This document presents Chebyshev's inequality and its applications. It begins with the statement of Chebyshev's inequality relating the probability that a random variable X deviates from its mean by a certain number of standard deviations. It then provides a proof of the inequality for both continuous and discrete random variables. Several examples are worked through applying Chebyshev's inequality to find bounds on probabilities for specific distributions. Problems include finding bounds on probabilities for uniform and exponential distributions and determining values for which probabilities exceed thresholds.

1. Chebyshev’s Inequality and Its Applications
8 July 2021
Presented by Mr. Pradip Panda

2. Statement:
If X is a random variable with mean µ and finite variance σ2,
then for any positive number k,
  2
1 k
k
X
P 

 

  )
1
(
1
, 2
k
k
X
P
or 


 

8 July 2021
Presented by Mr. Pradip Panda

3. Proof:
X is a continuous random variable with p.d.f. f(x).
 2
2
)
(X
E
X
E 







 dx
x
f
x )
(
)
( 2










k
dx
x
f
x )
(
)
( 2










k
k
dx
x
f
x )
(
)
( 2








k
dx
x
f
x )
(
)
( 2









k
dx
x
f
x )
(
)
( 2








k
dx
x
f
x )
(
)
( 2
8 July 2021
Presented by Mr. Pradip Panda

4. For the first integration x ≤ µ - kσ or, (x- µ)2 ≥ k2σ2
For the second integration x ≥ µ + kσ or, (x- µ)2 ≥ k2σ2
2
 







k
dx
x
f
k )
(
2
2







k
dx
x
f
k )
(
2
2
 


 k
X
P
k 

 .
2
2
  2
1 k
k
X
P 


 

    1
, 




 


 k
X
P
k
X
P
Since
   



 k
X
P
k
X
P 





 1 )
1
(
1 2
k

8 July 2021
Presented by Mr. Pradip Panda

5. X is a discrete random variable with p.m.f. f(x).
 2
2
)
(X
E
X
E 


  )
(
2
x
f
x
x
 
 
  )
(
2
x
f
x
k
x







   )
(
2
x
f
x
k
x








  )
(
2
x
f
x
k
x















k
x
x
f
k )
(
2
2
  2
1 k
k
X
P 


 

8 July 2021
Presented by Mr. Pradip Panda

6. then
c
by
replaced
is
k
If 
  2
2
c
c
X
P 
 


   2
2
1 c
c
X
P
and 
 



  2
)
(
)
( c
X
Var
c
X
E
X
P 



  2
)
(
1
)
( c
X
Var
c
X
E
X
P
and 



8 July 2021
Presented by Mr. Pradip Panda

7. Problem:
Let f(x) = 5/x6 for x ≥ 1 and 0 otherwise. What bound does Chebyshev`s inequality give for the probability
P(X ≥ 2.5)? For what value of a can you say P(X ≥ a) ≤15%?
Solution:









1
4
4
5
x 4
5




1
5
5
dx
x



1
6
2
2 5
.
)
( dx
x
x
X
E









1
3
3
5
x 3
5




1
4
5
dx
x
)
(
)
(
)
( 2
2
X
E
X
E
X
Var 
 2
)
4
5
(
3
5


48
5
48
75
80






1
6
5
.
)
( dx
x
x
X
E
8 July 2021
Presented by Mr. Pradip Panda

8. By Chebyshev`s inequality,
  2
1
k
k
X
P 

 

2
1
48
5
4
5
k
k
X
P 








2
1
48
5
4
5
48
5
4
5
k
k
X
P
k
X
P 


















 
5
.
2
, 
X
P
Here
10
25
48
5
4
5
, 
 k
So
4
5
4
5
10
25
48
5



 k
8 July 2021
Presented by Mr. Pradip Panda

9. 15
5
48
4
5


 k
 2
15
1
48
5
*
15
4
5
48
5
*
15
4
5


















 X
P
X
P
   
 2
15
1
0
5
.
2 



 X
P
X
P
 
15
1
5
.
2 

 X
P   0
0 

X
P

8 July 2021
Presented by Mr. Pradip Panda

10.  
20
3
%
15 

 a
X
P
For
20
3
1
2


k 3
20

 k
20
3
48
5
*
3
20
4
5
48
5
*
3
20
4
5

















 X
P
X
P
20
3
12
5
12
25















 X
P
X
P
12
25

a
8 July 2021
Presented by Mr. Pradip Panda

11. Problem:
Let f(x) be the uniform distribution on 0 ≤ x ≤ 10 and 0 otherwise. Give a bound using Chebyshev`s for
P(2 ≤ x ≤ 8). Calculate the actual probability. How do they compare?
Solution: 5
2
10
0
)
( 


X
E  
3
25
12
100
12
0
10
)
(
2




X
Var
and
By Chebyshev`s inequality,
2
1
1
3
5
.
5
k
k
X
P 









2
1
1
3
5
5
3
5
5
k
k
X
k
P 












8 July 2021
Presented by Mr. Pradip Panda

12. 5
3
3
2
3
5
5
, 


 k
k
Here
27
25
1
3
5
.
5
3
3
5
3
5
.
5
3
3
5 











 X
P
 
27
2
8
2 


 X
P
8 July 2021
Presented by Mr. Pradip Panda

13.  
8
2 
 X
P
 
5
3
10
6
10
1
10
1 8
2
8
2



  x
dx
Therefore, actual probability is 3/5 which is much more than
2/27.
8 July 2021
Presented by Mr. Pradip Panda

14. Problem:
Let f(x) = e.ex for x ≤ -1 and 0 otherwise. Give a bound using Chebyshev`s for P(-4 ≤ x ≤ 0). For what a
P(X ≥ a) ≥ 0.99?
Solution:





1
.
)
( dx
e
x
e
X
E x
1
















    dx
dx
e
x
dx
d
dx
e
x
e x
x
 1




 x
x
e
xe
e
2
1
1




 

ee
ee 8 July 2021
Presented by Mr. Pradip Panda

15. 




1
2
2
.
)
( dx
e
x
e
X
E x
1
2
2
















    dx
dx
e
x
dx
d
dx
e
x
e x
x
 1
2
.
2
.





 dx
e
x
e
x
e x
x
5
)
2
(
2
1 



1
4
5
)
(
)
(
)
( 2
2




 X
E
X
E
X
Var
8 July 2021
Presented by Mr. Pradip Panda

16.   2
1
1
.
)
(
k
k
X
E
X
P 


 
By Chebyshev`s inequality,
  2
1
1
)
2
(
k
k
X
P 




  2
1
1
2
2
k
k
X
k
P 






 If k= 2 is taken then
 
4
3
4
1
1
0
4 




 X
P
8 July 2021
Presented by Mr. Pradip Panda

17.   2
1
1
2
2
k
k
X
k
P 







.
10
,
99
.
1
1 2
taken
is
k
k
get
To 


  99
.
0
8
12 


 X
P
  99
.
0
12 


 X
P
12


a
8 July 2021
Presented by Mr. Pradip Panda

18. Generalised Form of Bienayme-Chebyshev’s
Inequality and Its Applications
8 July 2021
Presented by Mr. Pradip Panda

19. Statement:
If g(X) is a non-negative function of a random variable X, then
for every k > 0,
    k
X
g
E
k
X
g
P )
(
)
( 

8 July 2021
Presented by Mr. Pradip Panda

20. Proof:
X is a continuous random variable with p.d.f. f(x).
Let S be the set of all X where g(X) ≥ k, i.e., S = {x: g(x) ≥ k}.
 
 



S
k
X
g
P
S
X
P
x
dF
So )
(
)
(
)
(
,
  



 )
(
)
(
)
( x
dF
x
g
X
g
E


S
x
dF
x
g )
(
)
( 

c
S
x
dF
x
g )
(
)
(
8 July 2021
Presented by Mr. Pradip Panda

21. 

S
x
dF
x
g )
(
)
( 

S
x
dF
k )
(
   
k
X
g
P
k
x
dF
k
X
g
E
S



  )
(
.
)
(
)
(
   
k
X
g
E
k
X
g
P
So )}
(
{
)
(
, 

8 July 2021
Presented by Mr. Pradip Panda

22. X is a discrete random variable with p.m.f. f(x).
  

x
x
f
x
g
X
g
E )
(
)
(
)
(

 

c
S
S
x
f
x
g
x
f
x
g )
(
)
(
)
(
)
(


S
x
f
x
g )
(
)
(
 
k
X
g
P
k
x
f
k
S


  )
(
.
)
(
   
k
X
g
E
k
X
g
P
So )}
(
{
)
(
, 

8 July 2021
Presented by Mr. Pradip Panda

23.     then
k
by
replaced
is
k
and
taken
is
X
X
E
X
X
g
If 2
2
2
2
)
(
)
( 





 
   
2
2
2
2
2
2
}
{ 


 k
X
E
k
X
P 



2
2
2
2
1
k
k




  2
1 k
k
X
P 


 

Which is the Chebyshev’s inequality .
8 July 2021
Presented by Mr. Pradip Panda

24. 0
)
( 
 k
any
for
then
taken
is
X
X
g
If
  k
X
E
k
X
P 

Which is the Markov’s inequality .
then
k
by
replaced
is
k
and
taken
is
X
X
g
If r
r

)
(
  r
r
r
r
k
X
E
k
X
P 

Which is the generalised form of Markov’s
inequality .
8 July 2021
Presented by Mr. Pradip Panda

25. 8 July 2021
Presented by Mr. Pradip Panda