Post on 12-Mar-2020
Probability and Random Processes
Unit-I
Random Variables
1. (a) Prob. Mass function (or) Prob. Function (p.m.f)
The numbers P(x ), i =1, 2,....i satisfies
(i) P(x ) 0i
1
(ii) P(x ) 1i
i
Note:
If 1 2 3 4 5, , , , .X a a a a a then
1 4 1 2 3P( ) P(X = )+P(X = ) P(X = )a X a a a a
(b) Probability density function (p.d.f)
The function ( )f x satisfies
(i) f(x) 0
(ii) ( ) 1f x dx
Note:
P(a )X b ( )
d
c
f x dx where ( , )c d is a Common region to ( , )a b & range of ' 'x in ( )f x .
2. Distribution function (or) Cumulative distribution Function:
( ) ; if X is discrete R.V
[ ]
( ) ; if X is continuous R.V
i
i
x x
x
P x
F X
f x dx
Note: P( ) ( ) ( )a X b F b F a
3. Moments:
(i) ' Moment about origin ( ) (OR) Expectationr
thr
'
( ) ; if X is discrete R.V
[ ] (1)
( ) ; if X is continuous R.V
r
xr
rr
x P x
X
x f x dx
(ii) Mean ( ) [ ]X E X
22Variance = Var(X) = E[X ] - ( )E X
(iii) Moment about mean ( )th
rr X
( ) ( ) ; if X is discrete R.V
( )
( ) ( ) ; if X is continuous R.V
r
xr
rr
x X P x
E x X
x X f x dx
4. Moment generating function (M.G.F):
( ) ; if X is discrete R.V
( )
( ) ; if X is continuous R.V
tx
xtx
Xtx
e P x
M t E e
e f x dx
Note: (i)
0
( ) ( ) ------> (2)n
n
xn
t
dE x M t
dt
(ii) ( ) of in M (t) -----> (3)!
nn
x
tE x co efficient
n
(iii) ( ) ( )bt
ax b xM t e M at
So we can calculate E(x ) ( . ., mean, variance) using equation (1) or (2) or (3)n i e
5. Standard distributions:
(I-Discrete case)
I. Binomial distribution:
(Probability of x success in n trails)
P(X = x) = n , 0,1,2.....x
x n x
C p q x n
(i) Mean = np
(ii) Variance = npq
(iii) Mgf = M (t) = n
t
X pe q
II. Poisson distribution:
(It is a limiting case of binomial distribution)
eP(X = x) = , >0; 0,1,2.....
!
x
xx
(i) Mean =
(ii) Variance =
(iii) ( 1)Mgf = M (t) = e
te
X
III. Geometric distribution (Probability of success in xth trail)
1P(X = x) = , 1,2..... 0<p 1xp q x
(i) Mean = 1
p
(ii) Variance = 2
q
p
(iii) M (t) =1
t
X t
pe
qe
(iv) Memory less property (or) lack of memory property
/ ----- (4)P X m n X m P X n
(II-Continuous case)
a. Uniform or rectangular distribution
1, a < x < b
( )
0 , otherwise
f x b a
(i) Mean =2
b a
(ii) Variance =
2( )
12
b a
(iii)
M (t) =( )
tb ta
X
e e
t b a
b. Exponential distribution
, x 0, > 0( , )
0 , otherwise
xef x
(i) Mean =1
(ii) Variance = 2
1
(iii) M (t) = , ( )
X tt
(iv) Memory less property (4)
c. Erlang(or) Gamma distribution (particular form, a =1) 1
, a 0, > 0, 0 < x < ( ) ( 1)!
0 , otherwise
axa e x
f x
(i) Mean =a
(ii) Variance = 2a
(iii) M (t) =X
a
a t
d. Normal distribution 2
1
21( ) , 0, < , x <
2
x
f x e
(i) Mean =
(ii) Variance = 2
(iii)
2
2M (t) =
tt
X e
6. Hints and formulas:
0
!x ne x dx n
0
( )x
nn x n x
C
x
a b n a b
2 3
1 .....2! 3!
xx xx e
x n xC Cn n
lim 1
n
x
n
xe
n
' "
1 2 ........u dv uv u v u v
1
dx = & dx = + c 1
mx nmx ne x
e c xm n
!
!( )!xC
nn
x n x
( ) ( )E aX b aE X b
2( ) ( )Var aX b a Var x
Note:
i. Calculation of probability distribution ( ), ( ) from cumulative distribution F(x)P x f x
1 If X is discrete, then P F Fi i iX x X x X x
If X is continuous, then f(x) = F[x] d
dx
ii. Properties of cumulative distribution:
( ) 0F
( ) 1F
F(x) is non-decreasing function of x
iii. 2E ax = a E x , E b = b and Var(ax)= a Var(x), Var(b) =0
iv. Normal Distribution:
(a) If Z = X
then the Normal distribution becomes standard normal distribution
with pdf
2
21
( ) , - < x< , with = 0 & = 12
Z
Z e
(b) Properties of Probability of standard normal curve:
We know that, (0 ) 0.5P Z
(0 ) ( ) 0.5P Z a P a Z
( ) 0.5 (0 )P a Z P Z a
Problems:
1. A random variable ‘x’ has the following probability distribution
x -2 -1 0 1 2 3
P(x) 0.1 k 0.2 2k 0.3 3k
1. Find k
2. P( x < 2), P( -2 < x < 2)
3. Cdf of x
4. Mean of x
Solution:
a. To Find k
We know,3
1
2
( ) 1i
P x
0.1 + k +0.2 + 2k + 0.3 + 3k = 1
0.6 + 6k = 1
6k = 0.4
4 2 k =
3060k
Probability distribution is
x -2 -1 0 1 2 3
P(x) 330
230
630
430
930
630
b. P( x < 2) = P ( x = -2, x = -1, x = 0, x = 1)
= P (x = -2) + P (x = -1) + P (x = 0) + P (x = 1)
= 3 2 6 4
30 30 30 30
P (x < 2) = 15 1
P(x< 2) = 30 2
P(-2 < x < 2) = P (x = -1, x = 0, x = 1)
= P (x = -1) + P (x = 0) + P (x = 1)
= 2 6 4
30 30 30
P (-2 < x < 2) = 12 2
P( 2 < x < 2) = 30 5
c. Cdf: Cumulative distribution
x -2 -1 0 1 2 3
F(x) 3
30 5
30 11
30 15
30 24
30 30
30
( ) ( )F X x P X x
d. Mean = E(x) 3
2
E(x) = ( ) ( 2) ( 2) ( 1) ( 1) (0) ( 0) (1) ( 1)x
x P x P x P x P x P x
(2) ( 2) (3) ( 3)P x P x
3 2 4 9 6
2 1 1 2 330 30 30 30 30
32( )
30E x
2. A random variable ‘x’ has the following probability distribution
x 0 1 2 3 4 5 6 7
P(x) 0 k 2k 2k 3k 2k 22k
27k k
5. Find k
6. 1.5 4.5
2
xP
x
7. 1
Smallest ' ' show that p (x ) >2
Solution:
a. To Find k
7
1
0
( ) 1i
P x
2 2 20 2 2 3 2 7 1k k k k k k k k
210 9 1 0k k
(10 1)( 1) 0k k
10 1 ( ) 1 0k or k
1
k = -1 gives P(x) value negative which is meaningless10
k
Probability distribution is
x 0 1 2 3 4 5 6 7
P(x) 0 110
210
210
390
1100
2100
17100
b.
(1.5 4.5) ( 2)1.5 4.5 ( ) ----(1) P
2 ( 2) ( )
P x xx P A BAPBx P x P B
P (1.5 4.5) ( 2) ( 2, 3, 4) ( 3, 4, 5, 6, 7)x x P x x x x x x x x
= P(x = 3, x = 4)
= P(x = 3) + P(x = 4)
2 3 =
10 10
5
P (1.5 4.5) ( 2) ----- (2)10
x x
P(x > 2) = P(x = 3, x = 4, x = 5, x = 6, x = 7)
= P(x = 3) + P(x = 4) P(x = 5) P(x = 6) + P(x = 7)
2 3 1 2 17 70 =
10 10 100 100 100 100
7P(x > 2) = ---- (3)
10
Substitute 2, 3 in 1
51.5 4.5 510 =
72 710
xP
x
c. 1
Smallest ' ' show that p (x ) >2
If = 0, then P(x 0) P(x = 0) = 0
1 1 = 1, then P(x 1) P(x = 0) + P(x = 1) = 0+
10 10
1 2 3 = 2, then P(x 2) P(x = 0) + P(x = 1)+P(x = 2) = 0+
10 10 10
1 2 2 5 = 3, then P(x 3) P(x = 0) + P(x = 1)+P(x = 2)+P(x = 3) =
10 10 10 10
5 3 8 1 = 4, then P(x 4) P(x = 3) + P(x = 4) =
10 10 10 2
1Smallest ' ' such that P(x ) is 4
2
3. A random variable ‘x’ has the density function 2, - < x<
( ) 1
0 , otherwise
k
f x x
Find (a) k (b) Distribution function or CDF (c) P (x > 0)
Solution:
a. To find k
We know that ( ) 1f x dx
1
2 2 2
1 11 tan
1
dx xk dx
x a x a a
1 1tan 1 tan ( )=2
k x
1 1 1tan tan ( ) 1 tan ( )=2
k
12 2
k
2
1 1 f(x) = ;
(1 )k x
x
b. F(x):
( ) ( ) [For CDFupper limit must be x]
x
F x f x dx
2
1
(1 )
x
dxx
11 = tan ( )
x
x
1 11 = tan ( ) tan ( ) In equation 1, x = 0x
1 11 1( ) tan ( ) ----- (1) F(0) = tan (0)
2 2F x x
1 =
2
1 F(0) =
2
c. ( 0) (0 ) P(a < x < b) = F (b) - F(a)P x P x
= F( ) - F(0) F( ) = 1
1 = 1 -
2
1( 0)
2P x
4. A continuous random variable has PDF2( ) , x 0xf x kx e , Find k , mean and variance.
Solution:
i. First we have to find the value of ‘k’
We know that
( ) 1f x dx
2
0 0
1 ! ----- (1)
n
x n xk x e dx x e dx n
k(2!) = 1 { by equation 1 }
1 k =
2
21( ) , 0 x
2
xf x x e
ii. We know that
( ) ( )r rE x x f x dx
2
0
1 =
2
r xx x e dx
( 2)
0
1 =
2
r xx e dx
1( ) ( 2)! ----- (2) { by using equation 1}
2
rE x r
Substitute r = 1 in equation 2 Substitute r = 2 in equation 2
23! 6 4! 24E(x) = 3 ----- (3) E(x ) = 12 ----- (4)
2 2 2 2
Mean = 3
22Var(x) = E(x ) ( )E x
2 = 12 - (3) { by using equation 3 and 4}
Var(x) = 3
5.The CDF of a continuous random variable x is
2
2
0 ; x < 0
1x ; 0 x <
2( )
3 11 (3 ) ; x <3
25 2
1 ; x 3
F x
x
iii. Find PDF of x
iv. 1
Evaluate P(|x| 1), P 4 using PDF and CDF3
x
Solution:
I. PDF of x :
We know that ( ) ( )d
f x F xdx
12 ; 0 x<
2( )
6 1(3 ) ; 3
25 2
x
f x
x x
f(x) is defined in x (0,3)
II. Using PDF,
(| | 1 ) ( 1 1)P x P x
1
0
P(-1 x 1) = ( ) dxf x
1/2 1
0 1/2
= ( ) dx ( ) dx f x f x
1/2 1
0 1/2
6 = 2 dx (3-x) dx
25x
12
1/22
01/2
6 1 6 1 3 1 = 3 3
25 2 4 25 2 2 8
xx x
25 27 52 =
4 25 4 25 4 25
13P(-1 x 1) (| | 1)
25P x
3
1/3
1P x 4 = ( ) dx
3f x
1/2 3
1/3 1/2
= ( ) dx ( ) dx f x f x
1/2 3
1/3 1/2
6 = 2 dx (3-x) dx
25x
32
1/22
1/31/2
6 1 1 6 9 3 1 = 3 9
25 2 4 9 25 2 2 8
xx x
5 27 32 =
36 36 36
1 8P x 4
3 9
Using CDF: F(x) ( ) ( ) ( )P a x b F b F a
(| | 1) ( 1 1)P x P x 1
1 x <32
x
= F(1) - F(-1) 23
( ) 1 (3 )25
F x x
13 = - 0
25
23 (1) 1 (3 1)
25F
13(| | 1)
25P x
12 = 1
25
1 14 (4)
3 3P x F F
13
F(1) = 25
1 =1 -
9 1 x < 0, F(x) = 0x
F(-1) = 0
1 84
3 9P x
4 x 3, F(x) = 1x
F(4) = 1
1 1
0 x < 3 2
x
2( ) xF x
1 1
F = 3 9
6. The probability density function of a random variable ; 0 < x < 1
( )(2 ) ; 1 x 2
xF x
k x
v. Find the value of k
vi. Find, P 0.2 1.2x
vii. 0.5 1.5
Find, P1
x
x
viii. Find CDF
Solution:
i. Find the value of ‘k’
We know that
( ) 1 In f(x); x (0,2)f x dx
2
0
( ) 1 f x dx
1 2
0 1
(2 ) 1x dx k x dx
1 22 2
0 1
2 12 2
x xk x
1 14 2 2 1
2 2k
; 0 < x < 1 1 1 k = 1 ( )
(2 ) ; 1 x 22 2
xk F x
x
ii. 0.2 1.2 (1.2) 0.2P x F F ( ) ( ) ( )P a x b F b F a
= 0.68 - 0.02
2
1.2 1 x 2 F(x) = 2x 12
x
0.2 1.2 0.66P x 1.44
(1.2) 2.4 12
F
0.68
iii. 0.5 1.5
P1
x
x
2
0.2 0 x < 1, F(x) = 2
x
(0.5 1.5) ( 1)
( 1)
P x x
P x
2(0.2)(0.2) 0.02
2F
1 1.5
(1 )
P x
P x
2
1,1.5 1 x 2, F(x) = 2x 12
x
(1.5) (1)
( ) (1)
F F
F F
1 1(1) 2 1
2 2F
0.88 0.5
1 0.5
2.25(1.5) 3 1
2F
= 0.88
( ) 1F
0.5 1.5 P 0.76
1
x
x
iv. CDF: F(x)
x
a
F x f x dx ; a least value of range of ' 'x in f x (i.e) 0a
In f(x), x (0,2) F(x 0) = 0 &F(x >2) = 1 ----(1)
0
( ) ( ) dx
x
F x f x
In interval 0 < x < 1
0 0
( ) ( ) dx dx
x x
F x f x x
2
0
= 2
x
x
2
F(x) = ; 0 < x < 1 ------(2)2
x
In interval 1 x 2 1
0 0 1
( ) ( ) dx ( ) dx ( ) dx
x x
F x f x f x f x
1
0 1
= dx + (2 ) dx
x
x x
12 2
0 1
= 22 2
x
x xx
21 1 F(x) = 2 2
2 2 2
xx
2
( ) 2 1; 1 x 2 ----- (3)2
xF x x
Combining equation 1, 2 and 3
2
2
0 ; x 0
x ; 0 x < 1
2, ( )
2 1 ; 1 x 22
1 ; x 2
CDF F xx
x
7.Find the moment generating function of the random variable ‘x’ having PDF
2e ; x > 0 ( ) 4
0 ; otherwise
xx
F x
Also find first four moments about the origin
Solution:
2Given, f(x) = e , 0 < x < 4
xx
( ) M (t) = E[e ]tx
Xi
= e f(x) dxtx
2
0
= e e4
x
tx xdx
1
2
0
1 = e
4
t x
x dx
2 2 1
0
1 1! 1 1 ! = = x e
4 4 ( )1 21
2 4
n ax
n
ndx
att
2
1 M (t) =
1 2X
t
( ) First 4 moments about origin are ii
2 3 4 E(x) ,E(x ),E(x ),E(x )
We know that
E(x ) = co-efficient of in M (t)!
rr
X
t
r
2 2 2 3
2
1 M (t) = 1 2 1 1 2 3 4 ........
1 2X t x x x x
t
2 3 4 = 1+2(2t)+3(2t) +4(2t) +5(2t) ........
2 3 4 = 1 + 4t + 12t +32t +80t +...........
2 3 4
M (t) 1 4 24 (32 6) (80 24) ......... --------(1)1 2! 3! 4!
X
t t t t
1
E(x) = co-efficient of in M (t)1!
X
t
E(x) = 4
22E(x ) = co-efficient of in M (t)
2!X
t
2E(x ) = 24
33E(x ) = co-efficient of in M (t)
3!X
t
3E(x ) = 192
44E(x ) = co-efficient of in M (t)
4!X
t
4E(x ) = 1920
8.Find mean, variance and MGF of binomial distribution
Solution:
P(X = x) = n ; p = P(Success), q = 1- p; 0,1,2.....x
x n x
C p q x n
( ) MGF i
M (t) = E[e ]tx
X
= e P(x)tx
0
= e nx
ntx x n x
C
x
p q
0
= n ( e )x
nt x n x
C
x
p q
M (t) ( e ) ------- (1) t n
X p q
( ) Mean E[x] ii
0
E(x) = M (t)X
t
d
dt
0
E(x) = ( e ) By equation 1t n
t
dp q
dt
1 0
0 = n ( e ) e t = 0 in e , 1t n t t
tp q p e
1 = n ( ) p + q = 1np q p
E(x) = np ----- (2)
22( ) Var(x) = E[x ] - E(x) ---- (3)ii
22
2
0
E(x ) = M (t)X
t
d
dt
2
2
= 00
= ( e ) ( e ) t n t n
tt
d d dp q p q
dt dt dt
1
= 0
2 1
= 0
2 1
( e ) e
= ( 1)( e ) e e ( e ) e
= n(n-1) p (p + q) p + np(p + q)
t n t
t
t n t t t n t
t
n n
d p q pn
dt u v
np n p q p np p q
2 2 2 2E(x ) = n p -np +np ----- (4)
Substitute equation 2, 4 in 3
2 2 2 2 2( ) = n p -np +np -n pVar x
= np(1-p)
( ) = npqVar x
9.Find mean, variance and MGF of Poisson distribution
Solution:
eP(X = x) = , 0,1,2.....
!
x
xx
( ) MGF i
M (t) = E[e ]tx
X
= 0
= e P(x)tx
x
0
e = e
!
xtx
x x
0
=e !
xt
x
e
x
0 1 2
M (t) e ........ 0! 1! 2!
t t t
X
e e e
1 2
2
= e 1 ........ 1 ..........1! 2! 1! 2!
t t
xe e x x
e
= e ete
( 1) M (t) = ete
X
( ) Mean = E(x)ii
= P(x) x
0 !
x
x
ex
x
1
0 ( 1)!
x
x
xe
x x
1
1
= e
( 1)!
x
x x
2
= e 1 ........ 1! 2!
= e e
E(x) = ----- (1)
22( ) Var(x) = E[x ] - E(x) ---- (2)ii
2 2E(x ) = P(x) x
2
0 0
e = e [ ( 1) ]
! !
x x
x x
x x x xx x
2 2
0 0
e = e ( 1)
( 1)( 2)! !
x x
x x
x x xx x x x
2
2
2 ( 2)!
x
x
e E xx
22 = e 1 ........ + by equation 1
1! 2!
2 = e e
2 2 ( ) = ------- (3)E x
Substitute equation 1, 3 in 2
2 2 Var(x) =
Var(x)
10. Find mean, variance and MGF of Geometric distribution
Solution:
1P(X = x) = , 1,2..... xp q x
( ) MGF i
M (t) = E[e ]tx
X
= 1
= e P(x)tx
x
1
1
= e tx x
x
p q
1
1
= tx x
x
p q e q
1
= x
t
x
pqe
q
2 3
........ t t tpqe qe qe
q
2
. 1 ........ t t tpqe qe qe
q
1(1 )t tpe qe
1 M (t) =
(1 )
t
X t
pe
qe
( ) Mean = E(x)ii
= P(x) x
1 1
1 1
= x x
x x
x p q p x q
2 = p (1 + 2q+3q +........)
2 = p (1 - q)
2 =
(1 )
p
q
2 =
p
p
1 E(x) = ----- (1)
p
22( ) Var(x) = E[x ] - E(x) ---- (2)ii
2 2E(x ) = P(x) ( 1) ( )x x x x P x
1 1
1 1
= ( 1) x x
x x
x x p q x p q
2 1
1 1
E(x ) = ( 1) ( )x
x x
p x x q xP x
2 3 = (1.2 2.3 3.4 .......) ( )p q q q E x
2 11.2 2.3 3.4 .....pq q q
p by equation 1
3 1
2(1 )pq qp
3 2
2
(1 )
pq p
q p
2
2 2
2 ( ) = ------- (3)
q pE x
p p
Substitute equation 1, 3 in 2
2 2 2 2
2 1 1 Var(x) =
q p q q p
p p p p
2 Var(x)
q
p
11.Find mean, variance and MGF of Uniform distribution
Solution:
1( ) ; x (a,b) f x
b a
( ) MGF i
M (t) = E[e ]tx
X
M (t) = e ( )tx
X f x dx
1 = e
b
tx
a
dxb a
1 = e
b
tx
a
dxb a
1 =
btx
a
e
b a t
M (t) ( )
tb ta
X
e e
t b a
( ) We know that [ ] = ( ) r rii E x x f x dx
1
= x
b
r
a
dxb a
11 =
1
br
a
x
b a r
1 11[ ] = ----- (1)
( )( 1)
r r rE x b ab a r
Mean = E(x)
2 21 = ( ) By substitute r = 1 in equation 1
( )(2)b a
b a
( )( ) E(x) =
2( )
b a b a
b a
( ) E(x) = ------ (2)
2
a b
22(iii) Var(x) = E[x ] - E(x) ---- (3)
2 3 31 E[x ] = ( ) By substitute r = 2 in equation 1
( )(3)b a
b a
2 21 = ( )( )
3( )b a b ba a
b a
2 22 ( )
E(x ) = ------ (4) 3
a ab b
Substitute 2, 4 in 3
2 2 2 2 2 2 2( ) ( ) 4 4 4 (3 6 3 ) Var(x) = - =
3 4 12
a ab b a b a ab b a ab b
2 22 =
12
a ab b
2( ) Var(x) =
12
a b
12.Find mean, variance and MGF of exponential distribution
Solution:
( ) ; x 0, 0 xf x e
( ) MGF i
M (t) = E[e ]tx
X
M (t) = e ( )tx
X f x dx
0
= etx xe dx
( )
0
= t xe dx
0
t xe
t
M (t) ( )
Xt
[ ] = ( ) r rNow E x x f x dx
0
= xr xe dx
0
[ ] = xr r xE x e dx
1
! =
r
r
![ ] = ----- (1)r
r
rE x
(ii) Mean = E(x)
Substitute r = 1 in equation 1 we get,
1
1! 1 1 E(x) = = E(x) = ------- (2)
22(iii) Var(x) = E[x ] - E(x) ---- (3)
Substitute r = 2 in equation 1 we get,
2
2
2! E(x ) = ------- (4)
Substitute 2, 4 in 3
2 2
2 1 Var(x) = -
2
1 Var(x) =
13.Find mean, variance and MGF of Gamma (or) Erlang distribution
Solution:
1
( ) > 0, x > 0 { a =1 case } ( 1)!
xe xf x
( ) MGF i
M (t) = E[e ]tx
X
M (t) = e ( )tx
X f x dx
1
0
= e( 1)!
xtx e x
dx
1 (1 )
0
1 = e
( 1)!
t xx dx
1
0
1 ( 1)! ! = e . Here n = -1, a = (1-t)
( 1)! (1 )
n ax
n
nx dx
t a
x
1 M (t) =
(1 )t
[ ] = ( ) r rNow E x x f x dx
1
0
= x( 1)!
xr e x
dx
1
0
1 = e
( 1)!
r xx dx
0
1 [ ] = ( 1)! ---- (1) !
( 1)!
r n xE x r x e dx n
(ii) Mean = E(x)
Substitute r = 1 in equation 1 we get,
! ( 1)! E(x) = E(x) =
( 1)! ( 1)!
( ) ------ (2)E x
22(iii) Var(x) = E[x ] - E(x) ---- (3)
Substitute r = 2 in equation 1 we get,
2 2( 1)! ( 1) ( 1)! E(x ) = = = ------- (4)
( 1)! ( 1)!
Substitute 2, 4 in 3
2 2 Var(x) =
Var(x) =
14.Find mean, variance and MGF of Normal distribution (ND)
Solution:
21
21: ( ) ( , ), ( , ), 0 ---- (1)
2
x
ND f x e x
( ) MGF i
First we find MGF of standard normal distribution (SND)
M (t) = E[e ]tz
Z
M (t) = e ( )tz
Z z dz
2
21
= e2
z
tz e dz
22
2 21
= dz2
tz z
e
2 2 212
21
= dz2
z tz t t
e
2
2 212
2 21
= dz2
tz tz t
e
2
21
2 21
= e dz2
tz t
e
22
22e
= dz dz = 2 2 2
tz t
z te y dy
2
22e
= 2 dy : , ; : , 2
t
ye z y
2
2
t
e
2xe dx
2
2M (t) = e --- (3) is the MGF of standard Normal distributiont
Z
( ) MGF of Normal Distribution:ii
x M (t) = M ( ) In equation 1 z= equation 2z
xt
z = e M ( ) From equation 2, z = t xt
2( )
2 = e x = t
te z
2( )
2x x M (t) = e ----- (4) M ( ) e M ( )
tt
bt
ax b t at
2
2x M ( ) = e by equation 3
t
t
2( )
2z M ( )
t
t e
( ) Meaniii
0
E(x) = ( )x
t
dM t
dt
2 2
2
0
= e by equation 4t
t
t
d
dt
2 2
2 x x2
0
E(x) = e d(e ) = e d(x) t
t
t
t
E(x) = ------(5)
22(iv) Var(x) = E[x ] - E(x) ---- (6)
22
2
0
E(x ) = ( )x
t
dM t
dt
2 2
2
0
= e t
t
t
d d
dt dt
2 2
22
0
= e d(uv) = u dv + v du t
t
t
dt
dt
2 2 2 2
2 2 22 2
0
= e ( ) + e t t
t t
t
t t
2 2 2E(x ) = + ------(7)
Substitute 5, 7 in 6
2 2 2 Var(x) = +
2 Var(x) =
15.Show that Poisson distribution is a limiting case of binomial
Proof:
Binomial Distribution P(X = x) = n ----- (1) , 0,1,2..... x
x n x
C p q x n
Limiting case:
!( ) np = n ; q = 1-p
!( )!Cr
ni
r n r
( ) n ii
From equation 1
!P(X = x) = (1 ) (i) p =
!( )!
x n xnp p
x n x n
! = 1 n ! = n (n-1) .... 2. 1
!( )!
x n xn
x n x n n
1( 1)( 2).......( 1)( ).........2.1=
!( )! 1
n
x
xx
n n n n x n x n
x n x nn
1( 1)( 2)....... ( 1) ( )!=
!( )! 1
n
x
xx
n n n n x n x n
x n x nn
1 11 2. n 1 1 ........... 11
P(X = x) = ! 1
x n
x
xx
xn
n n n n
x nn
11 21 1 ........... 11
P(X = x) = ! 1
n
x
x
xn n n n
xn
Taking limit n
P(X = x) is poisson distribution !
x
ex
16.State and prove memory less property of Geometric Distribution
/ m,n > 0 'x' is discreteP X m n X m P X n
Proof:
( )
/( )
P X m n X mP X m n X m
P X m
= ----- (1)
( )
P X m n
P X m
x -1Geometric Distribution: P(X = x ) = pq , x = 1, 2, ......... ------ (2)
P(x > m) = P( x = m+1) +P( x = m+2) P( x = m+3) ........
m m+1 m+2 = pq pq pq .......... by equation 2
m 2 -1 2 = pq (1 ......) (1- x) = 1 + x + x +........q q
1
1pq q
m -1 = pq p p = 1- q
mP(x > m) q ----- (3)
m+nSimilarly P(x > m+n) q ----- (4)
Substitute 3, 4 in 1
m+n m n
n
m m
q q .q/ q
q qP X m n X m
/ ( ) by using equation 3P X m n X m P x n
17.State prove forgetfulness property of exponential distribution
Statement:
/ ( ) , 0 & 'x' is continuous random variableP X m n X m P x n m n
gExponential Distribution ( ) x (0, ) ---- (1) xf x e
Now,
( )
/( )
P X m n X mP X m n X m
P X m
= ----- (2)
( )
P X m n
P X m
( ) ( )qP X m P m x
= ( )m
f x dx
= x
m
e dx
= x
m
e
-m = -e + e
-( ) e ----- (3) mP X m
- ( )Similarly ( ) e ----- (4)m nP X m n
Substitute 3,4 in 2
e e
/ ee
m nn
mP X m n X m
/ ( ) by using equation 3P X m n X m P x n
Normal Distribution
4. The weekly wages of 1000 workman are normally distributed around a mean of Rs. 70 with standard
deviation of Rs.5. Estimate the number of workers whose wages will be (i) between 69 and 72 (ii)
less than 69 (iii) more than 72
Solution:
Mean = = 70
----- (1) SD = = 5
69 72(i) P(69 < x < 72) = P
x
69 70 72 70 = P by equation 1 & Z =
5 5
xZ
= P ( 0.2 0.4)Z
= P ( 0.2 0) P (0 0.4) Z Z
= P (0 < z < 0.2) + P (0 0.4) Z
= 0.0793 + 0.1554
P(69 < x < 72) = 0.2347
For 1000 workers, N = P (69 < x < 72) 1000
= 0.2347 1000
N 235
App 235 workman getting wages between 69 & 72
69(ii) P(x < 69) = P
x
69 70 = P
5Z
= P ( 0.2)Z
= P ( 0.2) Z
= P (0.2 < z < )
= 0.5 - P (0 < z < 0.2)
= 0.5 - 0.0793
P( x <69) = 0.4207
For 1000 workers, N = P ( x < 69) 1000
= 0.4207 1000
N 421
Approximately 421 workman getting wages below 69
72(iii) P(x > 72) = P
x
72 70 = P
5Z
= P ( 0.4)Z
= P (0.4 ) Z
= 0.5 - P (0 < z < 0.4)
= 0.5 - 0.1559
P( x > 72) = 0.3446
For 1000 workers, N = P ( x > 72) 1000
= 0.3446 1000
N 344
Approximately 344 workman getting wages more than 72
19.The marks obtained by a number of students in a certain subject are approximately normally
distributed with mean 65 and standard deviation 5. If 3 students are selected at random from this group,
what is the probability that at least 1 of them would have scored above 75?
Solution:
P(At least 1 have scored above 75 from 3 students) ----- (1)
x 1 P(success) n =3
( 1) ? ----- (1a) x follows binomial distribution 0,1,2.....
B.D P(X = x) = nx
x n x
C
P xx n
p q
Where p = P(Success)
p = P(Students scored above 75)
p = P(x >75) 'x' follows normal distribution
75 = P
x
75 65 = P Given: Mean = = 65, SD = = 5
5Z
p = P 2 Z
= P (2 ) Z
= 0.5 - P (0 < z < 2)
= 0.5 - 0.4772 From standard normal table P(2) = 0.4772
p = 0.0228 ----- (2)
We know that q = 1- p = 1-0.0228
q = 0.9772 ----- (3)
From equation (1a), we have
( 1) 1 ( 1)P x P x
= 1 P(x = 0)
= 1 nx
x n x
C p q
0
0 3 0 = 1 3 (0.0228) (0.9772) by equation 1, 2 and 3C
3 = 1 (0.9772)
( 1) 0.0667 P x
20.In an engineering examination, a student is considered to have failed, secured second class, first class,
distinction according as he scores less than 45%, between 45% and 60%, between 60% and 75%, above
75% respectively. In a particular year 10% of the students failed in the examination and 5% of the
students got distinction. Find the percentage of students who have got first class, second class. (Assume
Normal distribution of marks)
Solution:
( ) 10% of students failed fail = Marks less than 45 ( x< 45) i
P(getting fail marks) = 0.1 10% = 0.1
P(x < 45) = 0.1
45 P 0.1
x
45 P = 0.1 Z
( 45) P = 0.1 Z
( 45) P < z < = 0.1
( 45) 0.5 - P 0 < z < = 0.1
( 45) P 0 < z < 0.5 - 0.1 = 0.4
( 45) = 1.28 ---- (1) By using standard normal table
P (0 < x < a) = 0.4
then a = 1.28
( ) 5% got distinction Distinction = Marks > 75 ( x > 75) ii
P(getting distinction) = 0.05
P(x > 75) = 0.05
75 P 0.05
x
75 P 0.05
x
75 P = 0.05 Z
75 P = 0.05 Z
75 0.5 - P 0 < z < = 0.05
75 P 0 < z < 0.5 - 0.05 = 0.45
75 = 1.65 ---- (2) By using standard normal table
P (0 < x < a) = 0.45
then a = 1.65
From equation 1
45 1.28 1.28 45 ----- (3)
(2) --> 75 1.65 1.65 75 ----- (4)
Solving equation 3 and 4
58.12, 10.24
( )Percentage of student who got 1st class (Marks (60,75)) iii
60 75P(60 < x < 75) = P
x
60 58.12 75 58.12 = P Z =
10.24 10.24
xZ
= P (0.18 1.65)Z
P(60 < x < 75) = P(0 < z < 1.65) - P(0 < z < 0.18)
= 0.4495 - 0.0714
P(60 < x < 75) = 0.3781
% of students who got 1st class = 38
( )Percentage of student getting 2nd class iv
(getting 2nd class) = P (45 < x < 60)P
45 60P(45 < x < 60) = P
x
45 58.12 60 58.12 = P Z =
10.24 10.24
xZ
= P (- 1.28 0.18)Z
= P (- 1.28 0) (0 0.18)Z P z
= P (0 < Z < 1.28) (0 0.18)P Z
= 0.3997 + 0.0714
P(45 < x < 60) = 0.4711
% of students who got 2nd class = 47
21.Let x and y be independent normal variates with mean 45 and 44, standard deviation 2 and 1.5
respectively. What is the probability that randomly chosen values of x and y differ by 1.5 or more?
Solution:
2
x xGiven: = 45, = 2 Var(x) = 4 Var =
y y = 44, = 1.5 Var(y) = 2.25
(i) P( 1.5) x - y = tx y
1 | | 1.5p x y 45 44 1t x y t
= 1 - P( 1.5) Var(t)= Var(x-y)= Var(x) +Var(y) t
= 1 - P(-1.5< t <1.5) = 4 + 2.25 = 6.25
t t
1.5 1.5 = 1 - P = 6.25, 2.5t t t
t t t
t
1.5 1 1.5 1P( 1.5) 1 P
2.5 2.5x y Z
= 1- P (- 1 0.2)Z
= 1- P (- 1 0) (0 0.2)Z P Z
= 1- P (0 1) (0 0.2)Z P Z
=1 - [0.3413+0.0793]
P( 1.5) = 0.5794x y
22.Find first four central moments of Normal distribution
Solution:
We know that rth central moment = E ( ) = r
rx x
E ( ) ( ) ( ) E(x) = f(x) dx r rx x f x dx x 22
2
1( )
221 1
= ( ) f(x) =2 2
xx
r
r x e e
22( 2 ) ; 2
2 2
r t xt e dt t dt dx
22
= ( ) dt x (- , ) then t (- , )
rr
r tt e
2
22 ( ) dt ----(1) 2 , 2 ( )
r
r rr t r
r t e t x t x
Case(i)
'r' is odd (i.e) r = 2n +1
2
2 1
2 122 1
2n +1
2equation 1 becomes, ( ) dt ( )dx = 0, 'f' is odd
nan
n t
a
t e f x
2n +1 0 ---- (2)
Case(ii)
'r' is even (i.e) r = 2n
22
2
2 21
nn t
n
nequation t e dt
2 2t y tdt dy
21/2 2n-1 2n2 1
= 2 dy t t dt = t dt 2
n nn yy e
1/2 2 1 2 (y )2
n ndyt dt
1/2 21
2
n ny dy t dt
21/2
2n
0 0
2 = dy dx = n!
n nn y n xy e x e
2
2n
2 1 = ! ------- (3)
2
n n
n
From equation 2, 3 we have
2n+1
2
2n
01
------- (4) != , n!= n(n-1)! 2 12 2 = !
2
n n
n
If n = 0 in equation 4 n =1 in equation 4 n = 2 in equation 4
1 3 5 0 = 0 = 0
2 4
2
2 4
32 !
2 1 2 = ! =
2
4
2
3 14 !
2 2 2 = =
2
42
2
2 3 = =
2
4
4 = 3
First five central moments of ND are
2 4
1 2 3 4 50; = ; = 0; = 3 ; = 0
23.If cumulative distribution is ( ) 1 (1 ) , 0xF x x e x (i) Find PDF (ii) Mean, Variance
Solution: , 2, 2xxe
24.If the density function of a continuous random variable is given by
; 0 x 1
( ) a ; 1 x 2
3 ; 2 x 3
ax
f x
a ax
(i) Find the value of a (ii) CDF of x
Solution:
2 21 1 3 5; F(x) = , 0 x 1; F(x) = - , 1 x 2; ( ) ;2 x 3
2 4 2 4 2 4 4
x x xa F x x
25.Find MGF of1
( )2
f x x ; x = 1, 2, 3….. Also find Mean and variance.
Solution: (2 )
t
t
e
e
26.Show that area bounded by above normal curve is 1
27.If ‘x’ has density function f(x) =2x, 0 < x < 1. Find PDF of 38y x
Solution:
1
31 2( ) , 0 y < 8
6f y
y
28.
21
If X N ( , ) obtain PDF of y = 2
x
Solution: 1
( ) , - < y < 2
yf y ey
Two Marks
1.Verify whether the function ; -1 < x < 1f x x is PDF or not?
Solution:
A function f(x) is PDF if it satisfies
( ) f(x) 0 xi
(ii) ( ) 1f x dx
( ) Given f(x) = | x |, clearly f(x) 0 ( 1,1)i x
f(x) 0
1
1 0
( ) ( ) | | ( ) 2 ( ) if f(x) is even
a a
a
ii f x dx x dx f x dx f x dx
1
0
= 2 | x | = x if x (0, a) x dx
12
0
= 22
x
( ) 1f x dx
Given function f(x) = | x | ; x (-1,1) is PDF
2. If variance of x is 2 and variance of y is 3 then find Var(3x + 4y)
Solution:
Given, Var(x) = 2, Var(y) = 3 2 2 2 2Var(3x + 4y) = (3) Var(x) + (4) var(y) Var(ax+by) = a Var(x) + b Var(y)
= 9 (2) + 16 (3)
Var(3x + 4y) 66
3.Write the properties of distribution Function
Solution:
(i) F(- ) = 0
(ii) F( ) = 1
(iii) P(a x b) = F(b) - F(a)
4..If x has mean 4 and variance 9 and y has mean -2 and variance 5, x and y are independent. Find 2E(xy), E(xy )
Solution:
Given E(x) = 4 ---- > (1) E(y) = -2 ------- > (3)
Var(x) = 9 Var(y) = 5
2 22 2 E(x ) - ( ) 9 E(y ) - ( ) 5E x E y
2 2 E(x ) 25 ---- (2) E(y ) 9 ---- (4)
Now,
( ) E(xy) = E(x) E(y) i
= (4) ( -2) by equation 1 and 3 x, y is independent E(xy) = E(x) E(y)
E(xy) = 8
2 2( ) E(xy ) = E(x) E(y ) ii
= (4) ( 9) by equation 1 and 4
2 E(xy ) =36
5. Let X be a random variable taking values -1, 0, 1 such that P(X = -1) = 2, P (X =0) = P(x = 1). Find
mean of 2x-5
Solution:
Let P(x = 0) = k then
P (x = -1) = 2k
P (x =1) = 2k
X -1 0 1
P(X) 2k k 2k
We know that ( ) 1P x
2k +k +2k = 1
1 k =
5
Substitute k value in the table
X -1 0 1
P(X) 2 1 2
5 5 5
,Now
E(2x- 5) = 2 E(x) - 5
1
1
= 2 ( ) 5 Mean ( )x
xp x xp x
= 2 ( 1) ( 1) 0 (0) (1) (1) - 5p P P
2 2 = 2 ( 1) (1) 5
5 5
E(2x- 5) 5
6. A Continuous random variable ‘x’ has PDF 2( ) , 0 1f x kx x Find ‘b’ such that P(x > b) = 0.05
Solution:
To find 'k' we know that ( ) 1f x dx
1
2
0
1k x dx
13
0
k 13
x
k = 3
2( ) 3 , 0 1f x x x
; P(x > b) = 0.05 Given
P(b < x < ) 0.05
1
( ) 0.05b
f x dx
1
2 3 0.05b
x dx
1
3 3 0.05 b 0.95b
x
1
3 b = (0.95)
0.9830b
7.For the following cumulative distribution function
0, x < 0
( ) x, 0 x 1
1, x > 1
F x
Then find
( ) ( 0.2) ( ) (0.2 0.5)i P x ii P x
Solution:
( ) P(x> 0.2) = P(0.2 < x < ) P(a x b)= F(b) - F(a)i
= F( ) - F(0.2) F( ) = 1
= 1 - 0.2 0.2 (0,1) F(x) = x
P(x> 0.2) 0.8 F(0.2) = 0.2
(ii) P(0.2 x 0.5) = F(0.5) - F(0.2) 0.5 (0,1) F(x) = x
= 0.5 - 0.2 F(0.5) = 0.5
P(0.2 x 0.5) = 0.3
8. If a random variable ‘x’ has MGF 3
( )3
xM tt
obtain standard deviation of x
Solution:
3 ( )
3xGiven M t
t
1 2 33
= 1 1 ....
3 13
x x x xt
1
= 13
t
2 3
( ) = 1 + .......3 9 27
x
t t tM t
21 2 ( ) = 1 + ....... ----- (1)
3 1! 9 2!x
t tM t
We know that E(x) = co-efficient of in ( ) 1!
x
tM t
1 E(x) = by equation 1
3
22We know that E(x ) = co-efficient of in ( )
2!x
tM t
2 2 E(x ) =
9
22 2 1
Standard deviation = = E(x ) ( )9 9
E x
1
3
9. IF the MGF of a uniform distribution for a random variable ‘x’ is
5 4
. Find mean t te e
t
Solution:
5 4
: M (t) = ----- (1) t t
x
e eGiven
t
We know that M (t) = ---- (2) MGF of uniform distribution ( )
bt at
x
e e
b a t
Comparing equation 1 and 2, we get
a = 4, b = 5
4 5Mean =
2 2
a b
9 E(x) =
2
10.Let x be a random variable with moment generating function
4
2 1M (t) =
81
t
x
e then find mean and
variance
Solution: 4
2 1Given, M (t) =
3
t
x
e
42 1
M (t) = ----- (1)3 3
t
x e
We know that for binomial distribution,
M (t) = ---- (2) n
t
x pe q
2 1Comparing equation 1 and 2 p = , , 4
3 3q n
2 2 1We know that mean = np = 4 variance = npq = 4
3 3 3
8 8 Mean = Variance =
3 9
11.If x and y are independent Poisson variates such that P(x = 2 ) = P (x= 3) and P (y = 1) = P ( y = 2).
Find variance of 2x – 3y
Solution:
1 2Let ' ', ' ' be Poission parameter for x, y respectively
Poission distribution P (X = x) = , x = 0, 1, 2, 3..... !
xe
x
Given
P(x = 2 ) = P (x = 3) P(y = 1 ) = P (y = 2)
1 1 2 22 3 1 2
1 1 2 2 = = 2! 3! 1! 2!
e e e e
2 3 2
1 1 2 2 3 = 2 =
3 2 2
1 1 2 23 0 2 0
2
1 1 2 2( 3) 0 ( 2) 0
1 2 =3 = 2
1 2We know that Variance = Var(x) = 3, ( ) = 2 Var y
2 2(2 3 ) (2) ( ) ( 3) ( )Var x y Var x Var y
= 4 3 + 9 2
(2 3 ) 30Var x y
12. Determine Binomial distribution if mean is 4 and variance is 3
Solution:
We know that mean = np = 4 ----- (1)
Variance = npq = 3 ----- (2)
Substitute equation 2 in 1
(4) q = 3 We know that p = 1- q
3
4q
3 11 ,
4 4p p
Substitute 1
4p in equation 1
14
4n
16n
Binomial distribution P(X = x) = n , 0,1,2..... x
x n x
C p q x n
161 3
P(X = x) = 16 , 0,1,2.....164 4x
x x
C x
13. The time in hours required to repair a machine is exponentially distributed with parameter 1
2 .
What is the probability that the required time exceeds 2 hours.
Solution:
Exponential distribution ( ) , x 0, > 0xf X x e
21 1
= , ( ) , x > 02 2
x
f X x e
P(x > 2) =P(2 < x < )
2
= ( )f x dx
22
2
2
1 1 = e 0
12 2
2
xx
ee dx
2
2
= x
e
1 = e e
1 P(x > 2)
e
14. Let ‘x’ be a uniform random variable over (-1,1). Find 3
| |4
P x
Solution:
1Uniform distribution ( ) , x (a,b)f x
b a
1 1Given x (-1,1) ( )
1 ( 1) 2f x
3 3| | 1 | |
4 4P x P x
3 3 = 1 -
4 4P x
3/4
3/4
= 1- ( )f x dx
3/4
3/4
1 = 1 -
2dx
3/4
3/4
1 = 1-
2x
1 3 3 = 1-
2 4 4
3 1| |
4 4P x
15. If ‘X’ is uniformly distributed in ,2 2
. Find PDF of y = tan x
Solution:
‘X’ is uniformly distributed in ,2 2
1
If x (a,b) then ( ) f xb a
1 f(x) = ------ (1) x ,
2 2
We know that ( ) ( ). ----- (2)dx
f y f xdy
: y = tan x Given
-1 x = tan (y)
2 2
1 1 ----- (3)
1 1
dx dx
dy y dy y
To find range of 'y':
y = tan x If x = 2
y
x = 2
y
( , ) ----- (4)y
Substitute equation 1,3,4 in 2
2
1f(y) = ; y ( , )
(1 )y
Discrete Distribution:
16. Binomial Distribution
P(X = x) = n , 0,1,2.....x
x n x
C p q x n
p + q = 1; p = P(Success)
17. Poisson distribution
eP(X = x) = ; 0,1,2..... = Poisson parameter
!
x
xx
18. Geometric distribution 1P(X = x) = , 1,2..... xp q x
Continuous distributions
19. Uniform distribution:
1( ) ; x (a,b)f x
b a
20. Exponential distribution
( ) , x >0, > 0xf x e
21. Gamma (or) Erlang distributions
1 1
( ) , a 0, > 0, 0 (or) f(x) = ( 1)! ( 1)!
ax xa e x e xf x x
22. Normal Distribution
21
21( ) , 0, < x < < <
2
x
f x e