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3. Random Variables
Let (Ω, F, P) be a probability model for an experiment, and X a function that maps every to a unique point the set of real numbers. Since the outcome is not certain, so is the value Thus if B is some subset of R, we may want to determine the probability of “ ”. To determine this probability, we can look at the set that contains all that maps into B under the function X.
,Ω∈ξ,Rx∈ ξ
.)( xX =ξ
BX ∈)(ξ
Ω∈= − )(1 BXA Ω∈ξΩ
ξ
R)(ξX
x
A
B
Fig. 3.1 PILLAI
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Obviously, if the set also belongs to the associated field F, then it is an event and the probability of A is well defined; in that case we can say
However, may not always belong to F for all B, thus creating difficulties. The notion of random variable (r.v) makes sure that the inverse mapping always results in an event so that we are able to determine the probability for any
Random Variable (r.v): A finite single valued function that maps the set of all experimental outcomes into the set of real numbers R is said to be a r.v, if the set is an event for every x in R.
)(1 BXA −=
)).((" )("event theofy Probabilit 1 BXPBX −=∈ξ (3-1)
)(1 BX −
.RB ∈
) ( ⋅XΩ
)(| xX ≤ξξ)( F∈
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Alternatively X is said to be a r.v, if where Brepresents semi-definite intervals of the form and all other sets that can be constructed from these sets by performing the set operations of union, intersection and negation any number of times. The Borel collection B of such subsets of R is the smallest σ-field of subsets of R that includes all semi-infinite intervals of the above form. Thus if X is a r.v, then
is an event for every x. What about Are they also events ? In fact with since and are events, is an event and hence is also an event.
ax ≤<−∞
ab > ? , aXbXa =≤<
bX ≤
bXabXaX ≤<=≤∩> aXaX c >=≤
)(| xXxX ≤=≤ξξ
FBX ∈− )(1
aX ≤
(3-2)
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Thus, is an event for every n. Consequently
is also an event. All events have well defined probability. Thus the probability of the event must depend on x. Denote
The role of the subscript X in (3-4) is only to identify the actual r.v. is said to the Probability Distribution Function (PDF) associated with the r.v X.
≤<− 1 aX
na
∩∞
=
==
≤<−
1
1 n
aXaXn
a
)(| xX ≤ξξ
.0)( )(| ≥=≤ xFxXP Xξξ (3-4)
)(xFX
(3-3)
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Distribution Function: Note that a distribution function g(x) is nondecreasing, right-continuous and satisfies
i.e., if g(x) is a distribution function, then
(i)
(ii) if then
and
(iii) for all x.
We need to show that defined in (3-4) satisfies all properties in (3-6). In fact, for any r.v X,
,0)( ,1)( =−∞=+∞ gg
,0)( ,1)( =−∞=+∞ gg
,21 xx < ),()( 21 xgxg ≤
),()( xgxg =+
(3-6)
)(xFX
(3-5)
PILLAI
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1)( )(| )( =Ω=+∞≤=+∞ PXPFX ξξ
.0)( )(| )( ==−∞≤=−∞ φξξ PXPFX
(i)
and
(ii) If then the subset Consequently the event since implies As a result
implying that the probability distribution function is nonnegative and monotone nondecreasing.(iii) Let and consider the event
since
,21 xx < ).,(),( 21 xx −∞⊂−∞
, )(| )(| 21 xXxX ≤⊂≤ ξξξξ
1)( xX ≤ξ .)( 2xX ≤ξ
( ) ( ) ),()()()( 2211 xFxXPxXPxF XX =≤≤≤= ξξ (3-9)∆ ∆
(3-7)
(3-8)
,121 xxxxx nn <<<<< −
.)(| kk xXxA ≤<= ξξ (3-10)
, )( )( )( kk xXxXxXx ≤=≤∪≤< ξξξ (3-11)PILLAI
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using mutually exclusive property of events we get
But and hence
Thus
But the right limit of x, and hence
i.e., is right-continuous, justifying all properties of a distribution function.
( ) ).()()()( xFxFxXxPAP XkXkk −=≤<= ξ (3-12)
,11 −+ ⊂⊂ kkk AAA
.0)(lim hence and lim1
===∞→
∞
=∞→ kk
kkkk
APAA φ∩ (3-13)
.0)()(lim)(lim =−=∞→∞→
xFxFAP XkXkkk
),()( xFxF XX =+
)(xFX
(3-14)
,lim +
∞→= xxkk
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Additional Properties of a PDF
(iv) If for some then
This follows, since implies is the null set, and for any will be a subset of the null set.
(v)
We have and since the two events are mutually exclusive, (16) follows.
(vi)
The events and are mutually exclusive and their union represents the event
0)( 0 =xFX ,0x . ,0)( 0xxxFX ≤= (3-15)
( ) 0)()( 00 =≤= xXPxFX ξ 0)( xX ≤ξ
)( ,0 xXxx ≤≤ ξ
).(1 )( xFxXP X−=>ξ (3-16)
, )( )( Ω=>∪≤ xXxX ξξ
. ),()( )( 121221 xxxFxFxXxP XX >−=≤< ξ (3-17)
)( 21 xXx ≤< ξ )( 1xX ≤ξ
. )( 2xX ≤ξ
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(vii)
Let and From (3-17)
or
According to (3-14), the limit of as from the right always exists and equals However the left limit value need not equal Thus need not be continuous from the left. At a discontinuity point of the distribution, the left and right limits are different, and from (3-20)
( ) ).()()( −−== xFxFxXP XXξ (3-18)
,0 ,1 >−= εεxx .2 xx =
),(lim)( )( lim00
εξεεε
−−=≤<−→→
xFxFxXxP XX (3-19)
).()( )( −−== xFxFxXP XXξ (3-20)
),( 0+xFX )(xFX 0xx →
).( 0xFX
)( 0−xFX ).( 0xFX )(xFX
.0)()( )( 000 >−== −xFxFxXP XXξ (3-21)PILLAI
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Thus the only discontinuities of a distribution function are of the jump type, and occur at points where (3-21) is satisfied. These points can always be enumerated as a sequence, and moreover they are at most countable in number. Example 3.1: X is a r.v such that Find Solution: For so that and for so that (Fig.3.2)
Example 3.2: Toss a coin. Suppose the r.v X is such that Find
)( xFX
0x
. ,)( Ω∈= ξξ cX ).(xFX
, )( , φξ =≤< xXcx ,0)( =xFX
.,TH=Ω.1)( ,0)( == HXTX
)(xFX
xc
1
Fig. 3.2
).(xFX
.1)( =xFX ,)( , Ω=≤> xXcx ξ
PILLAI
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Solution: For so that
•X is said to be a continuous-type r.v if its distribution function is continuous. In that case for all x, and from (3-21) we get
•If is constant except for a finite number of jump discontinuities(piece-wise constant; step-type), then X is said to be a discrete-type r.v. If is such a discontinuity point, then from (3-21)
, )( ,0 φξ =≤< xXx .0)( =xFX
3.3) (Fig. .1)( that so , , )( ,1
,1 )( that so , )( ,10=Ω==≤≥
−===≤<≤xFTHxXx
pTPxFTxXx
X
X
ξξ
).()( −−=== iXiXii xFxFxXPp (3-22)
)(xFX)()( xFxF XX =−
.0== xXP
)(xFX
)(xFX
x
Fig.3.3
1q
1
ix
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From Fig.3.2, at a point of discontinuity we get
and from Fig.3.3,
Example:3.3 A fair coin is tossed twice, and let the r.v Xrepresent the number of heads. Find Solution: In this case and
.101)()( =−=−== −cFcFcXP XX
.0)0()0( 0 qqFFXP XX =−=−== −
, ,,, TTTHHTHH=Ω).(xFX
.0)(,1)(,1)(,2)( ==== TTXTHXHTXHHX
3.4) (Fig. .1)()( ,2
,43,, )(,, )( ,21
,41)()( )( )( ,10
,0)()( ,0
=⇒Ω=≤≥
==⇒=≤<≤
===⇒=≤<≤
=⇒=≤<
xFxXx
THHTTTPxFTHHTTTxXx
TPTPTTPxFTTxXx
xFxXx
X
X
X
X
ξ
ξ
ξ
φξ
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From Fig.3.4,
Probability density function (p.d.f)
The derivative of the distribution function is called the probability density function of the r.v X. Thus
Since
from the monotone-nondecreasing nature of
.2/14/14/3)1()1(1 =−=−== −XX FFXP
)(xFX)( xf X
.
)()(dx
xdFxf XX =
,0)()(lim
)(0
≥∆
−∆+=
→∆ xxFxxF
dxxdF XX
x
X
(3-23)
(3-24)
),( xFX
)(xFX
x
Fig. 3.4
1
4/11
4/3
2
∆
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it follows that for all x. will be a continuous function, if X is a continuous type r.v. However, if X is a discrete type r.v as in (3-22), then its p.d.f has the general form (Fig. 3.5)
where represent the jump-discontinuity points in As Fig. 3.5 shows represents a collection of positive discrete masses, and it is known as the probability mass function (p.m.f ) in the discrete case. From (3-23), we also obtain by integration
Since (3-26) yields
0)( ≥xf X )( xf X
,)()( ∑ −=i
iiX xxpxf δ
).(xFX
(3-25)
.)()( duufxFx
xX ∫ ∞−= (3-26)
,1)( =+∞XF
,1)( =∫+∞
∞−dxxf x
(3-27)
ix
)(xfX
xix
ip
Fig. 3.5
)( xf X
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which justifies its name as the density function. Further, from (3-26), we also get (Fig. 3.6b)
Thus the area under in the interval represents the probability in (3-28).
Often, r.vs are referred by their specific density functions -both in the continuous and discrete cases - and in what follows we shall list a number of them in each category.
.)()()( )( 2
11221 dxxfxFxFxXxP
x
x XXX ∫=−=≤< ξ (3-28)
Fig. 3.6
)( xf X ),( 21 xx
)(xfX
(b)
x1x 2x
)(xFX
x
1
(a)1x 2x
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Continuous-type random variables
1. Normal (Gaussian): X is said to be normal or Gaussian r.v, if
This is a bell shaped curve, symmetric around the parameter and its distribution function is given by
where is often tabulated. Since depends on two parameters and the notation ∼will be used to represent (3-29).
.2
1)(22 2/)(
2
σµ
πσ−−= x
X exf (3-29)
,µ
,2
1)(22 2/)(
2∫ ∞−
−−
−
==x y
XxGdyexFσµ
πσσµ (3-30)
dyexG yx 2/2
21)( −
∞−∫=π
,( 2σµNX)(xfX
∆
)(xfX
µ ,2σ )
xµFig. 3.7 PILLAI
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2. Uniform: ∼ if (Fig. 3.8), ),,( babaUX <
≤≤
−=otherwise. 0,
, ,1)( bxa
abxf X(3.31)
3. Exponential: ∼ if (Fig. 3.9))( λεX
≥=
−
otherwise. 0,
,0 ,1)(
/ xexfx
X
λ
λ (3-32)
)(xfX
x
Fig. 3.9
)(xfX
xa b
ab −1
Fig. 3.8
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4. Gamma: ∼ if (Fig. 3.10)
If an integer
5. Beta: ∼ if (Fig. 3.11)
where the Beta function is defined as
),( βαGX )0 ,0( >> βα
≥Γ=
−−
otherwise. 0,
,0 ,)()(
/1
xexxf
x
X
βα
α
βα (3-33)
n=α )!.1()( −=Γ nn
),( baX β )0 ,0( >> ba
<<−
=−−
otherwise. 0,
,10 ,)1(),(
1)(
11 xxxbaxf
ba
X β (3-34)
),( baβ
∫ −− −=1
0
11 .)1(),( duuuba baβ (3-35)
x
)( xf X
x
Fig. 3.1110
)( xf X
Fig. 3.10
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6. Chi-Square: ∼ if (Fig. 3.12)
Note that is the same as Gamma
7. Rayleigh: ∼ if (Fig. 3.13)
8. Nakagami – m distribution:
),( 2 nX χ
)(2 nχ ).2 ,2/(n
≥=
−
otherwise. 0,
,0 ,)(22 2/
2 xexxf
x
X
σ
σ
(3-36)
(3-37)
,)( 2σRX
x
)( xf X
Fig. 3.12
)( xf X
xFig. 3.13
PILLAI
≥
Γ=−−
otherwise. 0,
,0 ,)2/(2
1)(
2/12/2/ xex
nxfxn
nX
22 1 /2 , 0( ) ( )
0 otherwiseX
mm mxm x e x
f x m− − Ω ≥ = Γ Ω
(3-38)
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9. Cauchy: ∼ if (Fig. 3.14)
10. Laplace: (Fig. 3.15)
11. Student’s t-distribution with n degrees of freedom (Fig 3.16)
( ) . ,1)2/(
2/)1()(2/)1(2
+∞<<∞−
+
Γ+Γ
=+−
tnt
nnntf
n
T π
,),( µαCX
. ,)(
/)( 22 +∞<<∞−−+
= xx
xf X µαπα
. ,21)( /|| +∞<<∞−= − xexf x
Xλ
λ
)( xf X
x
Fig. 3.14µ
(3-41)
(3-40)
(3-39)
x
)( xf X
Fig. 3.15t
( )Tf t
Fig. 3.16PILLAI
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12. Fisher’s F-distribution/ 2 / 2 / 2 1
( ) / 2
( ) / 2 , 0( ) ( / 2) ( / 2) ( )
0 otherwise
m n m
m nz
m n m n z zf z m n n mz
−
+
Γ +≥= Γ Γ +
(3-42)
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Discrete-type random variables
1. Bernoulli: X takes the values (0,1), and
2. Binomial: ∼ if (Fig. 3.17)
3. Poisson: ∼ if (Fig. 3.18)
.)1( ,)0( pXPqXP ==== (3-43)
),,( pnBX
.,,2,1,0 ,)( nkqpkn
kXP knk =
== − (3-44)
, )( λPX
.,,2,1,0 ,!
)( ∞=== − kk
ekXPkλλ (3-45)
k
)( kXP =
Fig. 3.171 2 n
)( kXP =
Fig. 3.18 PILLAI
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4. Hypergeometric:
5. Geometric: ∼ if
6. Negative Binomial: ~ if
7. Discrete-Uniform:
We conclude this lecture with a general distribution duePILLAI
(3-49)
(3-48)
(3-47)
.,,2,1 ,1)( NkN
kXP ===
),,( prNBX1
( ) , , 1, .1
r k rkP X k p q k r r
r−−
= = = + −
.1 ,,,2,1,0 ,)( pqkpqkXP k −=∞===
)( pgX
, max(0, ) min( , )( )
m N mk n k
Nn
m n N k m nP X k
−−
+ − ≤ ≤= = (3-46)
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PILLAI
to Polya that includes both binomial and hypergeometric asspecial cases.
Polya’s distribution: A box contains a white balls and b blackballs. A ball is drawn at random, and it is replaced along withc balls of the same color. If X represents the number of whiteballs drawn in n such draws, find the probability mass function of X. Solution: Consider the specific sequence of draws where kwhite balls are first drawn, followed by n – k black balls. The probability of drawing k successive white balls is given by
Similarly the probability of drawing k white balls
0, 1, 2, , ,X n=
2 ( 1)
2 ( 1)W
a a c a c a k cpa b a b c a b c a b k c
+ + + −=
+ + + + + + + − (3-50)
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PILLAI
followed by n – k black balls is given by
Interestingly, pk in (3-51) also represents the probability of drawing k white balls and (n – k) black balls in any otherspecific order (i.e., The same set of numerator and denominator terms in (3-51) contribute to all other sequences as well.) But there are such distinct mutually exclusivesequences and summing over all of them, we obtain the Polyadistribution (probability of getting k white balls in n draws)to be
nk
1 1
0 0( )
( 1) ( 1) ( 1)
.
k w
k n k
i j
b jca ica b ic a b j k c
b b c b n k cp pa b kc a b k c a b n c
− − −
= =
+++ + + + +
+ + − −=
+ + + + + + + −
=∏ ∏ (3-51)
1 1
0 0( )( ) , 0,1,2, , .
k n k
ki j
n nk k
b jca ica b ic a b j k cP X k p k n
− − − = =
+++ + + + += = = =∏ ∏
(3-52)
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PILLAI
Both binomial distribution as well as the hypergeometric distribution are special cases of (3-52).
For example if draws are done with replacement, then c = 0 and (3-52) simplifies to the binomial distribution
where
Similarly if the draws are conducted without replacement, Then c = – 1 in (3-52), and it gives
( ) , 0,1,2, ,k n knk
P X k p q k n −
= = = (3-53)
, 1 .a bp q pa b a b
= = = −+ +
( 1)( 2) ( 1) ( 1) ( 1)( )
( )( 1) ( 1) ( ) ( 1)nk
a a a a k b b b n kP X ka b a b a b k a b k a b n
− − − + − − + += =
+ + − + − + + − + − +
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which represents the hypergeometric distribution. Finally c = +1 gives (replacements are doubled)
we shall refer to (3-55) as Polya’s +1 distribution. the general Polya distribution in (3-52) has been used to study the spreadof contagious diseases (epidemic modeling).
1 1
1
( 1)! ( 1)! ( 1)! ( 1)!( )( 1)! ( 1)! ( 1)! ( 1)!
= .
nk
a k b n kk n k
a b nn
a k a b b n k a b kP X ka a b k b a b n
+ − + − −−
+ + −
+ − + + + − − + + −= =
− + + − − + + −
(3-55)
! !( )! !( )!( )!( )! ( )!( )! ( )!( )!
a bk n k
a bn
n a a b k b a b nP X kk n k a k a b b n k a b k
−+
+ − + −= = =
− − + − + + −
(3-54)
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