Let $\mathcal{P}:=\mathcal{P}(\mathcal{X})$ be the $n$-dimensional manifold of all (strictly positive) probability vectors (distributions) on $\mathcal{X}=\{x_0,\dots,x_n\}$,

i.e., each $p=(p(x_0),\dots,p(x_n))\in \mathcal{P}$ is such that $p(x_i)>0$ for all $i$ and $\sum_{i}p(x_i)=1$ and can be thought of a point in $\mathbb{R}^{n+1}$. $\mathcal{P}$ is an $n$-dimensional manifold.

Let $\mathcal{P}=\{p_{\xi}\}$, where $\xi=(\xi_1,\dots,\xi_n)$ is the (global) coordinate system.

A Riemannian metric $G(\xi) = [g_{i,j}(\xi)]$ is defined on $\mathcal{P}$, where \begin{eqnarray} g_{i,j}(\xi) & = & \sum_x \frac{\partial}{\partial\xi_i} (p_{\xi}(x))~ \frac{\partial}{\partial\xi_j}(\log p_{\xi}(x)). \end{eqnarray} An affine connection $\nabla$ is defined on $\mathcal{P}$, given by the Christoffel symbols \begin{eqnarray} \Gamma_{ij}^k({\xi}) & = & \sum_x \frac{\partial}{\partial\xi_k}(p_{\xi}(x))~\frac{\partial}{\partial\xi_i}\left(\frac{\partial}{\partial\xi_j}\log p_{\xi}(x)\right). \end{eqnarray} Suppose that $\gamma_t$ is a geodesic on $\mathcal{P}$. Having the metric and the connection coefficients on hand, can I then claim from the geodesic equation $\nabla_{\dot\gamma_t}\dot\gamma_t=0$ that the following must be true? \begin{eqnarray} \sum_x \frac{\partial}{\partial\xi_k} (p_{\xi}(x))~\frac{d^2}{dt^2}\left(\log \gamma_t(x)\right) = 0 \end{eqnarray}


From this article of Amari in Ann. of Statistics, I came to know that the geodesic equation (for this connection) is given by $\ddot l_t+i_t=0$, where $l_t=\log\gamma_t$, and $i_t=\sum_x \dot\gamma_t(x)\dot l_t(x)=0$. But he hasn't given any explanation how he obtained this. See Appendix of the paper. $\alpha=1$ corresponds to my question. Once this geodesic equation is obtained, my claimed equation is obvious. If anyone can help me derive this geodesic equation, it would be great. Thank you.


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