How to convert non-normal probabilistic constraints to deterministic ones for mathematical modelling?

Question

I am working on a chance-constrained optimisation model that takes into account uncertainty. I am aware of how to convert constraints that are of a probabilistic nature into the equivalent deterministic one, but only when the parameter follows a normal distribution. The majority of my constraints do follow a normal distribution, but I have a constraint that follows a gamma distribution with parameters $k = 3.11$ and $\theta = 9.86$. How would I then convert this to the equivalent deterministic form?

The formulation I used to convert the probabilistic constraints is as follows:

There are three coal types, $i = 1,2,3$

Each coal type's ash percentage, denoted by $a_i$, follows a normal distribution, $a_i \sim N(\mu_i,\sigma_{i}^2)$, $\forall i$. Each coal's ash percentage is independent.

The blend has a maximum allowable ash percentage of $10$, and $\alpha = 0.025$.

Let $x_i$ be the proportion of coal type $i$ to be used in the blend.

$$ P\bigg\{\sum_{i=1}^{3} a_ix_i \leq 10\bigg\} \geq 1 -\alpha, \quad x_i \geq 0 \quad \forall i $$ $$ P\Bigg\{\dfrac{\sum_{i=1}^{3} a_ix_i - \sum_{i=1}^{3} \mu_ix_i}{\sqrt{\sum_{i=1}^{3} \sigma_i^2x_i^2}} \leq \dfrac{10 - \sum_{i=1}^{3} \mu_ix_i}{\sqrt{\sum_{i=1}^{3} \sigma_i^2x_i^2}}\Bigg\} \geq 1 -\alpha $$

Where $\quad$ $\dfrac{\sum_{i=1}^{3} a_ix_i - \sum_{i=1}^{3} \mu_ix_i}{\sqrt{\sum_{i=1}^{3} \sigma_i^2x_i^2}}$ $\quad$ represents the standard normal variant with mean $0$ and variance of $1$.

The following inequality is then formed:

$$ \phi\Bigg(\dfrac{10 - \sum_{i=1}^{3} \mu_ix_i}{\sqrt{\sum_{i=1}^{3} \sigma_i^2x_i^2}}\Bigg) \geq \phi(K_{1-a}) $$

Where $\phi(K_{1-a}) = 1 - \alpha$ and $\phi()$ represents the standard normal cumulative distribution function. This gives the following deterministic constraint:

$$ \sum_{i=1}^{3}\mu_ix_i + K_{1-\alpha} *\sqrt{\sum_{i=1}^{3}\sigma_i^2x_i^2} \leq 10 $$

Answer 1

Your constraint says that the $(1-\alpha)$th fractile of the normal distribution must be $\le 10$. The particular normal distribution that you are bounding depends on the values of the decision variables. In general, the $(1-\alpha)$th fractile of a $N(\mu,\sigma^2)$ distribution is $\mu + z_{1-\alpha}\sigma$, hence the particular form in your constraint.

But any distribution has a $(1-\alpha)$th fractile. So your constraint just needs to say that the $(1-\alpha)$th fractile is $\le 10$. The fractile itself will depend on your distribution, as well as on the values of the decision variables.

Note that in general, the $(1-\alpha)$th fractile of a distribution is $F^{-1}(1-\alpha)$, where $F(\cdot)$ is the cdf.