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Let $b_{{\ell }_1{\ell }_2{\ell }_3}^1$ denote the template at unit $f_{\mathrm{NL}}$, and define $A_{123}=b_{{\ell }_1{\ell }_2{\ell }_3}^1{\mathcal{G}}_{m_1{m}_2{m}_3}^{{\ell }_1{\ell }_2{\ell }_3}$. All sums below are over ordered triples of the retained multipoles, with the monopole removed and $C_{\ell }>0$. A finite maximum multipole makes these expressions ordinary finite sums. Taking the expectation of the cubic numerator and using the template relation givesThus the full-sky cubic bispectrum estimator is unbiased forThe assumed template must have nonzero support, so $F>0$. Using the Gaunt sum rule, writeThe factor $1/6$ belongs to the ordered sum; an unordered triangle sum would instead use multiplicity factors for equal multipoles. This normalization is also the Gaussian Fisher information for the template amplitude.

For the cosmic variance, use the reality condition ${\Theta }_{\ell m}^{\ast }=(-1{)}^m{\Theta }_{\ell ,-m}$ and Gaussian covarianceThere are $15$ pairings of the six multipoles in the squared cubic numerator. Six pairings connect every factor in the first triple to a factor in the second. Each gives $\sum A_{123}^2/(C_{{\ell }_1}{C}_{{\ell }_2}{C}_{{\ell }_3})=6F$. The covariance phases cancel: nonzero Gaunt integrals have $m_1+m_2+m_3=0$, and simultaneous reversal of the three $m$ indices multiplies the Gaunt integral by $(-1{)}^{{\ell }_1+{\ell }_2+{\ell }_3}=1$.

The other nine pairings contain one contraction within each triple. To see why their sums vanish, contract two legs of a Gaunt integral and use the spherical harmonic addition theorem:The template and inverse-covariance weights are independent of $m$, so they do not spoil this cancellation. The only possible surviving unpaired mode is the monopole, and ${\Theta }_{00}=0$ removes it. This is monopole cancellation of internal cubic-estimator contractions. It explains why no linear correction is needed in this ideal full-sky isotropic problem; masks or anisotropic noise would spoil the argument.

The cubic numerator therefore has Gaussian variance $6(6F)=36F$. Dividing by $(6F{)}^2$ givesThe result concerns the Gaussian-limit covariance. Non-Gaussian connected four- and six-point terms can change the variance at finite amplitude. Unbiasedness uses the assumed linear template relation for the observed three-point function.