Question

The n candidates for a job have been ranked 1, 2, 3, . . . , n. Let X be the rank of a randomly selected candidate, so the X has the pmf p(x) =    1/n, if x = 1, 2, 3 . . . n, 0, otherwise. This is called the discrete uniform distribution. Compute E(X) and Var(X). (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 1

By definition of expectation,

`\displaystyle E[X]=\sum_xx\,P(X=x)=\sum_(x=1)^n\frac xn=(n(n+1))/(2n)=\boxed{\frac{n+1}2}`

and variance,

`V[X]=E[(X-E[X])^2]=E[X^2-2X\,E[X]+E[X]^2]=E[X^2]-E[X]^2`

Also by definition, we have

`E[f(X)]=\displaystyle\sum_xf(x)\,P(X=x)`

so that

`E[X^2]=\displaystyle\sum_(x=1)^n\frac{x^2}n=(n(n+1)(2n+1))/(6n)=\frac{(n+1)(2n+1)}6`

and finally,

`V[X]=\frac{(n+1)(2n+1)}6-\frac{(n+1)^2}4=\boxed{(n^2-1)/(12)}`

Answer 2

`(n^(2) - 1 )/(12)`

Explanation:

Data:

We collect the variables and simplify the result:

E[X] =
`\SIGMA \\`Σ x · p(x) =
`(1)/(n)`= ....

E[X²] =∑ x²· p(x) = ∑x²·
`(1)/(n)` = ....

Var [X] = E[X²] - E[X]² = ...

We then use the identities:

∑x =
`(n(n+1))/(2)` and ∑ x² =
`(n(n+1)(2n+1))/(6)`

simplifying the identities above gives:

`(n^(2-1) )/(12)`