If two sampling functions $p_{f}$ and $p_{g}$ are used to estimate the value of $∫f(x)g(x)dx$, then our Monte Carlo estimate becomes:

$n_{f}1 i=1∑n_{f} p_{f}(x_{i})f(x_{i})g(x_{i})w_{f}(x_{i}) +n_{g}1 j=1∑n_{g} p_{g}(y_{j})f(y_{j})g(y_{j})w_{g}(y_{j}) $where $n_{f}$ is the number of samples taken from a distribution method $p_{f}$, and $n_{g}$ is the number of samples taken from a distribution method $p_{g}$.

$w_{f}$ and $w_{g}$ are special weighting functions chosen such that the value of this estimator is the value of the integral above.

## Balance heuristic

Using the *balance heuristic* as $w_{f}$ and $w_{g}$, we get:

Adam also had this on the board, for some reason that I’ll probably discover eventually (it feels important):

$p_{f}(ω_{if})+p_{l}(ω_{if})f(ω_{if})⋅Li(ω_{if}) +p_{f}(ω_{il})+p_{l}(ω_{il})f(ω_{il})⋅Li(ω_{il}) $## Power heuristic

The *power heuristic* modifies the balance heuristic by squaring each term in $w_{f}$ and $w_{g}$, which further reduces the variance or something. So use that instead.