Chebyshev's inequality

Chebyshev`s inequality and its application

1. Presented by Mr. Pradp Panda. 8 July 2021

2. or, PIX - u|<ko]>1-«/r?)

Presented by Mr. Pradp Panda.

3. X is a continuous random variable with p.d.f. f(x).

o” = E[X - E(X)}

= J (x) f (xdx

u-ko
= [a fends + N 0 food + [1 red

u+ko

poko

Bef GAY FOX + | (4) far

u+ko

4. For the first integration x < U - ko or, (x- 1? > k20?

u-ko
o” 20" [finde +K’o? [ foods

00 urko

= Ko. PIX — p|>ko]

AP IX = |< ko]=1- PX - 12 ko] 21/10)

5. X is a discrete random variable with p.m. fix).
o? =E[X -E(xX)}

=e fo

„au

> Ylr-u) fo)

Er ulzko

PEN carole
LE a

6. If ko is replaced by c then

Bp Pix - 1 > c]<0?*/c* and PIX - 4 <c]>1-0°/c°
> PIX -E(x)]>c]<Var(x)/c?

and P|X - E(X)<c]>1-Var(X)/c?

7. Let f(x) = 5/28 for x 21 and O 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) s159%?

Solution: p(x) =[x2 dx = [Las =| | ms
8 el

Aye 4

25 25 = ws
E(X?) = |x’. dx dx -| | =
J x IF ae 1 3)

5 5.2 80-75 5
Var(X)=E(X’)-E’(X) ==-(=) = =
ar)= EOC)- EX) E D

8. By Chebyshev 's inequality,
ly Chebyshev *s inequality, Px -p|2ko]< 5

OA <4
48 | x

4

Presented by Mr Pradp Panda

9. "x 22415 FILE <>_Ji5*

48 4
=> P[X >2.5]+ P[X <0]< Gay

>P[x>25]<{ SPER ZOO
15

Presented by Mr Pradip Panda

10. For P[X 2a]<15% == =

nx25. o [5 ‚pl x <> ALLEN EN BE
4 3 48 4 3 48 | 20

11. Let f(x) be the uniform distribution on O < x < 10 and 0 otherwise. Give a bound using Chebyshev ’s for
P(2< x <8). Calculate the actual probability. How do they compare?

2

By Chebyshev 's inequality,
ly Chebyshev “s inequality, Alyx 3st |> m

5 5 1
1 == SH ES [2 ==

12. 343

Here, 5-2 k=2=k= 20

V3

393 5 393 5 25
e ASAS en

5 3 5 3 27)

po<sx<ssje À
24

Presented by Mr. Pradip Panda 8 July 2021

13. Presented by Mr Pradip Panda 8 July 2021

14. Let f(x) = e.e*for x< -l and O otherwise. Give a bound using Chebyshev 's for P(-4 < x < 0). For what a
P(X > a) > 0.997

-1
‘Solution: E(X)=e fueras

q sferar-[{ 2 | ala]

=e ee =-2

8 July 2021

15. Presented by Mr Pradp Panda 8 July 2021

Si
E(X’) =efxeax

= de fetax- ne eJeajas
Seal,
SI-ADI=5

Var(X) = E(X*)- E?(X)=5-4=1

CL

00

16. By Chebyshev ‘s inequality, Pix u E(X)| 2 ko]2 1- =

Pix -(-2)<k)21-%

PL aks xs-aehpei : 7 = = =

Presented by Mr Pradp Panda

8 July 2021

17. P[-12< X <8]> 0.99
> P[-12< X]>0.99

Presented by Mr Pradip Panda

18. Presented by Mr. Pradp Panda. 8 July 2021

19. Presented by Mr. Pradp Panda. 8 July 2021

20. 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}.

So, [dF (x) = P(X € S$) = P[g(X) 2k]
Ss

E[g(X)]= | g@)dF()

oo

s

21. Presented by Mr. Pradp Panda. 8 July 2021

22. X is a discrete random variable with p.m. f(x).
Elg(X)]= Let)

= Laos + 20/0

> 2 g(x) f(x)

>) f(x) =k.P[g(X) =k]

0, Pls(X)>&]<[Ets(X))/R]

23. If g(X)= [x -E(xX)f = [x = ul is taken and k is replaced by k°o° then
P(x - u) 240" |<[E1x - 10)? /4%0?]
> P| x - > k0]<1/x?

Which is the Chebyshev’s inequality .

24. If g(X)=|X| is taken then for any k >0
P[X|>k]< Eix|/k
Which is the Markov’s inequality .
If g(X)=|X|' is taken and k is replaced by k" then
e

Which is the generalised form of Markov’s
inequality .

p|x| >k"|< EIX

P

25. Thank you

Presented by tr. Pradip Panda