Could you please give a hint how to prove that the generalised normal distribution is normalised?

\begin{align} p(x\mid\mu,\alpha,\beta) = \frac{\beta}{2\alpha\Gamma(\frac{1}{\beta})}e^{-(|x-\mu|/\alpha)^\beta} \end{align}

Such that:

\begin{align} \int_{-\infty}^{+\infty} p(x\mid\mu,\alpha,\beta)\,dx = 1 \end{align}


closed as off-topic by Did, heropup, C. Falcon, Watson, user223391 Dec 26 '16 at 10:06

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    $\begingroup$ Hint: Using the change of variable $t=x^\beta$, one gets $$\int_0^\infty e^{-x^\beta}dx=\int_0^\infty e^{-t}\beta^{-1}t^{\beta^{-1}-1}dt=\beta^{-1}\Gamma(\beta^{-1})$$ $\endgroup$ – Did Dec 21 '16 at 22:46

Without the absolute value, this is often called a Generalized Gamma distribution, which gives a hint to the solution: a Gamma variate $X$ has been shifted to a mean $\mu$ (which will not change the normalization constant), rescaled by $\alpha$ (which divides the normalization constant by $\alpha$, as you can see), and then takes the $\beta \gt 0$ power. The absolute value introduces a mirror image of the distribution on the negative reals, whence we need to divide by $2$.

That shows we only need to check that the integral

$$I(\beta)=\int_0^\infty \exp(-u^\beta) du$$

equals $\frac{1}{\beta} \Gamma\left(\frac{1}{\beta}\right)$ and strongly suggests the substitution $v=u^\beta$, whence $$du = d(v^{1/\beta}) = \frac{1}{\beta}v^{1/\beta-1}dv$$ and

$$I(\beta) = \int_0^\infty \frac{1}{\beta}v^{1/\beta-1} e^{-v}dv = \frac{1}{\beta} \Gamma\left(\frac{1}{\beta}\right),$$



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