For $X \sim U(a,b)$, $V(X) = \frac{(b-a)^2}{12}$, but I have seen some sources have it as $\frac{1}{12}(b-a)(b-a+2)$. In fact, a question simply asked me to prove the that variance is $\frac{1}{12}(b-a)(b-a+2)$ given the uniform distribution.
I am able to successfully prove it to be $\frac{(b-a)^2}{12}$ but I'm not sure where that $2$ term can come from. When I look online, I actually see BOTH these on google images, but no explanation of the difference.
Can someone explain?
The difference between the two formulas arises because one is the variance of the continuous uniform distribution on the interval $[a,b]$, and the other is the variance of the discrete uniform distribution on the set $\{a, a+1, \ldots, b-1, b\}$.
The difference is between two different distributions
1.
$$X\sim U(a,b)$$
2.
$$Y\sim U\{a;b\}$$
is a continuous distribution and its variance is $\mathbb{V}[X]=\frac{(b-a)^2}{12}$
is a discrete distribution with variance $\mathbb{V}[Y]=\frac{n^2-1}{12}=\frac{(b-a)(b-a+2)}{12}$
If you want to prove $V(Y)$ the easiest way is to consider a shifted rv
$$Y\sim U\{1;n\}$$
and calculate
$$\mathbb{E}[Y]=\frac{1}{n}\sum_{i=1}^{n} i=\frac{n+1}{2}$$
$$\mathbb{E}[Y^2]=\frac{1}{n}\sum_{i=1}^{n} i^2=\frac{(n+1)(2n+1)}{6}$$
and thus
$$\mathbb{V}[Y]=\frac{(n+1)(2n+1)}{6}-\left(\frac{n+1}{2}\right)^2=\dots=\frac{n^2-1}{12}$$
this is enough for your proof being
$$n=b-a+1$$
and thus
$$n^2-1=(b-a)(b-a+2)$$
To derive the variance of a discrete uniform random variable $Y\sim U\{a,a+1,\dots,b\}$, note that if $V\sim U[0,1)$ and independent of $Y$, then $X=Y+V\sim U[a,b+1)$. By additivity of variance of independent random variables, we have $$ \operatorname{Var}X = \operatorname{Var}Y+\operatorname{Var}V $$ and since $\operatorname{Var}V=1/12$ and $\operatorname{Var}(X)=(b+1-a)^2/12$, $$ \operatorname{Var} Y=\frac{(b+1-a)^2-1}{12}. $$
