Transcript Continuous random variables

CIS 2033 based on
Dekking et al. A Modern Introduction to Probability and Statistics. 2007
Instructor Longin Jan Latecki
Chapter 5: Continuous Random Variables
Probability Density Function of X
A random variable X is continuous if for some function ƒ: R  R and for any numbers a
and b with a ≤ b,
P(a ≤ X ≤ b) = ∫ab ƒ(x) dx
The function ƒ has to satisfy ƒ(x) ≥ 0 for all x and ∫-∞∞ ƒ(x) dx = 1.
Whereas Discrete Random Variables {px(a) =P(X=a)} map: Ω -- (X) -> R -- (px) -> [0,1]
Continuous Random Variables {P(a ≤ X ≤ b)=∫ab ƒ(x) dx} map: Ω -- (X) -> R -- (ƒ) -> R
To approximate the probability density function at a point a, one must find an ε
that is added and subtracted from a and then the area of the box under the curve
is obtained by the following: (2ε) * ƒ(a). As ε approaches zero the area under the
curve becomes more precise until one obtains an ε of zero where the area under
the curve is that of a width-less box. This is shown through the following
equation.
P(a – ε ≤ X ≤ a +ε) = ∫a-εa+ε ƒ(x) dx ≈ 2ε*ƒ(a)
A few asides:
DISCRETE  NO DENSITY
CONTINUOUS  NO MASS
BOTH  CUMULATIVE DISTRIBUTION Ƒ(a) = P(X ≤ a)
P(a < X ≤ b) = P(X ≤ b) – P(X ≤ a) = Ƒ(b) – Ƒ(a)
Ƒ(b) = ∫-∞b ƒ(x) dx and ƒ(x) = (d/dx) Ƒ(x)
*How the Distribution Function relates to the Density Function*
Uniform Distribution U(α,β)
•
A continuous random variable has a uniform distribution on the interval [α,β] if its
probability density function ƒ is given by ƒ(x) = 0 if x is not in [α,β] and,
ƒ(x) = 1/(β-α) for α ≤ x ≤ β
This simply means that for any x in the interval of alpha to beta has the same
probability and anything not in the interval is zero as shown in the figure below.
•
The cumulative distribution is given by F(x) = 0 if x < α,mF(x) = 1 if x > β,
and F(x) = (x − α)/(β − α) for α ≤ x ≤ β
Exponential Distribution Exp(λ)
Intuitively: a continuous version of the geometric distribution.
A continuous random variable has an exponential distribution with
parameter λ if its probability density function ƒ is given by
ƒ(x) = λe-λx for x ≥ 0
The Distribution function ƒ of an Exp(λ) distribution is given by
Ƒ(a) = 1 – e-λa for a ≥ 0
P(X > s + t | x > s) = P(x > s + t)/P(x>s)
= (e-λ(s+t))/(e-λs) =e-λt= P(X > t)
This simply means that s becomes the origin where t increases
therefore making s always less than t and the equation proven
true.
ƒ(x) = λe-λx for x ≥ 0
Ƒ(a) = 1 – e-λa for a ≥ 0
How much time will elapse before an earthquake occurs in a given region?
How long do we need to wait before a customer enters our shop?
How long will it take before a call center receives the next phone call?
How long will a piece of machinery work without breaking down?
Questions such as these are often answered in probabilistic terms using
the exponential distribution.
All these questions concern the time we need to wait before a certain event
occurs.
If this waiting time is unknown, it is often appropriate to think
of it as a random variable having an exponential distribution.
Roughly speaking, the time we need to wait before an event occurs
has an exponential distribution if the probability that
the event occurs during a certain time interval
is proportional to the length of that time interval.
More precisely, X has an exponential distribution
if the conditional probability
P(t  X  t  t | X  t )
is approximately proportional to the length Δt of the time interval
[t, t + Δt] for any time instant t .
In most practical situations this property is very realistic and this is
the reason why the exponential distribution is so widely used to
model waiting times.
From: http://www.statlect.com/ucdexp1.htm

Pareto Distribution Par(α)
Used for estimating real-life situations such as
• the number of people whose income exceeded level x,
• city sizes, earthquake rupture areas, insurance claims, and sizes of commercial
companies.
A continuous random variable has a Pareto distribution
with parameter α > 0 if its probability density function ƒ is given by
ƒ(x) = 0 if x < 1 and
f (x) 
for x ≥ 1

x  1
Normal Distribution N(μ,σ2)
Normal Distribution (Gaussian Distribution) with parameters μ and σ2 > 0 if its
probability density function ƒ is given by
1 x 
 (
1
f (x) 
e 2
 2

)2
for -∞ < x < ∞
*Where μ = mean and σ2 = standard deviation*
Distribution function is given by:

1 x 2
)

a
1  2(
F ( x)  
e
  2
dx
for -∞ < a < ∞
However, since ƒ does not have an antiderivative there is no explicit expression for Ƒ.
Therefore standard normal distribution where N(0,1) is given as follows, and the
distribution function is obtained similarly denoted by capital phi.
1
1  2x2
 (x) 
e
2
for -∞ < x < ∞
Normal Distribution
1 x  2
 (
)
2 
1
f (x) 
e
 2
a
1 x 2
)

1  2(
F ( x)  
e
  2
dx
Quantiles
Portions of the whole which increase from left to right, meaning
the 0th percentile is on the left hand side and the 100th
percentile is on the right side.
Let X be a continuous random variable and let p be a number
between 0 and 1. The pth quantile or 100pth percentile of
the distribution of X is the smallest number qp such that
Ƒ(qp) = P(X ≤ qp) = p
The median is the 50th percentile
Ƒ(qp) = P(X ≤ qp) = p
For continuous random variables qp is often easy to determine.
If F is strictly increasing from 0 to 1 on some interval (which may be infinite to
one or both sides), then
qp = Finv(p),
where Finv is the inverse of F.
What is the median of the U(2, 7) distribution?
The median is the number q0.5 = Finv(0.5).
You either see directly that you have got half of the mass to both sides
of the middle of the interval, hence
q0.5 = (2+7)/2 = 4.5, or you solve with the distribution function:
F(qp) = P(X ≤ qp) = p = 0.5
F(q) = 0.5
0.5
and F(x) = (x − α)/(β − α)
F(qp) = P(X ≤ qp) = p = 0.9
F(a) = 0.9
1  12 x2
f ( x)   ( x) 
e
2
a
0.9  F (a )  (a ) 
Matlab: a = norminv(0.9,0,1) => a =1.2816


1  12 x2
e dx
2