α) = (α − 1) . Γ(α − 1) α > 1
Γ(1) =
∫0
e−y
dy = 1 Γ(α) = (α − 1)! α > 0
Gamma Distribution

Gamma Distribution
De
fi
nition
• A random variable has a gamma distribution and it is referred to as a gamma
random variable if and only if its probability density is given by
Where and .
X
g(x, α, β) =
{
1
βαΓ(α)
xα−1
e− x
β for x > 0
0 elsewhere
α > 0 β > 0

Exponential Distribution
De
fi
nition
• A random variable has an exponential distribution and it is referred to as an
exponential random variable if and only if its probability density is given by
Where .
X
g(x; θ) =
{
1
θ
e− x
θ for x > 0
0 elsewhere
θ > 0

CHI-Square Distribution
De
fi
nition
• A random variable has a chi-square distribution and it is referred to as a chi-
square random variable if and only if its probability density is given by
The parameter is referred to as the number of degrees of freedom, or simply the
degrees of freedom.
Chi-Square distribution plays a very important role in sampling theory.
X
f(x, ν) =
{
1
2
ν
2
x
ν − 2
2 e− x
2 for x > 0
0 elsewhere
ν

Beta Distribution
De
fi
nition
• A random variable has a beta distribution and it is referred to as a beta
random variable if and only if its probability density is given by
Where and .
X
f(x; α, β) =
{
Γ(α + β)
Γ(α) . Γ(β)
xα−1
(1 − x)β−1
for 0 < x < 1
0 elsewhere
α > 0 β > 0

Normal Distribution
De
fi
nition
• A random variable has a normal distribution and it is referred to as a
normal random variable if and only if its probability density is given by
for
Where .
X
n(x, μ, σ) =
1
σ 2π
e−1
2 (X − μ
σ )2
−∞ < x < ∞
σ > 0

Normal Distribution
Graph
• The graph of a normal distribution, shaped like the cross section of a bell:

Normal Distribution
• We see that the two parameters of the normal distribution are and .
• in this formula is actually and in this formula is actually the square
root of .
• Now we show the formula of the normal distribution is a valid probability
density.
• It is obvious that the formula is positive. So we only need to show that the
total area under the curve is equal to 1.
μ σ
μ E(X) σ
Var(X)

Normal Distribution
If we replace then we will have
∫
∞
−∞
1
σ 2π
e− 1
2 (X − μ
σ )2
dx =
1
2π ∫
∞
−∞
e−1
2 z2
dz =
2
2π ∫
∞
0
e−1
2 z2
dz
y =
1
2
z2
2
2π ∫
∞
0
2
2
y−1
2 e−y
dy =
1
π
Γ(
1
2
) =
1
π
π = 1

Standard Normal Distribution
De
fi
nition
• De
fi
nition: The normal distribution with and is referred to as the
standard normal distribution.
μ = 0 σ = 1

• Standard Normal Distribution

Normal Distribution
Example
• Example: Find the probabilities that a random variable having the standard
normal distribution will take on value
• Less than 1.72
• Less than -0.88
• Between 1.30 and 1.75
• Between -0.25 and 0.45

Normal Distribution
Example
• Solution: Less than 1.72:
• We look up the entry corresponding to and add 0.5.
So we have 0.4573 + 0.5000 = 0.9573
z = 1.72

Example
• Solution: Less than 1.72:
• We look up the entry corresponding to and add 0.5.
So we have 0.4573 + 0.5000 = 0.9573
z = 1.72

Example
• Solution: b) Less than -0.88:
• We look up the entry corresponding to and subtract it from 0.5 so
we have 0.5000 - 0.3106 = 0.1894.
z = 0.88

Example
• Solution: b) Between 1.30 and 1.75
• We look up the entry corresponding to and and subtract
the second from the
fi
rst, so we have 0.4599 - 0.4032 = 0.0567.
z = 1.75 z = 1.30

Example
• Solution: b) Between -0.25 and 0.45
• We look up the entry corresponding to and and add them,
so we have 0.0987 - 0.1736 = 0.2723.
z = 0.25 z = 0.45

• If we need to
fi
nd a value of corresponding to a speci
fi
ed probability that
falls between values listed in the table, we always choose the value
corresponding to the value that comes closets to the speci
fi
ed probability.
• If it falls exactly midway we take the average.
z
z

Example
• Example: Use the table to
fi
nd the values of that correspond to entries of
• a) 0.3512
• b) 0.2533.
z

Example
fi
• a) 0.3512
• b) 0.2533.
• Solution: a) 0.3512 falls between 0.3508 and 0.3531, corresponding to
and and , and since 0.3512 is closer to 0.3508 than
0.3531, we choose .
z
z = 1.04 z = 1.05
z = 1.04

Example
fi
• a) 0.3512
• b) 0.2533.
• Solution: a) 0.3512 falls between 0.3508 and 0.3531, corresponding to
and and , and since 0.3512 is closer to 0.3508 than
0.3531, we choose .
• b) Since 0.2533 falls midway between 0.2517 and 0.2549, corresponding to
and , we choose
z
z = 1.04 z = 1.05
z = 1.04
z = 0.68 z = 0.69 z = 0.685

Theorem
• Theorem: If has a normal distribution with the mean and the standard
deviation , then
Has the standard normal distribution.
X μ
σ
Z =
X − μ
σ

Example
• Suppose that the amount of cosmic radiation to which a person is exposed
when
fl
ying by jet across the US is a random variable having a normal
distribution with a mean of mrem and a standard deviation of 0.59
mrem. What is the probability that a person will be exposed to more than 5.20
mrem of cosmic radiation on such a
fl
ight?
4.35

Example
when
fl
fl
ight?
• Solution: . Finding the corresponding value in the
table and subtracting it from 0.5000, we get 0.5000 - 0.4251 = 0.0749
4.35
z =
5.20 − 4.35
0.59
= 1.44

Example
when
fl
fl
ight?
• Solution: .
• Finding the corresponding value in the
• table and subtracting it from 0.5000,
• we get 0.5000 - 0.4251 = 0.0749
4.35
z =
5.20 − 4.35
0.59
= 1.44

Chebyshev’s Theorem
Motivation
• To demonstrate how or is indicative of the spread or dispersion of the
dispersion of the distribution of a random variable we prove a theorem that is
called Chebyshev’s Theorem after the nineteenth-century Russion
mathematician.
σ σ2

Theorem
• Chebyshev’s Theorem: If and are the mean and the standard deviation of
a random variable , then for any positive constant , the probability is at
least that will take on a value within standard deviation of the
mean; symbolically
μ σ
X k
1 −
1
k2
X k
P(|x − μ| < kσ) ≥ 1 −
1
k2
, σ ≠ 0

Theorem
• Chebyshev’s Theorem Proof: By de
fi
nition
.
Let’s divide this integral to 3 parts:
•
• Since the middle term is positive we can say:
•
σ2
= E[(X − μ)2
] =
∫
∞
−∞
(x − μ)2
. f(x)dx
σ2
=
∫
μ−kσ
−∞
(x − μ)2
. f(x)dx +
∫
μ+kσ
μ−kσ
(x − μ)2
. f(x)dx +
∫
∞
μ+kσ
(x − μ)2
. f(x)dx
σ2
≥
∫
μ−kσ
−∞
(x − μ)2
. f(x)dx +
∫
∞
μ+kσ
(x − μ)2
. f(x)dx

•
Chebyshev’s Theorem Proof:
• Since for or it follows that
•
and hence
•
which means
• And so :
σ2
≥
∫
μ−kσ
−∞
(x − μ)2
. f(x)dx +
∫
∞
μ+kσ
(x − μ)2
. f(x)dx
(x − μ)2
≥ k2
σ2
x ≤ μ − kσ x ≥ μ + kσ
σ2
≥
∫
μ−kσ
−∞
k2
σ2
. f(x)dx +
∫
∞
μ+kσ
k2
σ2
. f(x)dx
1
k2
≥
∫
μ−kσ
−∞
f(x)dx +
∫
∞
μ+kσ
f(x)dx P(|X − μ|) ≥ kσ) ≤
1
k2
P(|X − μ|) < kσ) ≤ 1 −
1
k2

• For instance, the probability is at least that a random variable
will take on a value within two standard deviations of the mean,
• The probability is at least that a random variable will take on a
value within three standard deviations of the mean,
• And the probability is at least that a random variable will take
on a value within
fi
ve standard deviations of the mean.
• The probability given by Chebushev’s theorem is only a lower bound.
1 −
1
s2
=
3
4
X
1 −
1
32
=
8
9
X
1 −
1
52
=
24
25
X

Example
• Example: If the probability density of is given by
•
• Find the probability that it will take on a value within two standard deviations
of the mean and compare this probability with the lower bound provided by
Chebyshev’s theorem.
X
f(x) =
{
630x4
(1 − x)4
for 0 < x < 1
0 elsewhere

Example
• Example: If the probability density of is given by
•
• Find the probability that it will take on a value within two standard deviations
of the mean and compare this probability with the lower bound provided by
Chebyshev’s theorem.
• Solution: We can see that and , so that or 0.15.
• The probability that will take on a value within two standard deviations of
the mean is the probability that it will take on a value between 0.20 and 0.80
X
f(x) =
{
630x4
(1 − x)4
for 0 < x < 1
0 elsewhere
μ =
1
2
σ2
=
1
44
σ =
1
44
X

Example
• Solution: We can see that and , so that or 0.15.
• The probability that will take on a value within two standard deviations of
the mean is the probability that it will take on a value between 0.20 and 0.80
•
• The probability is 0.96 is a much stronger statement than the probability is at
least 0.75 which is provided by Chebyshev’s Theorem.
μ =
1
2
σ2
=
1
44
σ =
1
44
X
P(0.20 < X < 0.80) =
∫
0.80
0.20
630x4
(1 − x)4
dx = 0.96

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