I am currently working on a model in my free time to see if I can predict the stock market. My methodology is that I first want to take the data and subtract out the function:
$Value(x)=A\left(1+\frac{B*x}{Volume(x)}\right)^{Volume(x)}+C$
Where A, B and C are constants. How do you go about fitting a curve with constants tucked in a non-linear fashion? Specifically, assuming you have the data for what $Values(x)$ and $Volume(x)$ are, what is the method to fit this curve?
Note: I understand that $e^x$ is equal to $\left(1+\frac{x}{n}\right)^n$ as $n$ goes to infinity. I am trying to see what happens when you don't assume that the stock is traded at infinite intervals since the noted function describes interest traded continuously.
Forgetting the stock market, suppose that you have $n$ data points $(x_i,y_i,z_i)$ and you want to fit the model $$z=A\left(1+\frac {B\, x} y \right)^y+C$$ in the least square sense.
Suppose that you give $B$ an arbitrary value. For this value, define $$t_i=\left(1+\frac {B \,x_i} {y_i} \right)^{y_i}$$ making the model $$z=A t+C$$ which is more than simple and a very basic linear regression gives the sum of squares $SSQ$ which is a function of $B$.
Try different values of $B$ and plot $SSQ$ as a function of it until you notice a minimum. At this point, you have at least consistent estimates of the parameters and you can start a full nonlinear regression.
If this works for the stock market, please let me know.
For illustration purposes, I took the following data set $$\left( \begin{array}{ccc} x & y & z \\ 1 & 0.10 & 249 \\ 2 & 0.10 & 252 \\ 3 & 0.10 & 255 \\ 4 & 0.20 & 259 \\ 5 & 0.30 & 264 \\ 6 & 0.40 & 272 \\ 7 & 0.70 & 281 \\ 8 & 0.90 & 293 \\ 9 & 1.30 & 307 \\ 10 & 1.80 & 325 \end{array} \right)$$ and applied the method proposed above. For the area of interest, the results of the first step are $$\left( \begin{array}{cccc} B & SSQ & A & C \\ 0.10 & 24.7654 & 65.0836 & 178.329 \\ 0.11 & 16.9369 & 58.1372 & 185.946 \\ 0.12 & 11.7504 & 52.3874 & 192.321 \\ 0.13 & 8.82305 & 47.5551 & 197.738 \\ 0.14 & 7.83916 & 43.4414 & 202.402 \\ 0.15 & 8.53552 & 39.9007 & 206.461 \\ 0.16 & 10.6906 & 36.8239 & 210.027 \\ 0.17 & 14.1165 & 34.1279 & 213.187 \\ 0.18 & 18.6525 & 31.7481 & 216.007 \\ 0.19 & 24.1604 & 29.6337 & 218.540 \\ 0.20 & 30.5204 & 27.7441 & 220.829 \end{array} \right)$$ So, as estimates for the nonlinear regression are $A=43.44$, $B=0.14$, $C=202.40$. The final result is $$z=43.2065 \left(1+\frac{0.14062 \,x}{y}\right)^y+202.67$$ which would produce for the predicted $z$'s $$\{249.842,252.063,253.639,259.136,264.73,270.686,282.558,292.311,307.23,324.805\}$$