I'm reading Bayesian data analysis. On page 85 it is stated that "In the d dimensional normal distribution, the logarithm of the density function is a constant plus a $χ^2_d$ distribution divided by −2." Can someone clarify how to derive this?
For example, let's use $d = 1$. Then the logarithm of the density function of the normal distribution is: $$ \begin{align} ln(p(\theta)) &= ln\left({\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {\theta-\mu }{\sigma }}\right)^{2}}\right) \\ &= ln\left(\frac {1}{\sigma {\sqrt {2\pi }}}\right) -{\frac {1}{2}}\left({\frac {\theta-\mu }{\sigma }}\right)^{2} \\ &= C -{\frac {1}{2}}\left({\frac {\theta-\mu }{\sigma }}\right)^{2} \end{align} $$ So the above statement would imply that $\left({\frac {\theta-\mu }{\sigma }}\right)^{2}$ is a $χ^2_1$ distribution. However, when I look up that distribution I get: $$ \begin{align} χ^2_1(\theta) &= {\frac {1}{2^{d/2}\Gamma (d/2)}}\;\theta^{d/2-1}e^{-\theta/2} \\ &= {\frac {1}{2^{1/2}\Gamma (1/2)}}\;\theta^{1/2-1}e^{-\theta/2} \\ &= \frac{1}{\sqrt{2\pi}}\;\theta^{-1/2}e^{-\theta/2} \end{align} $$ I do not see how these two things are the same. Especially the exponent of theta seems to behave very differently from what is in essence a quadratic formula in theta above.