Markov's Inequality: Estimating P(x>=a) in (1,4)

[clipperdude21]

Markov's Inequality !

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
EX= (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

Was this right? Thanks!

 

[EnumaElish]

For (a), may need to formally show actual value < upper bound.

For (b), I'd check the second derivative, to be safe.

 

[tacman]

Yes, your approach is correct. Markov's inequality states that for a non-negative random variable X and any constant a>0, the probability of X being greater than or equal to a is less than or equal to the expected value of X divided by a. In this case, the expected value of X is 5/2, so the probability of X being greater than or equal to a is less than or equal to 5/2a.

To find the value of a that minimizes the difference between the bound and the exact probability, we can set the two expressions equal to each other and solve for a. This gives us (4-a)/3 = 5/2a, which can be rearranged to get a quadratic equation 2a^2 - 11a + 12 = 0. Solving for a, we get a = 2 or a = 6. However, since a must be between 1 and 4, the only valid solution is a = 2.

Therefore, when a = 2, the bound given by Markov's inequality is the closest to the exact probability. This makes sense intuitively, as the expected value of X being 2.5 is closer to the exact probability (4-a)/3 = 1/3 when a = 2, compared to when a = 6.