Suppose the trial function (our model) is the exact wave function. The action of the hamiltionan on the wave function $$ H\Psi = \mathrm{constant}\times \Psi, $$ The integral which defines various expectation values involving moments of the hamiltonian becomes then $$ \langle E^n \rangle = \langle H^n \rangle = \frac{\int d\boldsymbol{R}\Psi^{\ast}(\boldsymbol{R})H^n(\boldsymbol{R})\Psi(\boldsymbol{R})} {\int d\boldsymbol{R}\Psi^{\ast}(\boldsymbol{R})\Psi(\boldsymbol{R})}= \mathrm{constant}\times\frac{\int d\boldsymbol{R}\Psi^{\ast}(\boldsymbol{R})\Psi(\boldsymbol{R})} {\int d\boldsymbol{R}\Psi^{\ast}(\boldsymbol{R})\Psi(\boldsymbol{R})}=\mathrm{constant}. $$ This gives an important information: If I want the variance, the exact wave function leads to zero variance! The variance is defined as $$ \sigma_E = \langle E^2\rangle - \langle E\rangle^2. $$ Variation is then performed by minimizing both the energy and the variance.