Transcript Probability distribution

Probability distribution
Dr. Deshi Ye
College of Computer Science, Zhejiang University
[email protected]
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Outline
 Random variable
 The Binomial distribution
 The Hypergeometric Distribution
 The Mean and the Variance of the a
Probability distribution.
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Random variables
 We concern with one number or a few
number that associated with the outcomes
of experiments.
 EX. Inspection: number of defectives
road test: average speed and average
fuel consumption.
In the example of tossing dice, we are
interested in sum 7 and not concerned
whether it is (1,6) or (2, 5) or (3, 4) or (4,
3) or (5, 2) or (6, 1).
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Definition
 A random variable: is any function that
assigns a numerical value to each
possible outcome of experiments.
 Discrete random variable: only a finite
or a countable infinity of values.
Otherwise, continuous random
variables.
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Probability distribution
 The probability distribution of the random variable: is
the probabilities that a random variable will take on
any one value within its range.
 The probability distribution of a discrete random
variable X is a list of possible values of X together
with their probabilities
f ( x)  P[ X  x]
The probability distribution always satisfies the
conditions
f ( x)  0, and  f ( x)  1
x
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Checking probability distribution
x
Prob.
0
.26
1
.5
2
.22
3
.02
Another
1) f(x) = (x-2)/2, for x=1,2, 3, 4
2) H(x) =x2/25, x=0,1,2,3,4
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Probability histogram & bar chart
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Bar chart
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Histogram
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Histogram
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Cumulative distribution
 F(x): value of a random variable is
less than or equal to x.
F ( x)  P( X  x)   f (t )
tx
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EX.
x
Prob.
0
.26
1
.76
2
.98
3
1.0
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Binomial Distribution
 Foul Shot: 1. Min Yao (Hou) .862.
2. O’Neal Shaquille .422
 The Question is: what is the probability
of them in two foul shots that they get
2 points, respectively?
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Binomial distribution
 Study the phenomenon that the
probability of success in repeat trials.
 Prob. of getting x “success” in n trials,
 otherwords, x “success” and n – x
failures in n attempt.
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Bernoulli trials
 1. There are only two possible outcomes for
each trial.
 2. The probability of success is the same for
each trial.
 3. The outcomes from different trials are
independent .
 4’. There are a fixed number n of Bernoulli
trials conducted.
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 Let X be the random variable that
equals the number of success in n
trials. p and 1- p are the probability of
“success” and “failure”, the
probability of getting x success and nx failure is
p (1  p)
x
n x
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Def. of Binomial Dist.
 #number of ways in which we can
select the x trials on which there is to
be a success is  n 
 
 x
 Hence the probability distribution of
Binomial is
 n x
b( x; n, p)    p (1  p) n x
 x
x  0,1,2,, n
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Expansions
n
 n x
n x
( p  1  p)     p (1  p)   b( x; n, p)
x 0  x 
x 0
n
n
n
 
 x
Binomial coefficient
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Table 1
 Table 1: Cumulative Binomial
distribution
x
B( x; n, p)   b(k ; n, p)
k 0
b( x; n, p)  B( x; n, p)  B( x  1; n, p)
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EX Solve
 Foul shot example:
Here n=2, x=2, and
p=0.862 for Yao, and p=0.422 for
Shaq.
0
1
Yao
.02
.24
Shaq.
.33
.49
2
.74
.18
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Bar Chats
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Minitab for Binomial
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Skewed distribution
Positively
skewed
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Hypergeometric Distr.
 Sampling with replacement
 Sampling without replacement
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Hyergeometric distr.
 Sampling without replacement
 The number of defectives in a sample of n
units drawn without replacement from a lot
containing N units, of which a are defectives.
Here: population is N, and a are total
defectives
Sampling n units, what is probability of x
defectives are found?
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formulations
 Hypergeometric distr.
 a  N  a 
 

x  n  x 

h( x; n, a, N ) 
N
 
n 
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Discussion
 Is Hypergeometric distribution a
Bernolli trial?
 Answer: NO!
The first drawing will yield a defect
unit is a/N, but the second is (a 1)/(N-1) or a/(N-1).
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EX
 A shipment of 20 digital voice
recorders contains 5 that are
defective. If 10 of them are randomly
chosen for inspection, what is the
probability that 2 of the 10 will be
defective?
 Solution: a=5, n=10, N=20, and x=2
h(2;10,5,20)  0.348
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Expectation
 Expectation: If the probability of
obtaining the amounts
a1, a2 ,, or ak are p1, p2 ,, and pk
then the mathematical expectation is
E  a1 p1  a2 p2    an pn
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Motivations
 The expected value of x is a weighted
average of possible values that X can
take on, each value being weighted
by the probability that X assumes it.
 Frequency interpretation
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4.4 The mean and the variance
 Mean and variance: two distributions
differ in their location or variation
 The mean of a probability distribution
is simply the mathematical
expectation of a random variable
having that distribution.
 Mean of discrete probability
distribution
   x  f ( x)
all x
Alternatively,
E (X )  
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EX
 The mean of a probability distribution
measures its center in the sense of an
average.
 EX: Find the mean number of heads
in three tosses.
 Solution: The probabilities for 0, 1, 2,
or 3 heads are 1/8, 3/8, 3/8, and 1/8
1
8
3
8
3
8
1
8
  0   1  2   3  
3
2
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Mean of Binomial distribution
 Contrast: please calculate the
following
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 xb( x;4,0.5)  ?
x 0
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 xb( x;16, 0.2)  ?
x 0
16
 xb( x;16, 0.8)  ?
x 0
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Mean of b()
 Mean of binomial distribution:
  n p
n
Proof.
n!
 x
p x (1  p ) n  x
x!(n  x)!
x 0
n
 np b( x  1; n  1; p )
x 1
m
 np b( y; m, p )
y 0
 np
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Mean of Hypergeometric Distr.
a
  n
N
Proof.
 a  N  a 
 

x
n

x

h( x; n, a, N )   
N
 
n 
n
   xh( x; n, a, N )
x 0
Similar proof or using the following hints:
 m  s
 m  s
 
  


r 0  r  k  r 
k

k
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EX
 5 of 20 digital voice records were
defectives ,find the mean of the prob.
Distribution of the number of
defectives in a sample of 10 randomly
chose for inspection.
 Solution: n=10, a= 5, N=20. Hence
  n
a
 2.5
N
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Expectation of a function of random
variable
 Let X denote a random variable that
takes on any of the values -1,0,1
respective probabilities
 P{x=-1}=0.2, P{x=0}=0.5,
P{x=1}=0.3
 Compute E[X2]
 Answer = 0.5
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Proposition
 If X is a discrete random variable that
takes on one of the value of xi, with
respective to probability p(xi), then for
any real-valued function g,
E ( g ( x))   g ( xi ) p( xi )
i
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Variance of probability
 Variance of a probability distribution
f(x), or that of the random variable X
which has that probability distribution,
as
   ( x   ) f ( x)
2
2
all x
We could also denote it as
D(X )
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Standard deviation
 Standard deviation of probability
distribution

 (x  )
2
f ( x)
all x
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Relation between Mean and
Variance
D( X )  E[(x   ) ]
2
 E[ x  2 x   ]
2
2
 E ( X )  ( E ( X ))
2
2
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Ex.
 Find the variance of the number of
heads in four tosses.
 Solution:
  4
1
2
1
4
6
 (1  2) 2   (2  2) 2 
16
16
16
4
2
2 1
 (3  2)   (4  2) 
16
16
1
 2  ( 0  2) 2 
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Variance of binomial distr.
 Variance of binomial distribution:
  n  p  (1  p)
2
Proof. Detailed proof after the section of disjoint probability distribution
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Some properties of Mean
 C is a constant, then E(C) = C.
 X is a random variable and C is a constant
E(CX) = CE(X)
 X and Y are two random variables, then
E(X+Y) = E(X)+E(Y)
If X and Y are independent random variables
E(XY) = E(X)E(Y)
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Variance of hypergeometric distr.
a
a N n
  n (1  )(
)
N
N N 1
2
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K-th moment
 K-th moment about the origin
k   x f ( x)
k
all x
 K-th moment about the mean
k   ( x   ) f ( x)
k
all x
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Case study
 Occupancy Problem
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Homework
 problems
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