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:
[tex]p(x) = \left \{ {{\frac{1}{n}\ \ x=1,2,3...,n} \atop {0\ \ \ Otherwise}} \right.[/tex]
(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 [tex]\frac{n(n +1)}{2}[/tex], whereas the sum of their squares is [tex]\frac{n(n +1)(2n+1)}{6}[/tex]
Answer:
[tex]E(x) = \frac{n+1}{2}[/tex]
[tex]Var(x) = \frac{n^2 -1}{12}[/tex] or [tex]Var(x) = \frac{(n+1)(n-1)}{12}[/tex]
Step-by-step explanation:
Given
PMF
[tex]p(x) = \left \{ {{\frac{1}{n}\ \ x=1,2,3...,n} \atop {0\ \ \ Otherwise}} \right.[/tex]
Required
Determine the E(x) and Var(x)
E(x) is calculated as:
[tex]E(x) = \sum \limits^{n}_{i} \ x * p(x)[/tex]
This gives:
[tex]E(x) = \sum \limits^{n}_{x=1} \ x * \frac{1}{n}[/tex]
[tex]E(x) = \sum \limits^{n}_{x=1} \frac{x}{n}[/tex]
[tex]E(x) = \frac{1}{n}\sum \limits^{n}_{x=1} x[/tex]
From the hint given:
[tex]\sum \limits^{n}_{x=1} x =\frac{n(n +1)}{2}[/tex]
So:
[tex]E(x) = \frac{1}{n} * \frac{n(n+1)}{2}[/tex]
[tex]E(x) = \frac{n+1}{2}[/tex]
Var(x) is calculated as:
[tex]Var(x) = E(x^2) - (E(x))^2[/tex]
Calculating: [tex]E(x^2)[/tex]
[tex]E(x^2) = \sum \limits^{n}_{x=1} \ x^2 * \frac{1}{n}[/tex]
[tex]E(x^2) = \frac{1}{n}\sum \limits^{n}_{x=1} \ x^2[/tex]
Using the hint given:
[tex]\sum \limits^{n}_{x=1} \ x^2 = \frac{n(n +1)(2n+1)}{6}[/tex]
So:
[tex]E(x^2) = \frac{1}{n} * \frac{n(n +1)(2n+1)}{6}[/tex]
[tex]E(x^2) = \frac{(n +1)(2n+1)}{6}[/tex]
So:
[tex]Var(x) = E(x^2) - (E(x))^2[/tex]
[tex]Var(x) = \frac{(n+1)(2n+1)}{6} - (\frac{n+1}{2})^2[/tex]
[tex]Var(x) = \frac{(n+1)(2n+1)}{6} - \frac{n^2+2n+1}{4}[/tex]
[tex]Var(x) = \frac{2n^2 +n+2n+1}{6} - \frac{n^2+2n+1}{4}[/tex]
[tex]Var(x) = \frac{2n^2 +3n+1}{6} - \frac{n^2+2n+1}{4}[/tex]
Take LCM
[tex]Var(x) = \frac{4n^2 +6n+2 - 3n^2 - 6n - 3}{12}[/tex]
[tex]Var(x) = \frac{4n^2 - 3n^2+6n- 6n +2 - 3}{12}[/tex]
[tex]Var(x) = \frac{n^2 -1}{12}[/tex]
Apply difference of two squares
[tex]Var(x) = \frac{(n+1)(n-1)}{12}[/tex]
