Lecture 7. Distributions. Probability
density and cumulative distribution
functions.
Poisson and Normal distributions.
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•We remind here some facts about the distributions
of the discrete and continuous random variable, and
also introduce some new concepts.
The discrete random distribution can be characterized by a
probability function p(xi) assigning the probabilities to all possible
values xi of a random variable X. The probability function should satisfy
the following equations :
p( xi )  0
 p( xi )  1
i
P (E ) 
( 7.1)
 p( xi )
i E
For those familiar with the concept of “-function”, this relation can be presented as
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Example:
Suppose that a coin is tossed twice, so that the sample space is
={HH,HT,TH,TT}.
Let X represent a number of heads that can come up. Find the
probability function p(x).
Assuming that the coin is fair, we have P(HH)=1/4, P(HT)=P(TH)=1/4,
P(TT)=1/4;
Then, P(X=0)=P(TT)=1/4; P(X=1)=P(HTTH)=1/4+1/4=1/2.
P(X=2)= ¼.
The probability function is thus given by the table:
x
0
1
2
p(x)
1/4
1/2
1/4
The graphical example presented below is a little bit of a stretch
and should be used with some care. Why? (Hint: see (7.4))
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~p(xi)
0.3
1
2
3
4
5
6
7
8
Possible values of the random variable, xi
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10 11
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7.1 Poisson distributiona.
Definition
Poisson distribution is one of the most important discrete distributions. Its
probability function is
 x
P(x)  e 
x!
(7.3)
Poisson distribution is a limiting case of the Binomial distribution P(pn,n), with
parameters pn and n such that
pn 0, n , pn n  
In other words, if we have a large number of independent events with small
probability, then number of occurrences has approximately Poisson distribution.
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Examples with Poisson distribution
1. Suppose that the probability of a defect in a foot of magnetic tape is
0.002. Use the Poisson distribution to compute the probability that 1500
feet roll will have no defects .
Exp[  ] x
P [ x _ .  _] :
;
x!
p  0.002; n  1500;   p n  0.002  1500  3;
p[0]  P [0, 3] // N  0.0498
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Exp[  ] x
P [ x _ .  _] :
;
x!
p  0.002; n  1500;   p n  0.002  1500  3;
p[0]  P [0, 3] // N  0.0498
This example helps to describe the PD in a new way by noticing that L (I
use it here instead of Lambda) is the expected (average) value of the
defects in 1500 feet of the tape.
In other words, the PD gives the probability of n events happening
in some particular setting if the average number of events, L, is
known.
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Example 2
An airline company sells 200 tickets for a plane with 198
seats, knowing that a probability that a passenger will not
show up for the flight is 0.01. Use the Poisson
approximation to compute the probability that they will
have enough seats for all the passengers that will show up.
Solution.
p=0.01, L=0.01*200=2 – the average number (out of 200
passengers ) that won’t show up for the flight.
p[x]= Exp[-2]2x/x!;
P[more than 1 person won’t show up] = 1-p[0]-p[1]=
1 – 3 Exp[-2] = 0.594.
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Example 3:
(working in groups)
10% of the tools produced in a certain manufacturing process
turns out to be defective. Find a probability that in a sample of
ten tools selected at random, exactly 2 will be defective, by
using (a) binomial and (b) Poisson distribution.
Open a Mathematica file, and find the probability using (a) and
(b).
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Continuous distribution. Probability density function
(PDF).
Remember:
For a continuous variable we must assign to each outcome a
probability p(x )=0. Otherwise, we would not be able to fulfill the
requirement 7.3.
A random variable X is said to have a continuous distribution
with density function f(x) if for all a b we have
b
P (a  X  b )   f ( x )dx
( 7.4)
a
The analogs of Eqs. 7.2 and 7.3 for the continuous distributions would be
(7.5)
 f (x)  1

P (E )   f ( x )dx
E
( 7.6)
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P(E) is a probability that X belongs to E.
f(x)
P(a<X<b)
a
b
Geometrically, P(a<X<b) is the area under the curve f(x) between a and b.
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Examples:
1. The uniform distribution on (a,b):
We are picking a value at random from (a,b).
1
,
 b a
f (x)  
0

axb
(7.7)
otherwise
By direct integration you can verify that (7.7) satisfies the condition (7.5).
We can now find PDF which describes the experiment with a spinner Lecture
1). In this case b-a=2, and
1
,
 2
f ( )  
0

0  x  2
otherwise
The probability that the arrow will stop in the rage
between  and  +  equals /2.
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2. The exponential distribution
  e  x ,
f ( x )  
0

x0
otherwise
(7.8)
Those who know how to integrate can verify that (7.8) satisfies (7.5)
(the total area under the curve f(x) equals 1.
Note: In Matematica, the integral of a function f[x] (notice that […]
rather than (…) is used) can be found as:
Integrate[f[x],{x,x1,x2}] , Shift+Enter.
Here x1 and x2 are the limits of integration.
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A typical example of the exponential distribution results from the
discussion of the waste products of the nuclear power plant. If at time t=0
there are N(0) identical unstable particles, and the number of particles dN(t)
decaying in time dt is proportional to dt and to the number of particles, then
we have
dN(t)= - N(t)dt
This is so called differential equation. Here is how it is solved with
Mathematica.
DSolve[{n’[t] + G n[t] == 0, n[0] == n0},n[t],t];
n[t]-> n0 Exp[-Gt];
As a result we came up with the exponential distribution.
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Let’s introduce the “half-time” T , such that N(T)=N0/2.
Then we find : T=ln2=0.693.
3. The standard normal distribution
f(x)=(2)-1/2 exp(-x2/2)
(7.12)
A.
Using Mathematica, check that this PDF satisfies the normality condition
(7.5). Make a plot of (7.12).
If a random variable y is related to x as y=ax+b, how the distribution
function f(y) looks like? (we assume that x is distributed according to (7.12).
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More generally, X is said to have a normal (,2)
distribution if it has density function
f(x)=(2 2)-1/2 exp[-(x- )2/2 2]
(7.12’)
B. Try to analyze, assigning different numeric values to  and 2 how
they affect the shape of f(x). For instance, how the parameters for the
green and red curves are related? Green and blue?
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Probability distribution function
( also called “cumulative distribution function”= CDF)
1.
Continuous random variable
From the “outside”, random distributions are well described by the probability
distribution function (we will use CDF for short) F(x) defined as
x
F ( x )  P(  X  x )   f ( y )dy

(7.13)
This formula can also be rewritten in the following very useful form:
P(a  X  b)  F (b)  F (a)
(7.14)
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To see what the distribution functions look like, we return to our examples.
1. The uniform distribution (7.7):
1
,
 b a
f (x)  
0

ax b
otherwise
Using the definition (7.13) and Mathematica, try to find F(x) for the
uniform distribution. Prove that
F(x)=0 for x a; (x-a)/(b-a) for a  x  b; 1 for x>b.
Draw the CDF for several a and b. Consider an important special case
a=0, b=1. How is it related to the spinner problem? To the balanced die?
2. The exponential distribution (7.8):
  e  x ,
f ( x )  
0

x0
otherwise
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Use Mathematica to prove that
F(x)= 0 for x  0; 1-exp(-x) for x >0. (7.15)
“Lack of memory” for the exponential distribution
Suppose that X has an exponential distribution (7.8). The probability that
the event (such as the radioactive decay) did not happen in t units of time
is P(X>t) = 1-F(x). According to (7.15) it results in P(X>t)= exp(-t) .
Let’s find now a probability that we will have to wait some additional
time s given that we have been waiting t units of time:
P(T>t+s|T>t) = P(T > T+s)/P(T > t) = exp[-(t+s)]/ exp[-t)]=
exp[-s].
As we see, the result depends only on s and does not depend on the
previous waiting time. The probability you must wait additional s units of
time is the same as if you had not been waiting at all.
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The standard normal distribution
Using Mathematica and Eq. (7.12), find F[x] for the snd.
Use NIntegrate[f[t],{t,-,x}] and Plot[…] functions.
2. CDF for discrete random variables
For discrete variables the integration is substituted for summation:
F ( x )  P( X  x )   p( x )
i
x x
i
(7.16)
It is clear from this formula that if X takes only a finite number of values,
the distribution function looks like a stairway.
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F(x)
p(x4)
1
p(x3)
p(x2)
p(x1)
x1
x2
x3
x4
x
Draw F(x) for the example in page 7.
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C.
•
Suppose that a pair of fair dice are to be tossed, and let the random
variable denote the sum of the points. Obtain F(x) for this variable.
•
For the standard normal distribution, find the interval (-a,a) such that
P(-a<x<a) = 0.95. Use Mathematica.
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Home assignment is in three blue areas marked as A., B. C.
and D.
D.
•
(1) Find the constant c such that the function f(x) = c x2 for 0<x<3, 0
otherwise is a density function and (b) compute P(1<X<2).
(use 7.4, 7.5 and Mathematica).
•
(2) Suppose X has density function f(x)=x/2 for 0<0<2, 0 otherwise. Find
(a) distribution function, (b) P(X<1), (c) P(X>3/2).
•
(3) Let X has exponential distribution with parameter . Using
Mathematica, find P(X> /2).
•
(4) Read the problem 4.22 in Schaum’s P&S and find the error in our
solution of this problem ( the previous class ). It looks the results with BD
and PD are much closer to each other.
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