How would we go about finding the expectation and variance of the r$^{\text{th}}$ order statistic $X_{(r)}$ from a random sample $X_1, \ldots, X_n$ from a uniform distribution with density function
$$f(x)=1 \quad \quad \text{for} \ x \in (0,1)$$
and $f(x)=0$ otherwise?
Intuitively we can see the expectation is $\frac{r}{n+1}$ though I'm keen to see how that is derived from simply knowing the pdf of the distribution.
Best Answer
Suppose you know that $$ a\Gamma(a) = \Gamma(a+1) \tag{1} $$ and that $$ \int_0^1 x^{\alpha-1} (1-x)^{\beta-1}\,dx = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}. \tag{2} $$ From $(2)$, we get the probability density function of the Beta distribution with parameters $\alpha$ and $\beta$: $$ \int_0^1 f(x)\,dx = \int_0^1 \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} x^{\alpha-1}(1-x)^{\beta-1}\,dx=1. $$ Now we want the first and second moments $\mathbb{E}(X)$ and $\mathbb{E}(X^2)$ of a random variable $X$ with this density. $$ \begin{align} \mathbb{E}(X) & = \int_0^1 x f(x)\, dx = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \int_0^1 x \cdot x^{\alpha-1} (1-x)^{\beta-1} \, dx \\ \\ & = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \int_0^1 x^{(\alpha+1)-1} (1-x)^{\beta-1} \, dx. \tag{3} \end{align} $$ The integral identity in $(2)$ holds if $\alpha$ is any number at all; therefore it holds for $\alpha+1$: $$ \int_0^1 x^{(\alpha+1)-1} (1-x)^{\beta-1} \, dx = \frac{\Gamma(\alpha+1)\Gamma(\beta)}{\Gamma((\alpha+1)+\beta)}. $$ It follows that the product in $(3)$ is $$ \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \cdot \frac{\Gamma(\alpha+1)\Gamma(\beta)}{\Gamma((\alpha+1)+\beta)} $$
So $\Gamma(\beta)$ cancels, and $(1)$ can be applied to change $\displaystyle\frac{\Gamma(\alpha+1)}{\Gamma(\alpha)}$ to $\alpha$, and to change $\displaystyle\frac{\Gamma(\alpha+\beta)}{\Gamma((\alpha+1)+\beta)}$ to $\displaystyle\frac{1}{\alpha+\beta}$. Therefore the expression in $(3)$ simplifies to $$ \frac{\alpha}{\alpha+\beta}. $$ A similar technique differing only in details finds $\mathbb{E}(X^2)$.
Once you have $\mathbb{E}(X)$ and $\mathbb{E}(X^2)$, you can use $\operatorname{var}(X) = \mathbb{E}(X^2) - (\mathbb{E}(X))^2$. There's some algebraic simplifying to be done after that. Remember that the variance should be symmetric in $\alpha$ and $\beta$, so if what you get is not symmetric, there's a mistake.