Question

Let \sigma =1 and consider the special case of only two observations (n=2). Write down a formula for the mean squared error \mathbb {E}[(\hat{\Theta }_1-\Theta _1)^2], as a function of t_1 and t_2. Enter t_1 for t_1 and t_2 for t_2.

\mathbb {E}[(\hat{\Theta }_1-\Theta _1)^2]=\quad
unanswered

Consider the "experimental design" problem of choosing when to make measurements. Under the assumptions of the previous part, and under the constraints 0\leq t_1,t_2 \leq 10, find the values of t_1 and t_2 that minimize the mean squared error associated with the MAP estimator.

t_1=\quad

t_2=\quad

Answer 1

To find a formula for the mean squared error \mathbb {E}[(\hat{\Theta }_1-\Theta _1)^2] for the special case of only two observations (n=2), we need to consider the MAP estimator. The MAP estimator is given by:

\hat{\Theta}_1 = \frac{t_1}{t_1 + t_2} \cdot x_1 + \frac{t_2}{t_1 + t_2} \cdot x_2

where x_1 and x_2 are the observed values. Assuming that the true value of \Theta_1 is a constant, we can calculate the mean squared error as:

\mathbb {E}[(\hat{\Theta }_1-\Theta _1)^2] = \mathbb {E}[\left(\frac{t_1}{t_1 + t_2} \cdot x_1 + \frac{t_2}{t_1 + t_2} \cdot x_2 - \Theta_1\right)^2]

Since the observations are assumed to be independently and identically distributed with a normal distribution, we can substitute the expectation with the true values:

\mathbb {E}[(\hat{\Theta }_1-\Theta _1)^2] = \left(\frac{t_1}{t_1 + t_2} \cdot \Theta_1 + \frac{t_2}{t_1 + t_2} \cdot \Theta_2 - \Theta_1\right)^2

So the formula for the mean squared error is:

\mathbb {E}[(\hat{\Theta }_1-\Theta _1)^2] = \left(\frac{t_1}{t_1 + t_2} \cdot \Theta_1 + \frac{t_2}{t_1 + t_2} \cdot \Theta_2 - \Theta_1\right)^2

And t_1 and t_2 are the values that we need to find to minimize this mean squared error.