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Independence means $\mathrm{Pr}(X\in A,Y\in B)=\mathrm{Pr}(X\in A)\mathrm{Pr}(Y\in B)$ for all Borel sets $A,B$. For variables with a joint density this is equivalent to factorization of that density into the two marginal densities almost everywhere.

The inverse transformation is $x=uv$, $y=(1-u)v$, on $0<u<1$, $v>0$. Its absolute Jacobian determinant is $v$. Multiplying the independent gamma densities by it givesThe density is zero outside that support. Integrating in $v$ with the gamma integral givesIt is a beta density. Its normalization also proves the beta integral ${\int }_0^1{u}^{\alpha -1}(1-u{)}^{\beta -1}du=\Gamma (\alpha )\Gamma (\beta )/\Gamma (\alpha +\beta )$. The joint density consequently factors as this normalized beta density timesThus $V$ is gamma with shape $\alpha +\beta$ and rate $\lambda$, and $U,V$ are independent.

Finally $X=UV$. All relevant first moments are finite, and independence gives $\mathbb{E}X=\mathbb{E}U\mathbb{E}V$, while $\mathbb{E}V=\mathbb{E}X+\mathbb{E}Y>0$. HenceThis identity follows from the particular beta-gamma independence here; a general ratio of random variables would not permit moving expectation through division.

The same beta integral with one additional power of $y$ givesSubtracting the square of the mean yieldsAt $r=1$ and $r=n$, therefore,Their equality also follows by reflecting every uniform observation via $y\mapsto 1-y$.

For nonnegative integers $j,k$, repeated integration by parts gives the beta integral ${\int }_0^1{y}^j(1-y{)}^{k}dy=j!k!/(j+k+1)!$. Multiplying the stated density by $y$ and using this formula yieldsCancellation gives