Download Lecture 5

PHP2510: Principles of Biostatistics & Data Analysis
Lecture V: Contiuous Random Variables
PHP 2510 – Lec 5: continuous RV
1
Consider a random variable X that can take values 0 and 1 with
equal probability. What distribution does X have?
What if X can take values 0,.5, and 1, each with equal probability?
P (X = 0) = P (X = .5) = P (X = 1) = 1/3
What if X can take values 0, .1, .2, ..., .9, 1.0?
P (X = 0) = P (X = .1) = P (X = .2) = . . . = P (X = .9) = P (X = 1) = 1/11
What if X can take any value between 0 and 1, each with same
chance?
PHP 2510 – Lec 5: continuous RV
2
0.8
0.0
0.4
0.8
0.4
0.0
0.5
1.0
1.5
−0.5
0.0
0.5
1.0
1.5
−0.5
0.0
0.5
1.0
1.5
−0.5
0.0
0.5
1.0
1.5
0.4
0.0
0.0
0.4
0.8
0.0
0.8
−0.5
PHP 2510 – Lec 5: continuous RV
3
For continuous random variables, we use probability density
function (PDF) instead of probability mass function to describe the
distribution.
For discrete random variables we have the sum of all probability
mass being 1.
X
P (X = x) = 1
x∈Ω
For continuous random variables we replace the “sum” with
integration such that the “area under the curve” is 1. For Unif(0,1),

 1 for 0 ≤ x ≤ 1
f (x) =
 0
otherwise
Z
∞
f (x)dx = 1
−∞
PHP 2510 – Lec 5: continuous RV
4
In general, for uniform distribution between a and b Unif(a, b),

 1
for a ≤ x ≤ b
b−a
f (x) =
 0
otherwise
Z
∞
f (x)dx = 1
−∞
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).
A continuous random variable X can take values over a continuum.
In many cases this will be an interval on the real line
R = {−∞, +∞}.
PHP 2510 – Lec 5: continuous RV
5
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
F (k)
= P (X < k) = P (X ≤ k) =
Z
k
f (x) dx
−∞
These ideas are best illustrated with examples.
PHP 2510 – Lec 5: continuous RV
6
0.06
For example, Unif(5,25) has probability density function:
0.04
P (X < 10) =
Z
∞
f (x)dx
0.02
−∞
=
Z
10
0.00
5
0
5
10
15
20
25
.05dx = 5 × .05 = .25
30
P (13 < X < 17) =
Z
∞
f (x)dx
−∞
=
Z
17
13
.05dx = (17 − 13) × .05 = .20
R 10
Notice that P (X = 10) = 10 .05dx = 0. In fact, the probability that a
continuous random variable X takes on any particular value is 0.
PHP 2510 – Lec 5: continuous RV
7
Example.
Suppose X has a uniform distribution on the interval [0, 5]. Find
the following:
• P (X < 2)
• P (1.5 < X < 2)
• The cumulative density function (CDF)
F (x) = P (X < x) =
Z
x
f (u)du
−∞
PHP 2510 – Lec 5: continuous RV
8
Example.
0.20
0.15
f.x
0.20
0.15
f.x
0.25
P(X<2)
0.25
Uniform (0,5)
0
1
2
3
4
5
0
1
2
x
3
4
5
3
4
5
x
P(X<k)
P(X<k)
0.0
0.15
0.2
0.4
0.20
f.x
0.6
0.8
0.25
1.0
P(1.5<X<2)
0
1
2
3
x
PHP 2510 – Lec 5: continuous RV
4
5
0
1
2
k
9
Uniform distribution
f (x) =
P (X < 2) =
1
( if x in [0,5])
5
Z
2
2
1
dx =
5
5
Z
2
.5
1
dx =
= .1
5
5
0
P (1.5 < X < 2) =
1.5
F (k)
= P (X ≤ k) =
Z
k
0
k
1
dx =
5
5
What is the average value of this uniform R.V.?
PHP 2510 – Lec 5: continuous RV
10
Normal distribution
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....).
The probability density function (pdf) of normal distribution is a
bell shaped curve. Two parameters determine the distribution, one
for the location (mean) and one for its spread (variance). We will
formally introduce the definition of mean and variance for a
random variable later.
PHP 2510 – Lec 5: continuous RV
11
Normal distribution
Properties of normal distribution
• Takes values from −∞ to ∞:
Ω = (−∞, ∞)
• symmetrical about its mean.
• Mean=median=mode
• Like all other distributions, area under the pdf curve is 1
Density of normal distribution N (µ, σ 2 )
0
2σ
4σ
µ
PHP 2510 – Lec 5: continuous RV
12
Normal distribution
All normal distributions have the same shape:
0.6
0.2
µ
−4
−2
0
2
4
2σ
4σ
0.0
0.0
4σ
−6
0.4
Probability
0.4
2σ
0.2
Probability
0.6
0.8
mean=0, sd=2
0.8
mean=0, sd=1
6
−6
−4
−2
2
4
2
4
6
0.8
0.6
0.4
2σ
0.2
0.4
Probability
0.8
0.6
mean=1, sd=.5
0.2
Probability
mean=−1, sd=1.5
µ
0
2σ
4σ
0.0
0.0
4σ
µ
−6
−4
−2
0
2
PHP 2510 – Lec 5: continuous RV
4
6
µ
−6
−4
−2
0
6
13
Normal distribution
0.0
0.1
0.2
0.3
0.4
Calculating Probabilities: The probability that X falls in any
interval (a, b) is the area under the curve between a and b.
a
µ
P (a ≤ X ≤ b) =
PHP 2510 – Lec 5: continuous RV
b
Z
a
b
(x−µ)2
1
−
√ e 2σ2 dx
σ 2π
14
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
2
1
z
f (z) = √ exp −
2
2π
• Tail areas are probabilities
2
Z ∞
z
1
√ exp −
dz
P (Z > z) =
2
2π
z
• Tail areas are difficult to compute by hand but exist in tables
(Table A1, p. 744 in SMMR, Table 2 in Rice.)
PHP 2510 – Lec 5: continuous RV
15
Normal distribution
0.0
0.1
0.2
0.3
0.4
Table for Standard normal distribution probabilities: The
integration is not easy to compute by hand but many books
provide tables. In your textbook, the area under the curve for the
upper tail is given. That is, you can look up P (Z > z0 ) for any z0
0
PHP 2510 – Lec 5: continuous RV
z0
16
0.0
0.1
0.2
0.3
0.4
Normal distribution
0
z0
PHP 2510 – Lec 5: continuous RV
Notice that not all normal tables
give you P (Z > z0 ). Some give
you lower tail P (Z < z0 ) (Rice
Table 2), some give you
P (0 < Z < z0 ), some may even
give you P (−z0 < Z < z0 ). It is
important to read the
description.
17
Normal distribution
0.3
0.4
The symmetry of normal distribution allows you to calculate
P (a ≤ Z ≤ b) for any interval provided the table.
0.0
0.1
0.2
P (Z > z0 ) = P (Z < −z0 )
0
z0
0.4
−z0
0.2
0.3
For any z0 > 0
0.0
0.1
P (0 < Z < z0 ) = .5 − P (Z > z0 )
−z0
0
z0
PHP 2510 – Lec 5: continuous RV
18
Normal distribution
0.2
0.3
0.4
For example, to calculate
P (−2.02 < Z < 1.53),
0.0
0.1
You can calculate
P (−2.02 ≤ Z ≤ 1.53) =
−2.02
0
1.53
we read off the table
1.53 −→ P (Z > 1.53) = .063
2.02 −→
P (Z > 2.02)
= P (Z < −2.02)
=
P (−2.02 ≤ Z ≤ 0) + P (0 ≤ Z ≤ 1.53)
=
= (.5 − .022) − (.5 − .063) = .915
OR
P (−2.02 ≤ Z ≤ 1.53) =
=
P (Z ≤ 1.53) − P (Z ≤ −2.02)
=
= (1 − .063) − (.022) = .915
= 0.022
PHP 2510 – Lec 5: continuous RV
19
Normal distribution
What about non-standard normal distributions?
Recall that all normal distributions have the same shape and all
necessarily have total area under the curve =1. Therefore we can
”shift” and ”scale” any normal distribution:
If a random variable X follows normal distribution with mean µ
and standard deviation σ, then
X −µ
Z=
σ
has standard normal distribution with mean 0 and standard
deviation 1.
PHP 2510 – Lec 5: continuous RV
20
Normal distribution
Example: Suppose infant birthweight follow a normal distribution with mean
3000 gram and standard deviation 1000 gram.
• What is the probability of an infant weighing more than 5000 g?
P (X > 5000)
=
P(
5000 − 3000
X − 3000
>
) = P (Z > 2) = 0.0228
1000
1000
• What is the probability of an infant weighing less than 3500g?
P (X < 3500)
=
=
X − 3000
3500 − 3000
<
)
1000
1000
P (Z < .5) = 1 − P (Z > .5) = 1 − .3085 = .6915
P(
• What is the probability of an infant weighing between 2500 and 4000g?
P (2500 < X < 4000)
PHP 2510 – Lec 5: continuous RV
=
4000 − 3000
2500 − 3000 X − 3000
<
)
1000
1000
1000
P (−.5 < Z < 1)
=
1 − P (Z > 1) − P (Z < .5)
=
1 − .1587 − .3085 = .5328
=
P(
21
Normal distribution
0e+00
2e−04
4e−04
Now a more challenging problem: What is the range of birthweight for the
middle 80% of babies?
80%
10%
10%
a
µ=3000
b
That is, we want to find out x such that P (−a < X < b) = 80% and a, b are
symmetric around mean 3000
Since P (a < X < b) = .8, P (X > b) + P (X < a) = .2, symmetry of normal
distribution implies P (X > b) = .1.
From the normal table we find P (Z > 1.28) = .100. We also know
x−3000
x−3000
>
)
=
P
(Z
>
) = .100
P (X > x) = P ( X−3000
1000
1000
1000
Thus x−3000
= 1.28 =⇒ x = 4280
1000
Can you find a?
PHP 2510 – Lec 5: continuous RV
22
Normal distribution
In general, if a random variable X has normal distribution with
mean µ and standard deviation σ, then Y = aX + b is also
normally distributed, with mean aµ + b and standard deviation aσ
PHP 2510 – Lec 5: continuous RV
23
Standard normal distribution
Example – distribution of heights
Let X denote height in cm of men randomly sampled from some
population. Suppose the mean height is 173cm and the standard
deviation is 6.25cm.
µ = 172.5, σ 2 = 6.252
We write
0.00
0.01
0.02
0.03
0.04
0.05
0.06
X ∼ N (172.5, 6.252 ).
140
150
160
170
180
190
200
Height
PHP 2510 – Lec 5: continuous RV
24
Standard normal distribution
The probability that a randomly chosen men
• is taller than 180cm
>
P (X > 180) = P ( X−172.5
6.25
180−172.5
)
6.25
= P (Z > 1.2) = 0.115
• is shorter than 180cm
P (X < 180) = · · · = 0.885
• has height between 165 and 175 cm
<Z<
P (165 < X < 175) = P ( 165−172.5
6.25
P (−1.2 < Z < 0.40) = 0.54
175−172.5
)
6.25
=
• What is the 90th percentile for heights?
Find q such that
P (X < q) = P (
q − 172.5
q − 172.5
X − 172.5
<
) = P (Z <
) = .9
6.25
6.25
6.25
We know P (Z < 1.28) = 0.9 from the table, so
q = 1.28 × 6.25 + 172.5 = 181
PHP 2510 – Lec 5: continuous RV
25
Standard normal distribution
Now consider Y be the height measured in inches when men wear
shoes that add 2-inch to their height.
1inch = 2.54cm
so
Y = X/2.54 + 2
The mean and standard deviation for Y:
172.5/2.54 + 2 = 69.91339
6.25/2.54 = 2.46063
Y ∼ N (69.91339, 2.460632 )
• some one is taller than 180cm means, when he wears shoes that
add 2 inches, he would be taller than (180/2.54 + 2 = 72.86614)
inches
72.86614−69.91339
>
) = P (Z >
P (Y > 72.86614) = P ( Y −69.91339
2.46063
2.46063
1.20)
PHP 2510 – Lec 5: continuous RV
26
Standard normal distribution
• The probability that a randomly chosen man is less than 65
inches when he wears shoes that add 2 inches, P (Y < 65)
is the same as
the probability that a randomly chosen man is less than
(65-2)*2.54=160.02cm.
P (Y < 65) = P (X < 160.02)
• both statements above mean that this man is 1.9968 Standard
Deviation lower than the average
(65 − 69.91339)/2.46063 = (160.02 − 172.5)/6.25 = −1.9968
PHP 2510 – Lec 5: continuous RV
27
Standard 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.
PHP 2510 – Lec 5: continuous RV
28
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. Or equivalently, by parameter
λ = 1/θ, which represents the rate. Its probability density function
(PDF) and cumulative density function (CDF) are
f (x)
F (x)
1 −x/θ
e
= λe−λx
θ

Z x
 1 − e−λx
=
f (u)du =

−∞
0,
=
PHP 2510 – Lec 5: continuous RV
x≥0
x<0
29
Exponential distribution
CDF
1.0
2.0
PDF
0.8
pexp(x, 2)
0.4
0.6
0.0
0.0
0.2
0.5
dexp(x, 2)
1.0
1.5
λ=
2
1
0.5
0
2
4
x
6
8
PHP 2510 – Lec 5: continuous RV
10
0
2
4
x
6
8
10
30
Exponential distribution
Memory less property of exponential distribution
Exponential distribution is often used to model waiting times. If
r.v. T describes the time till when an event happens, and T follows
exponential distribution, then
P (T > t + s|T > s)
=
=
=
P (T > t + s&T > s)
P (T > s)
P (T > t + s)
P (T > s)
1 − F (t + s)
1 − F (s)
1 − e−λ(t+s)
=
1 − e−λ(s)
= e−λt
= P (T > t)
PHP 2510 – Lec 5: continuous RV
31
Exponential distribution
Given that one has waiting for time s, the probability that the
event will happen after a period of length t is the same as the
unconditional probability. Relationship between Poisson and
exponential distribution:
If events occur in time as a Poisson process with mean λ. Let X(t)
represent the number of events happened between (t0 , t0 + t)
P (T > t) = P (X(t) = 0) = e−λt
PHP 2510 – Lec 5: continuous RV
32