The vertical coordinate (“height") of an object in free fall is described by an equation of the form
x(t)=θ0+θ1t+θ2t2,
where θ0, θ1, and θ2 are some parameters and t stands for time. At certain times t1,…,tn, we make noisy observations Y1,…,Yn, respectively, of the height of the object. Based on these observations, we would like to estimate the object's vertical trajectory.
We consider the special case where there is only one unknown parameter. We assume that θ0 (the height of the object at time zero) is a known constant. We also assume that θ2 (which is related to the acceleration of the object) is known. We view θ1 as the realized value of a continuous random variable Θ1. The observed height at time ti is Yi=θ0+Θ1ti+θ2t2i+Wi,i=1,…,n, where Wi models the observation noise. We assume that Θ1∼N(0,1), W1,…,Wn∼N(0,σ2), and all these random variables are independent.
In this case, finding the MAP estimate of Θ1 involves the minimization of
θ21+1σ2∑i=1n(yi−θ0−θ1ti−θ2t2i)2
with respect to θ1.
Carry out this minimization and choose the correct formula for the MAP estimate, θ^1, from the options below.
θ^1=∑ni=1ti(yi−θ0−θ2t2i)σ2θ^1=∑ni=1ti(yi−θ0−θ2t2i)σ2+∑ni=1t2iθ^1=∑ni=1ti(yi−θ0−θ2t2i)σ2+∑ni=1θ2t2inone of the above
The formula for θ^1 can be used to define the MAP estimator, Θ^1 (a random variable), as a function of t1,…,tn and the random variables Y1,…,Yn. Identify whether the following statement is true.
The MAP estimator Θ^1 has a normal distribution.
True - unanswered
Let σ=1 and consider the special case of only two observations (n=2). Write down a formula for the mean squared error E[(Θ^1−Θ1)2], as a function of t1 and t2. Enter 't1' for t1 and 't2' for t2.
E[(Θ^1−Θ1)2]=- unanswered
Consider the “experimental design" problem of choosing when to make measurements. Under the assumptions of part (3), and under the constraints 0≤t1,t2≤10, find the values of t1 and t2 that minimize the mean squared error associated with the MAP estimator.
t1= - unanswered
t2=- unanswered
1.Carry out this minimization and choose the correct formula for the MAP estimate, θ^1, from the options below.
(second choice) θ^1=∑ni=1ti(yi−θ0−θ2t2i)σ2+∑ni=1t2i
4. t1 and t2 = 10
IF YOU ARE READING THIS. PLEASE, PROVIDE YOUR ANSWERS TO THE WHOLE PROBLEM SET
We can recognize, that our mean squared error = var(Θ), so our final answer will be:
1/(1+t1^2+t2^2)
2) True
i found now t2 = 10
Problem 4. Trajectory estimation, Part II
Answer 1:
1/(1+t_1^2+t_2^2)
Why did the scarecrow win an award?
Because he was outstanding in his field!
To find the MAP estimate of Θ1, we need to minimize the given expression:
θ21 + 1/σ2 * Σ[ (yi - θ0 - θ1ti - θ2t2i)^2 ].
To minimize the expression, we differentiate it with respect to θ1 and set it equal to 0:
d/dθ1 [θ21 + 1/σ2 * Σ( yi - θ0 - θ1ti - θ2t2i)^2 ] = 0.
Expanding the expression and simplifying, we get:
2θ1 - 2/σ2 * Σ( ti(yi - θ0 - θ1ti - θ2t2i)) = 0.
Rearranging the terms, we have:
θ1 = 1/σ2 * Σ( ti(yi - θ0 - θ2t2i)).
Therefore, the correct formula for the MAP estimate, θ^1, is:
θ^1 = Σ( ti(yi - θ0 - θ2t2i))/σ2.
Now, let's move on to the next question.