Discrete uniform distribution prrof

[synkk]

Hello, I'm currently in high school and going over discrete uniform distribution, and we've come across this formula. I'm curious if anyone could show me how the formula is true, as when I asked my teacher he just said that it'll confuse the class and we don't need to know why it's true.

If anyone could show me a proof or something i'd be very grateful :)

 

[Ray Vickson]

Hello, I'm currently in high school and going over discrete uniform distribution, and we've come across this formula. I'm curious if anyone could show me how the formula is true, as when I asked my teacher he just said that it'll confuse the class and we don't need to know why it's true.

If anyone could show me a proof or something i'd be very grateful :)

You can do it directly if you know formulas for $\sum_{k=1}^n k \text{ and } \sum_{k=1}^n k^2,$ and these can be found on-line, for example. Another way is to prove the results by induction (although I don't know if you have studied that, yet).

Let's just do it directly for the E(X). The probability mass function is $p(k) = \Pr \{X=k\} = 1/n,$ for k = 1, 2, ..., n . The expected value is *defined* as $$E(X) = 1\cdot p(1) + 2 \cdot p(2) + 3 \cdot p(3) + \cdots + n \cdot p(n) = \frac{1}{n}[1 + 2 + \cdots + n].$$ This last summation is $$1+2+ \cdots +n = \frac{n(n+1)}{2},$$ so we we get the stated result.

Getting $\text{Var}(X)$ is more complicated, but you can use the easily-proven fact that $\text{Var}(X) = E(X^2) - (EX)^2,$ and so reduce the problem to finding $$E(X^2) = p(1) \cdot 1^1 + p(2) \cdot 2^2 + \cdots + p(n) \cdot n^2 = \frac{1}{n} [ 1^2 + 2^2 + \cdots n^2].$$

RGV