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

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This discussion focuses on applying Markov's Inequality to estimate the probability P(X ≥ a) for a uniformly distributed random variable X over the interval (1, 4). The user calculated the expected value EX as 2.5 and derived the upper bound for the probability as 5/(2a). The exact probability was computed as (4-a)/3. The user also sought to minimize the difference between the bound and the exact probability, finding that the optimal value of a is approximately 2.738. The need for a formal verification of the upper bound against the actual value was highlighted.

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

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

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