Exercise 4.9 - Sensor fusion with known variances in 1d

Answers

Denate the two observed datasets by $Y^{(1)}$ and $Y^{(2)}$ , with size $N_{1}, N_{2}$ , the likelihood is:

\begin{align} p (Y^{(1)}, Y^{(2)} | μ) = & \prod_{n_{1} = 1}^{N_{1}} p (Y_{n_{1}}^{(1)} | μ) \cdot \prod_{n_{2} = 1}^{N_{2}} p (Y_{n_{2}}^{(2)} | μ) \\ \propto & \exp {A \cdot μ^{2} + B \cdot μ}, \end{align}

where we have dropped terms independent from $μ$ and used:

\begin{align} A = & - \frac{N_{1}}{2 v_{1}} - \frac{N_{2}}{2 v_{2}}, \\ B = & \frac{1}{v_{1}} \sum_{n_{1} = 1}^{N_{1}} Y_{n_{1}}^{(1)} + \frac{1}{v_{2}} \sum_{n_{2} = 1}^{N_{2}} Y_{n_{2}}^{(2)} . \end{align}

Differentiate the likelihood w.r.t. $μ$ and set it to zero, we have:

μ_{MLE} = - \frac{B}{2 A}

The conjugate prior of this model must have a form proportional to $\exp {A \cdot μ^{2} + B \cdot μ}$ , so it is a normal distribution:

p (μ | a, b) \propto \exp {a \cdot μ^{2} + b \cdot μ} .

The posterior distribution is:

p (μ | Y) \propto \exp {(A + a) \cdot μ^{2} + (B + b) \cdot μ} .

Hence we have the MAP estimation:

μ_{MAP} = - \frac{B + b}{2 (A + a)} .

It is noticable that the MAP converges to ML estimation when observation times grow:

μ_{MAP} \to μ_{ML} .

The posterior distribution is another normal distribution, with:

σ_{MAP}^{2} = - \frac{1}{2 (A + a)} .

For non-informative prior, we have $a = b = 0$ so $p (μ | a, b)$ is uniform in the domain, then the MAP estimation is the same as MLE.

solour_lfq

2021-03-24 13:42