Question

A random variableX= {0, 1, 2, 3, ...} has cumulative distribution function.a) Calculate the probability that 3 ≤X≤ 5.b) Find the expected value of X, E(X), using the fact that. (Hint: You will have to evaluate an infinite sum, but that will be easy to do if you notice that

Answer 1

a) P ( 3 ≤X≤ 5 ) = 0.02619

b) E(X) = 1

Explanation:

Given:

- The CDF of a random variable X = { 0 , 1 , 2 , 3 , .... } is given as:

`F(X) = P ( X =< x) = 1 - (1)/((x+1)*(x+2))`

Find:

a.Calculate the probability that 3 ≤X≤ 5

b) Find the expected value of X, E(X), using the fact that. (Hint: You will have to evaluate an infinite sum, but that will be easy to do if you notice that

Solution:

- The CDF gives the probability of (X < x) for any value of x. So to compute the P ( 3 ≤X≤ 5 ) we will set the limits.

`F(X) = P ( 3=<X =< 5) = [1 - (1)/((x+1)*(x+2))]\limits^5_3\\\\F(X) = P ( 3=<X =< 5) = [-(1)/((5+1)*(5+2)) + (1)/((3+1)*(3+2))}\\\\F(X) = P ( 3=<X =< 5) = [-(1)/((42)) + (1)/((20))}]\\\\F(X) = P ( 3=<X =< 5) = 0.02619`

- The Expected Value can be determined by sum to infinity of CDF:

E(X) = Σ ( 1 - F(X) )

`E(X) = (1)/((x+1)*(x+2)) = (1)/((x+1)) - (1)/((x+2)) \\\\= (1)/((1)) - (1)/((2))\\\\= (1)/((2)) - (1)/((3)) \\\\= (1)/((3)) - (1)/((4))\\\\= ............................................\\\\= (1)/((n)) - (1)/((n+1))\\\\= (1)/((n+1)) - (1)/((n+ 2))`

E(X) = Limit n->∞ [1 - 1 / ( n + 2 ) ]

E(X) = 1