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PHP 2510
Continuous random variables
Important continuous distributions
• Uniform
• Exponential
• Normal
• chi-square
Density functions
Calculations using the normal distribution
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Continuous random variables
A continuous random variable X can take values over a continuum.
In many cases this will be an interval on the real line
R = {−∞, +∞}.
The distribution of a continuous random variable is characterized
by a density function f (x). The density function characterizes
relative frequency.
Unlike a mass function, a density function can take values greater
than 1; therefore it should not be interpreted as a probability
(more on this later).
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Properties of density functions
Let f (x) be the density function for a random variable X. Then
1. Probabilities are calculated in terms of area under the curve
Z k2
f (x) dx
P (k1 < X < k2 ) =
k1
2. The density function integrates to 1 over the interval R
Z ∞
f (x) dx = 1
−∞
3. The cumulative distribution function is
Z
F (k) = P (X < k) =
k
f (x) dx
−∞
These ideas are best illustrated with examples.
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Uniform distribution
The uniform distribution takes values in an interval [a, b]. Each of
the values has the same relative frequency.
The density function is
f (x) =
1
b−a
for values in the interval [a, b].
For values outside the interval, f (x) = 0.
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Example. Suppose X has a uniform distribution on the interval
[0, 5]. Find the following:
• P (X < 2)
• P (1.5 < X < 2)
• The CDF
• the average value E(X)
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0.20
0.15
f.x
0.25
Uniform (0,5)
0
1
2
3
4
5
x
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0.20
0.15
f.x
0.25
P(X<2)
0
1
2
3
4
5
x
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0.20
0.15
f.x
0.25
P(1.5<X<2)
0
1
2
3
4
5
x
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0.0
0.2
0.4
P(X<k)
0.6
0.8
1.0
P(X<k)
0
1
2
3
4
5
k
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Uniform distribution
f (x) =
1
5
Z
P (X < 2)
2
1
2
dx =
5
5
2
.5
1
dx =
= .1
5
5
=
0
Z
P (1.5 < X < 2) =
1.5
Z
F (k)
= P (X ≤ k) =
0
k
1
x
=
5
5
What is the average value of this uniform R.V.?
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Exponential distribution
The exponential distribution is useful for modeling waiting times
on a continuous scale. It is characterized by a parameter θ, which
represents the average waiting time. Its density is
f (x) =
1 −x/θ
e
θ
A more useful function is the survivor function, which captures the
proportion remaining at time t
S(t) =
=
=
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P (X > t)
Z ∞
1 −x/θ
e
dx
θ
t
e−t/θ
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0.5
1.0
f1.x
1.5
2.0
Exponential Density Functions
0.0
0.2
0.4
0.6
0.8
1.0
x
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0.6
0.2
0.4
S1.x
0.8
1.0
Exponential Survivor Fns
0.0
0.2
0.4
0.6
0.8
1.0
x
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Normal distribution
The normal (Gaussian) distribution is useful for describing the
behavior of a continuous random variable.
• Sometimes the normal model is used to describe a single
random variable such as age, weight, blood pressure, IQ score,
etc.
• Distribution is symmetric and shaped like a bell (hence ‘bell
shaped curve’)
• It also can be used to describe the distribution of certain
summary statistics, such as a sample mean. In fact the normal
distribution is used far more often for applications of this type,
than as a model for a single random variable.
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Density function of the normal distribution
A density function represents relative frequency of individual values
of a random variable, and does not represent probability per se.
A random variable X having a normal distribution is characterized
by its mean µ and variance σ 2 . Its density function is given by the
function
·
¸
2
1
(x − µ)
√
f (x) =
exp −
2σ 2
σ 2π
(notice here that π represents the constant 3.14159....).
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0.2
0.0
0.1
f.x
0.3
0.4
N(0,1) density
−4
−2
0
2
4
x
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0.4
0.0
0.2
f2.x
0.6
0.8
Normal Density, Sigma=1.5 and 0.5
−4
−2
0
2
4
x
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0.2
0.0
0.1
f.x
0.3
0.4
P(X>2)
−4
−2
0
2
4
x
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Some properties of the normal distribution
Mean and variance
Z
∞
E(X) =
x f (x) dx = µ
−∞
∞
Z
var(X) =
(x − µ)2 f (x) dx = σ 2
−∞
(No need to prove. )
Linear function of a normal random variable is normal. That is, if
X ∼ N (µ, σ 2 ), then aX + b also follows a normal distribution.
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Standard normal distribution
The standard normal distribution has mean zero and variance one.
A random variable having standard normal distribution is
usually denoted by Z ∼ N (0, 1).
Its density function is
½
2
1
z
f (z) = √ exp −
2
2π
Tail areas are probabilities
Z
P (Z > z) =
z
∞
½
¾
2
1
z
√ exp −
2
2π
¾
dz
Tail areas are difficult to compute by hand but exist in tables
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(Table A1, p. 744 in ABM), or can be computed using Stata
etc.
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Some probabilities associated with Z
Areas in the right hand tail (from Table A.1)
P (Z > 0)
=
0.500
P (Z > 1)
=
0.159
P (Z > 2)
=
0.223
Can use symmetry to calculate other probabilities
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P (Z < 0)
= ···
P (Z < −1)
= ···
P (Z < 1)
= ···
P (−1 < Z < 2)
= ···
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Standardizing a normal random variable
If a random variable X is normally distributed with mean µ and
variance σ 2 , then
X −µ
Z=
σ
has a standard normal distribution with mean zero and variance
one.
This allows us to calculate probabilities associated with any normal
distribution.
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Example – distribution of heights
(From book). Let X denote height in cm of men randomly
sampled from some population. Suppose the mean height is
173 and the standard deviation is 6.25.
µ =
σ2
=
173
6.252
We write
X ∼ N (173, 6.252 ).
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0.06
0.05
0.04
0.03
0.02
0.01
0.00
140
150
160
170
180
190
200
Height
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Computations with the normal distribution
To compute probabilities associated with X, must first convert to
standard normal deviate Z. The standard normal deviate is
tabulated in Appendix A.1.
Suppose X ∼ N (µ, σ 2 ). Then
X −µ
∼ N (0, 1)
Z=
σ
Consequence:
µ
P (X > x) = P
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x−µ
Z>
σ
¶
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Examples
Suppose X ∼ N (173, 6.252 ). Find the following.
P (X > 180) = · · · = P (Z > 1.20) = 0.115
P (X < 180) = · · · = 0.885
P (165 < X < 175) = · · · = P (−1.20 < Z < 0.40) = 0.54
What is the 90th percentile for heights? Ans = 181.
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