We have a set of unidimensional data, $X_1, . . . , X_n$. : The data are drawn from a uniform distribution on the interval $[a, b]$. This model has two positive real parameters, a and b, such that $0 < a < b$. The idea is that we think about the joint distribution of MLE's $\hat a$ and $\hat b$. The joint pdf of the minimum and the maximum of a set of data drawn from pdf $f$, with cdf $F$, is:
$$f_{\hat a,\hat b}(x,y)=n(n-1)(F(y)-F(x))^{n-2}f(y)f(x)$$ Looking from here we can write the pdf down as (I hope I am not wrong here): $$f_{\hat a,\hat b}(x,y)=\frac{n(n-1)(y-x)^{n-2}}{(b-a)^{n}}$$My question is how we go about finding the expected value of mean of this estimator from here on.
I know that the result should be $$E[\hat \mu]=\frac{a+b}{2}$$ But I cannot reach that. I know how to get that for a pdf in one variable by integration. How do we approach this for two variables $x $ and $y$.
The best one can do with this not-a-real-question is perhaps to assume that one is interested in $$ \hat\mu=\frac{\hat a+\hat b}2. $$ Then, by definition, $$ E(\hat\mu)=\iint \frac{x+y}2\,f_{\hat a,\hat b}(x,y)\,\mathrm dx\mathrm dy. $$ Thus, the task is to check that, for every $a\lt b$, $$ E(\hat\mu)=\frac{a+b}2, $$ or, equivalently, that $$ \int_a^b\int_x^b(x+y)\,(y-x)^{n-2}\,\mathrm dy\,\mathrm dx=k\,(a+b)\,(b-a)^n, $$ where $$ k=\frac1{n(n-1)}. $$ The change of variable $x=a+(b-a)u$, $y=a+(b-a)v$, reduces this to $$ \int_0^1\int_u^1(2a+(b-a)(u+v))\,(v-u)^{n-2}\,\mathrm dv\,\mathrm du=k\,(a+b)=k\,(2a+(b-a)), $$ which holds for every $(a,b)$ if and only if $$ \int_0^1\int_u^1(v-u)^{n-2}\,\mathrm dv\,\mathrm du=k=\int_0^1\int_u^1(u+v)\,(v-u)^{n-2}\,\mathrm dv\,\mathrm du. $$ Can you check these two identities?