Calculating mean and variance of reparameterized distributions If I have an exponential distribution:
$$ x \sim \operatorname{Exponential}(1) $$
mean and variance can be expressed as:
$$ \lambda = 1 $$
$$ \text{variance} = \lambda^{-2}  $$
$$ \text{mean} = 1 / \lambda $$
How would the mean and variance change if the distribution was expressed with respect to a normal random variable with mean $= 1$?
$$ x \sim \operatorname{HalfNormal}(1, 3) $$
$$ y \sim \operatorname{Exponential}(x) $$
It doesn't seem like the mean of the exponential distribution would change but it does feel like variance should increase. How can you express this mathematically?
 A: The half-normal distribution is not parametrized by two parameters, but only one:  $$f_X(x) = \frac{\sqrt{2}}{\sigma \sqrt{\pi}} e^{-x^2/(2\sigma^2)}, \quad x \ge 0.$$  Thus its mean and variance are uniquely determined by $\sigma$.
If you use the folded normal distribution, you get two parameters, but I don't know if this is what you want to use.
In any case, the general approach to computing moments for a hierarchical model is to use the formulas for total expectation and total variance; namely, $$\operatorname{E}[Y] = \operatorname{E}[\operatorname{E}[Y \mid X]], \\ \operatorname{Var}[Y] = \operatorname{E}[\operatorname{Var}[Y \mid X]] + \operatorname{Var}[\operatorname{E}[Y \mid X]].$$  In the case that $$Y \mid X \sim \operatorname{Exponential}(X),$$ with your choice of parametrization such that $$\operatorname{E}[Y \mid X] = 1/X, \quad \operatorname{Var}[Y \mid X] = 1/X^2,$$ we then get $$\operatorname{E}[Y] = \operatorname{E}[1/X], \\ \operatorname{Var}[Y] = \operatorname{E}[1/X^2] + \operatorname{Var}[1/X].$$  For a half-normal $X$, the expectation of $1/X$ is not finite, thus the unconditional moments of $Y$ fail to exist.
