# How to generate a 'Discretized' beta distribution with mean and variance matching a 'Pure' beta distribution

Apologies for the rather vague title, I'm missing some terminology here. Feel free to correct me.

I would like to sample random variables from two distributions, $$Z$$, a Beta distribution with shape parameters $$\hat{a}$$ and $$\hat{b}$$, and $$XY$$ a product distribution generated by multiplying random variables sampled from a (different) Beta distribution ($$X$$, shape parameters $$a$$ and $$b$$) with independently sampled discrete random variables $$Y$$ equal to 1 with a probability $$c$$ and zero otherwise. The tricky part is I need the mean and variance of $$Z$$ and $$XY$$ to be identical.

Formally then (I'm not a mathematician so expect notation errors),

$$Z\sim B(\hat{a}, \hat{b})$$

$$X\sim B(a, b)$$

$$P(Y=1)=c,\ P(Y=0)=1-c$$

The following are the analytic expectation and variance of $$Z$$ and $$XY$$, the latter I get to using the expressions for E/var of a production distribution here:

$$E(Z)=\frac{1}{1+\hat{b}/\hat{a}}$$

$$Var(Z)=\frac{1}{4(2\hat{b}+1)}$$

$$E(XY)=\frac{c}{1+b/a}$$

$$Var(XY)=c\left(\frac{1}{4(2b+1)} + \frac{1-c}{(1+b/a)^2}\right)$$

Then setting $$E(Z)=E(XY)$$ and $$Var(Z)=Var(XY)$$ I aim to get expressions for $$a$$ and $$b$$ for known $$\hat{a}$$ and $$\hat{b}$$ and $$c$$. I arrive at the following:

$$a=\left(\frac{1}{4}\left(\frac{1}{4c(2\hat{b}+1)} - \frac{(1-c)}{(1-\hat{\beta})^2} \right)^{-1}-1\right)\frac{1}{2\hat{\beta}}$$

$$b=a\hat{\beta}$$

where

$$\hat{\beta}=(1+\hat{b}/\hat{a})c-1$$

Not super pretty in know.

The trouble is that when I compute $$a$$ and $$b$$ as above, the analytic expectation and variances of $$Z$$ and $$XY$$ don't match so clearly I have made a mistake somewhere in the analysis.

Can anyone suggest a correction or alternative approach to getting what I want?

Cheers