Update exercise 4.4

code/exercises-4-kalman.ipynb

@@ -52,7 +52,7 @@
    "source": [
     "### Gaussian prior\n",
     "\n",
-    "$$p(X_0) = \\mathcal{N}(\\mu_0, \\Sigma_0)$$"
+    "$$p(x_0) = \\mathcal{N}(x_0 | \\mu_0, \\Sigma_0)$$"
    ]
   },
   {
@@ -82,7 +82,7 @@
    "source": [
     "### Linear Gaussian transition model\n",
     "\n",
-    "$$p(X_t \\vert X_{t - 1}) = \\mathcal{N}(F X_{t - 1} + u, \\Sigma_x)$$"
+    "$$p(x_t \\vert x_{t - 1}) = \\mathcal{N}(x_t | F x_{t - 1} + u, \\Sigma_x)$$"
    ]
   },
   {
@@ -119,7 +119,7 @@
    "source": [
     "### Linear Gaussian sensor model\n",
     "\n",
-    "$$p(E_t \\vert X_t) = \\mathcal{N}(H X_t + v, \\Sigma_e)$$"
+    "$$p(e_t \\vert x_t) = \\mathcal{N}(e_t | H x_t + v, \\Sigma_e)$$"
    ]
   },
   {

exercises/e4.tex

@@ -326,9 +326,9 @@ \section{Super Spring Ultra Pro Max XXL}
         \end{itemize}
         In a \href{https://wikipedia.org/wiki/Multivariate_normal_distribution}{multivariate Gaussian distribution} $x = \mat{x_1 & x_2 & \dots & x_n}^T \sim \N(\mu, \Sigma)$, the first argument is the \emph{mean vector} and the second is the \emph{covariance matrix}. The mean vector is the vector of the variable means, \ie{} $\mu_i = \E\sbk{x_i}$. An element $\Sigma_{ij}$ of the covariance matrix is the covariance between the variables $x_i$ and $x_j$, \ie{} $\Sigma_{ij} = \E\sbk{(x_i - \mu_i) (x_j - \mu_j)}$. If $x_i$ is independent from $x_j$, their covariance is null by definition, \ie{} $\Sigma_{ij} = 0$. Interestingly, the diagonal elements $\Sigma_{ii}$ are the variable variances $\V\sbk{x_i} = \E\sbk{(x_i - \mu_i)^2}$.
 
-        In our case, the state $X_t$ is a 4-dimensional vector $\mat{p_t & \dot{p}_t & \ddot{p}_t & w_t}^T$ and according to the provided information, the prior is defined by
+        In our case, the state $X_t$ is a 4-dimensional vector $\mat{p_t & \dot{p}_t & \ddot{p}_t & w_t}^T$ and according to the provided information, the prior is defined\footnote{The stationarity of the wind implies that $\V\sbk{w_0} = \V\sbk{w_1} = \V\sbk{\alpha w_0 + \Delta w} = \alpha^2 \V\sbk{w_0} + \sigma_w^2$, and therefore, $\V\sbk{w_0} = \frac{\sigma_w^2}{1 - \alpha^2}.$} by
         \begin{equation*}
-            \mu_0 = \mat{10 \\ 0 \\ 0 \\ 0} \quad \text{and} \quad \Sigma_0 = \mat{0.5^2 & 0 & 0 & 0 \\ 0 & 0.1^2 & 0 & 0 \\ 0 & 0 & 0.1^2 & 0 \\ 0 & 0 & 0.0 & \frac{\sigma_w^2}{1 - \alpha^2}} .
+            \mu_0 = \mat{10 \\ 0 \\ 0 \\ 0} \quad \text{and} \quad \Sigma_0 = \mat{0.5^2 & 0 & 0 & 0 \\ 0 & 0.1^2 & 0 & 0 \\ 0 & 0 & 0.1^2 & 0 \\ 0 & 0 & 0 & \frac{\sigma_w^2}{1 - \alpha^2}} .
         \end{equation*}
         Then, our transition model is defined by
         \begin{equation*}