Given a state vector, $ x $, given by $ x = {[r, v, a]}^{T} $ (Range, Velocity, Acceleration) the Time to Hit is the the time which holds the following: $$ r + v {T}_{tth} + \frac{a {T}_{tth}^{2}}{2} = 0 $$ Now, given the State Vector covariance $ P $ what would the the Time to Hit Covariance?
If the Time to Hit was given by a regular function it would be easy to do given its Jacobian, yet how can it be for this kind of calculation (Implicit Function)?
I think I have the solution.
I'd be happy to hear others' thought.
Defining $ F \left(r, v, a, {T}_{tth} \right) = r + v {T}_{tth} + \frac{a {{T}_{tth}}^{2}}{2} $ which is the implicit function which connects all variables. Since we're dealing with non linear function the variance is given by:
$$ var \left( {T}_{tth} \right) = J P {J}^{T} $$
Where $ P $ is the state vector covariance at a given time and $ J = \left[ \frac{\partial {T}_{tth} }{\partial r} \frac{\partial {T}_{tth} }{\partial v} \frac{\partial {T}_{tth} }{\partial a} \right] $
Using Total Derivative Law on the Implicit Function: $$ \frac{\partial {T}_{tth} }{\partial r} = -\frac{{F}_{r}}{{F}_{{T}_{tth}}}, \frac{\partial {T}_{tth} }{\partial v} = -\frac{{F}_{v}}{{F}_{{T}_{tth}}}, \frac{\partial {T}_{tth} }{\partial a} = -\frac{{F}_{a}}{{F}_{{T}_{tth}}} $$
Where $ {F}_{x} $ is the partial derivative of $ F $ with respect to $ x $.
Each individual component would be:
$$ \frac{\partial {T}_{tth} }{\partial r}=-\frac{1}{v+aT_{tth}}, \frac{\partial {T}_{tth} }{\partial v}=-\frac{T_{tth}}{v+aT_{tth}}, \frac{\partial {T}_{tth} }{\partial a}=-\frac{T^2_{tth}}{2\left(v+aT_{tth}\right)} $$
Transform the state covariance matrix with $J P {J}^{T}$ and you'd have the right solution.
Given a diagonal state covariance matrix, we should have:
$$ P_{tth}=\frac{\sigma^2_r}{\left(v+aT_{tth}\right)^2}+\frac{T^2_{tth}\sigma^2_v}{\left(v+aT_{tth}\right)^2}+\frac{T^4_{tth}\sigma^2_a}{4\left(v+aT_{tth}\right)^2} $$
