T 1 t x ² t x ² t u c t t 1 2 1 m m t 1 t x ² t v

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t ± 1 T x ² t x ² t U / c t | t " 1 2 " 1 m * ' ± M * ' ! t ± 1 T x ² t v t 1   / c t | t " 1 2 What have we accomplished? Original formulation Gibbs sampled p * | 8 , y , 5   and p 8 | * , y , 5   problem was they were too correlated Augmented Kalman filter allowed us instead to sample from p * | y , 5   and then with usual Kalman filter we can draw from p 8 | * , y , 5   to get p 8 , * | y , 5  
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16 Note this also allows us to calculate the distribution of y without conditioning on * or 8 : p y | 5   ± p y | * ± 0 , 5   p * ± 0   p * ± 0 | y , 5   The first term we know from the basic Kalman filter: p y | * ± 0 , 5   ± ² t ± 1 T 2 = c t | t " 1 2   " 1/2 exp " y t " x t U * " h U 8 t 1   " h U 8 t 2   *   2 2 c t | t " 1 2 * ± 0 ± ² t ± 1 T 2 = c t | t " 1 2   " 1/2 exp " y t " h U 8 t 1     2 2 c t | t " 1 2 The second term we know by evaluating our prior at * ± 0 : p * ± 0   ± 2 =   " k /2 | M * | " 1/2 exp " 1/2   * " m *   U M * " 1 * " m *   * ± 0 ± 2 =   " k /2 | M * | " 1/2 exp " 1/2   m * U M * " 1 m *
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17 The third term we know by evaluating our posterior at * ± 0 : p * ± 0 | y , 5   ± 2 =   " k /2 | M * ' | " 1/2 exp " 1/2   * " m * '   U M * '" 1 * " m * '   * ± 0 ± 2 =   " k /2 | M * ' | " 1/2 exp " 1/2   m * ' U M * '" 1 m * ' Thus for m * ± 0 we have p y | 5   ± p y | * ± 0 , 5   p * ± 0   p * ± 0 | y , 5   - ² t ± 1 T 2 = c t | t " 1 2   " 1/2 exp " y t " h U 8 t 1     2 2 c t | t " 1 2 | M * | " 1/2 | M * ' | 1/2 exp 1/2
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