# Markov's Inequality Problem Help! MIDTERM MONDAY!

• Nov 17th 2007, 03:20 PM
clipperdude21
Markov's Inequality Problem Help! MIDTERM MONDAY!
Let X be uniformly distributed over (1,4)

(a) Use Markov's inequlaity to estimate P(x>=a) a is between 1 to 4 and compare this result to the exact answer.

(b) Find the value of a in (1,4) that minimizes the difference between the bound and the exact probability computed in (a).

For this question i used
http://upload.wikimedia.org/math/7/f...7c9cea7ccb.pngEX= (a+b)/2 since its uniformly distributed so i got EX=5/2 which means that the probability of X being greater than or equal to a is less than 5/2a. For the exact value I got the dist function of a uniform RV as being (x-a)/(b-a) so the F(x) should be (a-1)/4. The exact value is 1-(a-1)/4 so i got the exact value as being (4-a)/3.(b) I had (4-a)/3 <= 5/2a and then got them to one side took the derivative and set it equal to 0 and got 2.738

• Nov 17th 2007, 05:52 PM
kalagota
Quote:

Originally Posted by clipperdude21
Let X be uniformly distributed over (1,4)

(a) Use Markov's inequlaity to estimate P(x>=a) a is between 1 to 4 and compare this result to the exact answer.

(b) Find the value of a in (1,4) that minimizes the difference between the bound and the exact probability computed in (a).

For this question i used
http://upload.wikimedia.org/math/7/f...7c9cea7ccb.pngEX= (a+b)/2 since its uniformly distributed so i got EX=5/2 which means that the probability of X being greater than or equal to a is less than 5/2a. For the exact value I got the dist function of a uniform RV as being (x-a)/(b-a) so the F(x) should be (a-1)/4. The exact value is 1-(a-1)/4 so i got the exact value as being (4-a)/3.(b) I had (4-a)/3 <= 5/2a and then got them to one side took the derivative and set it equal to 0 and got 2.738