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This post is purely about checking my computation . I am getting a result I am not supposed to get , which means there must be a computation mistake but I really need help finding it . To answer this post, all you need to do is to follow through my calculation and find out the first mistake you see. (Update: the LaTeX newcommand issue and the font issue has been completely resolved.)
Fix positive integer $d$ and real number $\theta$. Define function $f: \mathbb{R}^d \to \mathbb{R}$ by setting, for every $x = (x_1, \cdots, x_d)$:
\begin{equation*} f(x) = C \exp \curly{- \sum_{i=1}^d (x_i-\theta)^4} \end{equation*}
Here $C$ is a constant such that $f$ becomes a probability density function with respect to the Lebesgue measure.
The integral calculator shows that:
\begin{equation*} \int_{\mathbb{R}} \exp(-x^4)\, dx = \frac{1}{2} \Gamma \bracket{\frac{1}{4}} \end{equation*} \begin{equation*} \int_{\mathbb{R}} x^3\exp(-x^4)\,dx = 0 \end{equation*} \begin{equation*} \int_{\mathbb{R}} x^2\exp(-x^4)\,dx = \frac{1}{2} \Gamma \bracket{\frac{3}{4}} \end{equation*} \begin{equation*} \int_{\mathbb{R}} x^4\exp(-x^4)\, dx = \frac{1}{8} \Gamma \bracket{\frac{1}{4}} \end{equation*} \begin{equation*} \int_{\mathbb{R}} x^6\exp(-x^4)\, dx = \frac{3}{8} \Gamma \bracket{\frac{3}{4}} \end{equation*}
Therefore, $C = \bracket{\frac{1}{2} \Gamma \bracket{\frac{1}{4}}}^{-d}$. Next, define:
\begin{equation*} l_{\theta}(x) = \ln (C) - \sum_{i=1}^d (x_i-\theta)^4 \end{equation*} \begin{equation*} \dot{l_{\theta}}(x) = 4 \sum_{i=1}^d (x_i-\theta)^3 \end{equation*} \begin{equation*} \ddot{l_{\theta}}(x) = - 12 \sum_{i=1}^d (x_i - \theta)^2 \end{equation*}
And we define the matrix:
\begin{equation*} M_{\theta} = \begin{bmatrix} \mathbb{E}_{\theta} \squared{\dot{l_{\theta}}^2} & \mathbb{E}_{\theta} \squared{\dot{l_{\theta}}\ddot{l_{\theta}}} \\ \mathbb{E}_{\theta} \squared{\dot{l_{\theta}}\ddot{l_{\theta}}} & \mathbb{E}_{\theta} \squared{\ddot{l_{\theta}}^2} - \iota_{\theta}^2 \end{bmatrix} \end{equation*}
Here $\iota_{\theta} = \mathbb{E}_{\theta} \squared{\dot{l_{\theta}}^2}$. Finally, the statistical curvature $\gamma_{\theta}$ is defined as:
\begin{equation*} \gamma_{\theta} = \sqrt{\frac{\det{M_{\theta}}}{\iota_{\theta}^3}} = \sqrt{\frac{\mathbb{E}_{\theta}\squared{\ddot{l_{\theta}}^2}}{\iota_{\theta}^2} - \frac{\mathbb{E}_{\theta}^2\squared{\dot{l_{\theta}} \ddot{l_{\theta}}}}{\iota_{\theta}^3}- 1} \end{equation*}
Our goal is to compute the curvature for this $f$. We will be repeatedly using the formula:
\begin{equation*} \bracket{\sum_{i=1}^da_i}^2 = \sum_{i=1}^da_i^2 + 2\sum_{i=1}^d \sum_{j=1}^{i-1} a_ia_j \end{equation*}
We have:
\begin{equation*} \begin{split} & \iota_{\theta} = \mathbb{E}_{\theta} \squared{\dot{l_{\theta}}^2} = \int_{\mathbb{R}^d} \bracket{4 \sum_{i=1}^d (x_i-\theta)^3}^2 C \exp{- \sum_{i=1}^d (x_i-\theta)^4} dx \\ & = 16 C \curly{\sum_{i=1}^d \int_{\mathbb{R}^d} x_i^6 \exp{- \sum_{i=1}^d x_i^4} dx + 2 \sum_{i=1}^d \sum_{j=1}^{i-1}\int_{\mathbb{R}^d} x_i^3x_j^3 \exp{- \sum_{i=1}^d x_i^4} dx } \\ & = 16 C d \bracket{\frac{1}{2} \Gamma \bracket{\frac{1}{4}}}^{d-1} \frac{3}{8} \Gamma \bracket{\frac{3}{4}} = 12d \bracket{\Gamma \bracket{\frac{1}{4}}}^{-1}\Gamma \bracket{\frac{3}{4}} \end{split} \end{equation*}
Clearly, $\mathbb{E}_{\theta} \squared{\dot{l_{\theta}} \ddot{l_{\theta}} } = 0$. Finally:
\begin{equation*} \begin{split} & \mathbb{E}_{\theta} \squared{\ddot{l_{\theta}}^2} = \mathbb{E}_{\theta} \squared{ \curly{- 12 \sum_{i=1}^d (x_i - \theta)^2 }^2 } \\ & = 144C \curly{\sum_{i=1}^d \int_{\mathbb{R}^d} x_i^4 \exp{-\sum_{i=1}^d x_i^4} dx + 2\sum_{i=1}^d \sum_{j=1}^{i-1} \int_{\mathbb{R}^d} x_i^2x_j^2 \exp{-\sum_{i=1}^d x_i^4} dx} \\ & = 144C \curly{d \bracket{\frac{1}{2} \Gamma\bracket{\frac{1}{4}}}^{d-1} \frac{1}{8} \Gamma \bracket{\frac{1}{4}} + d(d-1) \bracket{\frac{1}{2} \Gamma\bracket{\frac{1}{4}}}^{d-2} \bracket{\frac{1}{2} \Gamma \bracket{\frac{3}{4}}}^2} \\ & = 36d + 144d(d-1) \bracket{\Gamma\bracket{\frac{1}{4}}}^{-2}\bracket{\Gamma\bracket{\frac{3}{4}}}^{2} \end{split} \end{equation*}
We see that the last entry is:
\begin{equation*} \mathbb{E}_{\theta} \squared{\ddot{l_{\theta}}^2} - \iota_{\theta}^2 = 36d - 144d \bracket{\Gamma\bracket{\frac{1}{4}}}^{-2}\bracket{\Gamma\bracket{\frac{3}{4}}}^{2} \end{equation*}
Now we have a problem. The covariance matrix $M_{\theta}$ is positive-semidefinite. The statistical curvature should be a real number. However, my computation shows that this is not the case for $f$.
\begin{equation*} M_{\theta} = \begin{bmatrix} 12d \bracket{\Gamma \bracket{\frac{1}{4}}}^{-1}\Gamma \bracket{\frac{3}{4}} & 0 \\ 0 & 36d - 144d \bracket{\Gamma\bracket{\frac{1}{4}}}^{-2}\bracket{\Gamma\bracket{\frac{3}{4}}}^{2} \end{bmatrix} \end{equation*} \begin{equation*} \gamma_{\theta} = \frac{\sqrt{\bracket{\Gamma \bracket{\frac{1}{4}}}^2 - 4 \bracket{\Gamma \bracket{\frac{3}{4}}}^2}} {2 \Gamma \bracket{\frac{3}{4}}} \end{equation*}
What went wrong ?