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Actually, I wrote a paper in 1991 addressing this issue, available on Research Gate. In short, the mean and variance do not exist.

More precisely, if one defines a generalised inverse normal distribution by its density ($\alpha>1,\sigma>0$) $$f(z|\mu,\sigma,\alpha) = \dfrac{K(\alpha,\mu,\sigma)}{|z|^\alpha}\exp\left\{-(z^{-1}-\mu)^2/2\sigma^2\right\}$$ then $$K(\alpha,\mu,\sigma)^{-1}=\sigma^{\alpha-1}e^{-\mu^2/2\sigma^2}2^{(\alpha-1)/2}\Gamma((\alpha-1)/2){}_1F_1\left(\frac{1}{2}(\alpha-1);\frac{1}{2};\frac{\mu^2}{2\sigma^2}\right)$$ where ${}_1F_1$ is a confluent hypergeometric function. From this derivation of the normalising constant, one deduces that the mean only exists for $\alpha>2$ (while the inverse normal distribution corresponds to $\alpha=2$) and the variance only exists for $\alpha>3$. The mean is given by $$\mathbb{E}[X]=\frac{\mu}{\sigma^2}\,\frac{{}_1F_1\left(\frac{1}{2}(\alpha-3);\frac{3}{2};\frac{\mu^2}{2\sigma^2}\right)}{{}_1F_1\left(\frac{1}{2}(\alpha-1);\frac{1}{2};\frac{\mu^2}{2\sigma^2}\right)}$$

A much simpler argument as to why the mean does not exist is that, if $f$ is the normal $\text{N}(b,p)$ density, $$x^-1 f(x)\equiv \frac{e^{-b^2/2p}}{\sqrt{2\pi p}}x^{-1}$$ at zero, which is not integrable.

Actually, I wrote a paper in 1991 addressing this issue, available on Research Gate. In short, the mean and variance do not exist.

More precisely, if one defines a generalised inverse normal distribution by its density ($\alpha>1,\sigma>0$) $$f(z|\mu,\sigma,\alpha) = \dfrac{K(\alpha,\mu,\sigma)}{|z|^\alpha}\exp\left\{-(z^{-1}-\mu)^2/2\sigma^2\right\}$$ then $$K(\alpha,\mu,\sigma)^{-1}=\sigma^{\alpha-1}e^{-\mu^2/2\sigma^2}2^{(\alpha-1)/2}\Gamma((\alpha-1)/2){}_1F_1\left(\frac{1}{2}(\alpha-1);\frac{1}{2};\frac{\mu^2}{2\sigma^2}\right)$$ where ${}_1F_1$ is a confluent hypergeometric function. From this derivation of the normalising constant, one deduces that the mean only exists for $\alpha>2$ (while the inverse normal distribution corresponds to $\alpha=2$) and the variance only exists for $\alpha>3$. The mean is given by $$\mathbb{E}[X]=\frac{\mu}{\sigma^2}\,\frac{{}_1F_1\left(\frac{1}{2}(\alpha-3);\frac{3}{2};\frac{\mu^2}{2\sigma^2}\right)}{{}_1F_1\left(\frac{1}{2}(\alpha-1);\frac{1}{2};\frac{\mu^2}{2\sigma^2}\right)}$$

A much simpler argument as to why the mean does not exist is that, if $f$ is the normal $\text{N}(b,p)$ density, $$x^{-1} f(x)\equiv \frac{e^{-b^2/2p}}{\sqrt{2\pi p}}x^{-1}$$ at zero, which is not integrable.

Actually, I wrote a paper in 1991 addressing this issue, available on Research Gate. In short, the mean and variance do not exist.

More precisely, if one defines a generalised inverse normal distribution by its density ($\alpha>1,\sigma>0$) $$f(z|\mu,\sigma,\alpha) = \dfrac{K(\alpha,\mu,\sigma)}{|z|^\alpha}\exp\left\{-(z^{-1}-\mu)^2/2\sigma^2\right\}$$ then $$K(\alpha,\mu,\sigma)^{-1}=\sigma^{\alpha-1}e^{-\mu^2/2\sigma^2}2^{(\alpha-1)/2}\Gamma((\alpha-1)/2){}_1F_1\left(\frac{1}{2}(\alpha-1);\frac{1}{2};\frac{\mu^2}{2\sigma^2}\right)$$ where ${}_1F_1$ is a confluent hypergeometric function. From this derivation of the normalising constant, one deduces that the mean only exists for $\alpha>2$ (while the inverse normal distribution corresponds to $\alpha=2$) and the variance only exists for $\alpha>3$. The mean is given by $$\mathbb{E}[X]=\frac{\mu}{\sigma^2}\,\frac{{}_1F_1\left(\frac{1}{2}(\alpha-3);\frac{3}{2};\frac{\mu^2}{2\sigma^2}\right)}{{}_1F_1\left(\frac{1}{2}(\alpha-1);\frac{1}{2};\frac{\mu^2}{2\sigma^2}\right)}$$

Actually, I wrote a paper in 1991 addressing this issue, available on Research Gate. In short, the mean and variance do not exist.

More precisely, if one defines a generalised inverse normal distribution by its density ($\alpha>1,\sigma>0$) $$f(z|\mu,\sigma,\alpha) = \dfrac{K(\alpha,\mu,\sigma)}{|z|^\alpha}\exp\left\{-(z^{-1}-\mu)^2/2\sigma^2\right\}$$ then $$K(\alpha,\mu,\sigma)^{-1}=\sigma^{\alpha-1}e^{-\mu^2/2\sigma^2}2^{(\alpha-1)/2}\Gamma((\alpha-1)/2){}_1F_1\left(\frac{1}{2}(\alpha-1);\frac{1}{2};\frac{\mu^2}{2\sigma^2}\right)$$ where ${}_1F_1$ is a confluent hypergeometric function. From this derivation of the normalising constant, one deduces that the mean only exists for $\alpha>2$ (while the inverse normal distribution corresponds to $\alpha=2$) and the variance only exists for $\alpha>3$. The mean is given by $$\mathbb{E}[X]=\frac{\mu}{\sigma^2}\,\frac{{}_1F_1\left(\frac{1}{2}(\alpha-3);\frac{3}{2};\frac{\mu^2}{2\sigma^2}\right)}{{}_1F_1\left(\frac{1}{2}(\alpha-1);\frac{1}{2};\frac{\mu^2}{2\sigma^2}\right)}$$

A much simpler argument as to why the mean does not exist is that, if $f$ is the normal $\text{N}(b,p)$ density, $$x^-1 f(x)\equiv \frac{e^{-b^2/2p}}{\sqrt{2\pi p}}x^{-1}$$ at zero, which is not integrable.

Actually, I wrote a paper in 1991 addressing this issue, available on Research Gate. The mean and variance do not exist.

Actually, I wrote a paper in 1991 addressing this issue, available on Research Gate. In short, the mean and variance do not exist.

More precisely, if one defines a generalised inverse normal distribution by its density ($\alpha>1,\sigma>0$) $$f(z|\mu,\sigma,\alpha) = \dfrac{K(\alpha,\mu,\sigma)}{|z|^\alpha}\exp\left\{-(z^{-1}-\mu)^2/2\sigma^2\right\}$$ then $$K(\alpha,\mu,\sigma)^{-1}=\sigma^{\alpha-1}e^{-\mu^2/2\sigma^2}2^{(\alpha-1)/2}\Gamma((\alpha-1)/2){}_1F_1\left(\frac{1}{2}(\alpha-1);\frac{1}{2};\frac{\mu^2}{2\sigma^2}\right)$$ where ${}_1F_1$ is a confluent hypergeometric function. From this derivation of the normalising constant, one deduces that the mean only exists for $\alpha>2$ (while the inverse normal distribution corresponds to $\alpha=2$) and the variance only exists for $\alpha>3$. The mean is given by $$\mathbb{E}[X]=\frac{\mu}{\sigma^2}\,\frac{{}_1F_1\left(\frac{1}{2}(\alpha-3);\frac{3}{2};\frac{\mu^2}{2\sigma^2}\right)}{{}_1F_1\left(\frac{1}{2}(\alpha-1);\frac{1}{2};\frac{\mu^2}{2\sigma^2}\right)}$$