clipperdude21
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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 rightEX 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 <= 4/3For 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!
The answer in the book got something different than what i got so i wanted to see if i was rightEX 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 <= 4/3For 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!