Question

The n candidates for a job have been ranked 1, 2, 3,..., n. Let X 5 the rank of a randomly selected candidate, so that X has pmf p(x)551yn x51,2,3,...,n 0 otherwise (this is called the discrete uniform distribution). Compute E(X) and V(X) using the shortcut formula. [Hint: The sum of the first n positive integers is n(n 1 1)y2, whereas the sum of their squares is n(n 1 1)(2n 1 1)y6.]

Answer 1

Question:

The n candidates for a job have been ranked 1, 2, 3,..., n. Let x = rank of a randomly selected candidate, so that x has pmf:

`p(x) = \left \{ {{(1)/(n)\ \ x=1,2,3...,n} \atop {0\ \ \ Otherwise}} \right.`

(this is called the discrete uniform distribution).

Compute E(X) and V(X) using the shortcut formula.

[Hint: The sum of the first n positive integers is
`(n(n +1))/(2)`, whereas the sum of their squares is
`(n(n +1)(2n+1))/(6)`

Answer:

`E(x) = (n+1)/(2)`

`Var(x) = (n^2 -1)/(12)` or
`Var(x) = ((n+1)(n-1))/(12)`

Explanation:

Given

PMF

`p(x) = \left \{ {{(1)/(n)\ \ x=1,2,3...,n} \atop {0\ \ \ Otherwise}} \right.`

Required

Determine the E(x) and Var(x)

E(x) is calculated as:

`E(x) = \sum \limits^(n)_(i) \ x * p(x)`

This gives:

`E(x) = \sum \limits^(n)_(x=1) \ x * (1)/(n)`

`E(x) = \sum \limits^(n)_(x=1) (x)/(n)`

`E(x) = (1)/(n)\sum \limits^(n)_(x=1) x`

From the hint given:

`\sum \limits^(n)_(x=1) x =(n(n +1))/(2)`

So:

`E(x) = (1)/(n) * (n(n+1))/(2)`

`E(x) = (n+1)/(2)`

Var(x) is calculated as:

`Var(x) = E(x^2) - (E(x))^2`

Calculating:
`E(x^2)`

`E(x^2) = \sum \limits^(n)_(x=1) \ x^2 * (1)/(n)`

`E(x^2) = (1)/(n)\sum \limits^(n)_(x=1) \ x^2`

Using the hint given:

`\sum \limits^(n)_(x=1) \ x^2 = (n(n +1)(2n+1))/(6)`

So:

`E(x^2) = (1)/(n) * (n(n +1)(2n+1))/(6)`

`E(x^2) = ((n +1)(2n+1))/(6)`

So:

`Var(x) = E(x^2) - (E(x))^2`

`Var(x) = ((n+1)(2n+1))/(6) - ((n+1)/(2))^2`

`Var(x) = ((n+1)(2n+1))/(6) - (n^2+2n+1)/(4)`

`Var(x) = (2n^2 +n+2n+1)/(6) - (n^2+2n+1)/(4)`

`Var(x) = (2n^2 +3n+1)/(6) - (n^2+2n+1)/(4)`

Take LCM

`Var(x) = (4n^2 +6n+2 - 3n^2 - 6n - 3)/(12)`

`Var(x) = (4n^2 - 3n^2+6n- 6n +2 - 3)/(12)`

`Var(x) = (n^2 -1)/(12)`

Apply difference of two squares

`Var(x) = ((n+1)(n-1))/(12)`