# Thread: Chebyshev's Inequality Problem!! Help

1. ## Chebyshev's Inequality Problem!! Help

Let X be uniformly distributed over (-2,2). Use Chebyshev's inequality to estimate P(abs(x) >= 1) and compare to the exact value.

The answer in the book got something different than what i got so i wanted to see if i was right

EX of a uniformly distrubuted RV is (a+b)/2 so -2+2/0 =0
So i plugged into the formula.

P(abs(x)>=1 <= sigma^2

the variance of a uniformly RV is (b-a)^2/12 so (2-(-2))^2=14 and 14/12 is 4/3.

my first answer ======= P(abs(x)>=1 <= 4/3

For the exact part I used the P(X<=x) cdf of a unform RV and got F(1)=3/4. So 1-F(1) = 0.25 which is the probability we are looking for.

Did i do this right? Thanks!

2. Originally Posted by clipperdude21
Let X be uniformly distributed over (-2,2). Use Chebyshev's inequality to estimate P(abs(x) >= 1) and compare to the exact value.

The answer in the book got something different than what i got so i wanted to see if i was right

EX of a uniformly distrubuted RV is (a+b)/2 so -2+2/0 =0
So i plugged into the formula.

P(abs(x)>=1 <= sigma^2

the variance of a uniformly RV is (b-a)^2/12 so (2-(-2))^2=14 and 14/12 is 4/3.

my first answer ======= P(abs(x)>=1 <= 4/3

For the exact part I used the P(X<=x) cdf of a unform RV and got F(1)=3/4. So 1-F(1) = 0.25 which is the probability we are looking for.

Did i do this right? Thanks!
where did you get your equation?

anyways, i'll try to solve it using our way..
Chebyshev's Inequality. If X is a random variable such that Var X exists, then for all r>0,
$\displaystyle P[(x- \mu _x)^2 \geq k^2 \sigma _x ^2] \leq \frac{1}{k^2}$

you want to estimate $\displaystyle P[|x| \geq 1]$.

using what you got, E[X] = 0, Var X = 4/3
so,

$\displaystyle P[(x- \mu _x)^2 \geq k^2 \sigma _x ^2] = P\left[ {(x- 0)^2 \geq k^2 \left( {\frac{4}{3}}\right) }\right]$

$\displaystyle =P\left[ {(x - 0)^2 \geq k^2 \left( {\frac{4}{3}}\right) }\right]$

you need $\displaystyle k^2 = \frac{3}{4}$ so that

$\displaystyle P\left[ {(x - 0)^2 \geq \left( {\frac{3}{4}}\right) \left( {\frac{4}{3}}\right) }\right] = P[|x| \geq 1] \leq \frac{1}{k^2} = \frac{4}{3}$
(well, it is just the same.. Ü)
the exact part is:

$\displaystyle P[X \leq x] \implies P[|x| \geq 1] = 1 - P[|x| < 1]$

$\displaystyle = 1 - (P[x < 1]-P[x < -1]) = 1 + P[x < -1] - P[x<1]$ ----- i haven't solved it..