Thread: Best fit the measured data

1. Best fit the measured data

Estimate X and Y which best fit the following set of equations:
X*Y=Z
in which
X: n by 1 comlumn vector (unknown)
Y: 1 by 1 unknown constant
Z: n by 1 measured column vector (known)
n is constant and can be as large as expected.
Could you please tell me what method (either numerical or algebra form) I can use to solve the above-mention problem or any suggestion?
THanks a lot.

2. Originally Posted by tvauce
Estimate X and Y which best fit the following set of equations:
X*Y=Z
in which
X: n by 1 comlumn vector (unknown)
Y: 1 by 1 unknown constant
Z: n by 1 measured column vector (known)
n is constant and can be as large as expected.
Could you please tell me what method (either numerical or algebra form) I can use to solve the above-mention problem or any suggestion?
THanks a lot.
This is not a "best fit" problem. That's a statistics problem.

What are you solving for? X? Then
$X = \frac{Z}{Y}$

If you are solving for Y things get a bit trickier:
$Y = X^{T}Z$
where $X^{T}$ is the transpose of X.

-Dan

3. I'm sorry not stating the problem clearly.
In my problem, I need to solve for both X and Y. The number of unknowns is (n+1) while the number of equations is only n. That's the most challenging things in my problem.

4. Originally Posted by tvauce
I'm sorry not stating the problem clearly.
In my problem, I need to solve for both X and Y. The number of unknowns is (n+1) while the number of equations is only n. That's the most challenging things in my problem.
One thing you could do would be to absorb the unknown constant Y into the X vector:
$Y \left ( \begin{matrix} a \\ b \\ c \end{matrix} \right ) = \left ( \begin{matrix} d \\ e \\ f \end{matrix} \right )$

You can factor any scalar constant you wish out of your solution.

-Dan

5. Originally Posted by tvauce
Estimate X and Y which best fit the following set of equations:
X*Y=Z
in which
X: n by 1 comlumn vector (unknown)
Y: 1 by 1 unknown constant
Z: n by 1 measured column vector (known)
n is constant and can be as large as expected.
Could you please tell me what method (either numerical or algebra form) I can use to solve the above-mention problem or any suggestion?
THanks a lot.
So you have some measured data and you know the relationship is linear (rather than, say, quadratic or cubic).
In other words, your data is assumed to display an equation of the following form:

z = y*x + b

where b and y are constants. Your task now is to find the values of b and y which yield the best line that describes the data.

A first step might be to use a Linear Least-Squares approach, which minimizes the vertical distance between each measured data point and the resulting interpolating line. Other approaches are also available, depending upon what you mean by "best fit". MATLAB has several tools available. There is also an online tool available too:

Linear Least-Squares Data-Fitting Utility

Regards.

6. Best fit measured data

Thanks a lot for your time and consideration.

My problem is not estimating the two unknown constants Y and b since X is also unknown. I thus do not have a set of pair measured data (X,Z) to estimate Y and b, only Z is measured and available.
I have thought of the need for an iteration within each step. I have tried Recursive least square estimation method as well as Extended Kalman filter method but still have not find the answer.
If you have any other comments or suggestion, please let me know. Your time and kindness are highly appreciated.
Best regards

7. Originally Posted by tvauce
Thanks a lot for your time and consideration.

My problem is not estimating the two unknown constants Y and b since X is also unknown. I thus do not have a set of pair measured data (X,Z) to estimate Y and b, only Z is measured and available.
I have thought of the need for an iteration within each step. I have tried Recursive least square estimation method as well as Extended Kalman filter method but still have not find the answer.
If you have any other comments or suggestion, please let me know. Your time and kindness are highly appreciated.
Best regards
Now I am lost; I have no idea what you are trying to do.