Let ( x(i), b(i), y(i) ) be a 3-tuple of real numbers. i=1,...,n can be interpreted as discrete time, meaning that x(i+1) occurs after x(i).
x(i) and y(i) are observable, whereas b(i) is not observable.
The elements in the 3-tuple are related as such:
y(i) = a * x(i) + b(i)
where a is a constant.
Without knowing anything about the statistical distribution of b(i), can you compute a?
You can assume some sort of "continuity" on b, i.e. b(i+1) ~ b(i)
You have n equations with n + 1 unknowns (the n values of bi and the value of a). So all you need is one additional constraint and you have a simple, solvable system of linear equations. That constraint could be knowing, say, b0.
However, there is another possibility. You mentioned a form of "continuity", which can be made exact by requiring that the sum of the squared differences, (bi+1 − bi)2, be minimal. Then it becomes a simple optimization problem, which actually can be solved exactly. First,
bi = yi − axi.
Therefore,
bi+1 − bi = yi +1 − yi − a(xi +1 − xi),
and
(bi +1 − bi)2 = (yi +1 − yi)2 + a2(xi +1 − xi)2 − 2a(yi +1 − yi)(xi +1 − xi).
You can sum this equation for all values of i, then differentiate the right-hand side with respect to a to find the minimum. (It is guaranteed to be a minimum since the coefficient of a2 is a sum of squares, which will always be positive.)
Thank you. Re. the first option, I should have added that the functional form y(i) = a * x(i) + b(i) is only an educated guess that I am trying to calibrate with the data available. It is "exact" only in a statistical sense. However, the second option is quite helpful and I will look into it further. I appreciate your feedback.
Anthony -
It sounds like a "simple linear regression," which means you have a residual ("error" term ), an intercept and a slope, might meet your needs. However, it sounds like you might have a time series component and that could complicate the error structure. Still, simple linear regression might do here. It is generally written as y=a+bx+e, where e here is likely heteroscedastic. Note that the a and b are reversed from what you have.
I am not certain of some of the particulars of your situation, but if you have continuous data and can let the data determine estimates of both the slope and intercept, you could check out section 7.1 in the attached. Setting w=1/x is probably robust.
Perhaps I misunderstood your question?
Cheers - Jim
Article Properties of Weighted Least Squares Regression for Cutoff S...
IF you are looking at a time series, say ARIMA with an additional regressor, perhaps you could use a software package that allowed for that?
- Badges
- Science topic
- Similar topics
- Mathematics
- Statistics
- Statistical Modeling