The gamma distribution is defined as

\[\begin{align*} Gamma_X(x; \alpha, \beta)=\begin{cases} \frac{\beta^\alpha x^{\alpha-1}e^{-\beta x}}{\Gamma(\alpha)} &x>0\\ 0 &otherwise \end{cases} \end{align*}\]

and it has

\[\begin{align*} EX = \frac{\alpha}{\beta} && Var(X) = \frac{\alpha}{\beta^2} \end{align*}\]

We define the bigamma distribution as

\[\begin{align*} BiGamma_X(x; \alpha, \beta)=\begin{cases} \frac{1}{2}Gamma_X(x; \alpha, \beta) &x>0\\ \frac{1}{2}Gamma_X(-x; \alpha, \beta) &otherwise \end{cases} \end{align*}\]

which has mean zero and variance

\[\begin{align*} E_{Bigamma}X^2 = &\int_{-\infty}^\infty x^2 Bigamma(x)dx\\ = &\frac{1}{2}\int_{-\infty}^0 x^2 Gamma(-x)dx + \frac{1}{2}\int_{0}^\infty x^2 Gamma(x)dx\\ = &\int_{0}^\infty x^2 Gamma(x)dx\\ = &E_{Gamma}X^2 = \frac{\alpha(\alpha+1)}{\beta^2}\\\\ \end{align*}\]

so it has

\[\begin{align*} EX = 0 && Var(X) = E_{Gamma}X^2-(E_{Gamma}X)^2= \frac{\alpha(\alpha+1)}{\beta^2} \end{align*}\]

The slightly more general form

\[\begin{align*} BiGamma_X(x; \alpha, \beta, \delta)=\begin{cases} (1-\delta) Gamma_X(x; \alpha, \beta) &x>0\\ \delta Gamma_X(-x; \alpha, \beta) &otherwise \end{cases} \end{align*}\]

with $\delta \in [0, 1]$ has instead

\[\begin{align*} E_{Bigamma}X = &\int_{-\infty}^\infty x Bigamma(x)dx\\ = &\delta\int_{-\infty}^0 x Gamma(-x)dx + (1-\delta)\int_{0}^\infty x Gamma(x)dx\\ = &-\delta\int_{0}^\infty x Gamma(x)dx+ (1-\delta)\int_{0}^\infty x Gamma(x)dx\\ = &(1-2\delta)E_{Gamma}X = (1-2\delta)\frac{\alpha}{\beta}\\ \end{align*}\]

and

\[\begin{align*} E_{Bigamma}X^2 = &\int_{-\infty}^\infty x^2 Bigamma(x)dx\\ = &\delta\int_{-\infty}^0 x^2 Gamma(-x)dx + (1-\delta)\int_{0}^\infty x^2 Gamma(x)dx\\ = &\delta\int_{0}^\infty x^2 Gamma(x)dx+ (1-\delta)\int_{0}^\infty x^2 Gamma(x)dx\\ = &E_{Gamma}X^2 = \frac{\alpha(\alpha+1)}{\beta^2}\\ \end{align*}\]

so it has

\[\begin{align*} EX =& (1-2\delta)\frac{\alpha}{\beta} \\ Var(X) =& E_{Gamma}X^2-(E_{Gamma}X)^2= \frac{\alpha(\alpha+1)}{\beta^2}-(1-2\delta)^2\frac{\alpha^2}{\beta^2}\\ =&\frac{\alpha}{\beta^2}(\alpha+1-(1-2\delta)^2\alpha) \end{align*}\]

It therefore iterpolates between the $Gamma$ and the first definition of a zero mean $BiGamma$.

For higher moments:

\[\begin{align*} E_{Bigamma}X^n = &\int_{-\infty}^\infty x^n Bigamma(x)dx\\ = &\delta\int_{-\infty}^0 x^n Gamma(-x)dx + (1-\delta)\int_{0}^\infty x^n Gamma(x)dx\\ = &\delta(-1)^n \int_{0}^\infty x^n Gamma(x)dx+ (1-\delta)\int_{0}^\infty x^n Gamma(x)dx\\ = &(1-\delta + \delta(-1)^n)E_{Gamma}X^n \\ = &(1-\delta + \delta(-1)^n)\frac{\Gamma(n+\alpha)}{\beta^n\Gamma(\alpha)}\\ \end{align*}\]