The vertical coordinate (“height") of an object in free fall is described by an equation of the form
$x(t) = \theta _0 + \theta _1t + \theta _2 t^2,$
We assume that $\theta_0$ is a known constant. We also assume that $\theta_2$ is known. We view $\theta_1$ as the realized value of a continuous random variable $\Theta _1$.
In this case, finding the MAP estimate of $\Theta _1$ involves the minimization of
$$\theta _1^2+ \frac{1}{\sigma ^2}\sum _{i=1}^ n(y_ i-\theta _0-\theta _1 t_ i -\theta _2 t_ i^2)^2$$
with respect to $\theta_1$.
1)$$\hat{\theta }_1 = \frac{\sum ^ n_{i=1}t_ i(y_ i-\theta _0-\theta _2t^2_ i)}{\sigma ^2+\sum _{i=1}^ n t_ i^2}$$
2) Let$\sigma =1$, and consider the special case of only two observations($n=2$). Write down a formula for the mean squared error ${\bf E}[(\hat\Theta _1-\Theta _1)^2]$
my answer is $$1/(2(1+t_1^2+t_2^2))$$ I am not sure if it is right.
3) Consider the “experimental design" problem of choosing when to make measurements. Under the assumptions of part (1), 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.
I am really stuck here, any help will be appreciated.
Best Answer
2) $${\bf E}[(\hat\Theta _1-\Theta _1)^2] = {\bf E}[(\hat\theta _1-\Theta _1)^2] = var(\Theta)$$ For a normal distribution, we recognize that the variance is determined by the coefficient that comes next to a term of the form $\theta _1^2$. So what we're looking for is the constant that sits next to the $\theta _1^2$ term. We collect all of them, it is $1/2*(1+t _1^2+t_2^2)$. So the final answer is $$1/(1+t_1^2+t_2^2)$$
3)Because $0\leq t_1,t_2\leq 10$, in order to minimize the mean squared error, I just guess $t_1=t_2=10$ intuitively.