2
$\begingroup$

My set-up is the following, I have two variables $N$ and $TTR$, and I have these points for each variable:

$N$ = [35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50]

$TTR$ = [0.818, 0.812, 0.812, 0.804, 0.804, 0.798, 0.792, 0.793, 0.788, 0.784, 0.781, 0.778, 0.776, 0.771, 0.767, 0.767]

I want to fit these data to the following model:

$$TTR=D/N \cdot \left( \sqrt{1 + 2N/D} - 1\right)$$

where $D$ is the parameter that I want to find and is gives the best fit.

Based on the example I am taking this, D = 66 should be around the "optimal" parameter, plotting these results I get:

enter image description here

I am trying a brute force approach, where I vary D from 1 to 70 (since D is "limited" to this set of values), and finding the D with the least mean squared error, however I get $D_{optimal} = 28$, and the result of that is shown in the following figure:

enter image description here

which is not better than the real optimal value, at least visually. How can I address this problem?

$\endgroup$
5
  • 1
    $\begingroup$ The blue line in the first plot does not correspond to the values of TTR that you are giving. $\endgroup$
    – AugSB
    Jan 18, 2016 at 17:32
  • $\begingroup$ Then your brute force algorithm must be wrong. When I do that in matlab I get something like 63.04... as optimal value (in the least squares sense) and this also visually very good (much better also as 66). $\endgroup$
    – Elmar Zander
    Jan 18, 2016 at 17:40
  • $\begingroup$ It would be useful to give more details about what you are doing when trying a brute force approach and finding the D with the least mean squared error. Obviously, $D=28$ is not optimal. $\endgroup$
    – AugSB
    Jan 18, 2016 at 17:41
  • $\begingroup$ The blue line may vary, since is a mean value of 1000 runs, I forgot to mention that. However, the results that Elmar Zander make sense. How did you got that value? What I did was the following: Varied D from 1 to 70, then applied the formula for the N I had, and then compute mean squared error and pick the D which had the minimal value. $\endgroup$
    – dpalma
    Jan 18, 2016 at 17:54
  • $\begingroup$ Thanks to everyone, the way I was fitting the curve was obviously wrong. I did apply least square fitting and got a result very close to what Elmar found, thanks to everyone $\endgroup$
    – dpalma
    Jan 18, 2016 at 18:25

1 Answer 1

Reset to default
1
$\begingroup$

There are many ways of finding the optimal value. One is to express the value $D$, making a distribution $D=D(TTR,N)$ and then use some other technique of averaging. Basically you could take simply the average value but you can do a little bit better if you observe your data better.

You start: $TTR=\frac{D}{N}(\sqrt{1+\frac{2N}{D}}-1)$ and after some work you have $$D(TTR,N)=-\frac{TTR^2\cdot N}{2(TTR-1)}$$

If you calculate this for your data, you have got the distribution of the values for the assumed distribution. I have got 63.0875 for the average value, and with that the error is below 0.004 for each point.

Another way is to use the above relation and create a linear dependency

$$TTR^2\cdot N=-D(TTR,N)\cdot(2(TTR-1))$$

Now you can use various methods for linear interpolation and obtain $D(TTR,N)$. There are other methods but this should help.

You should set the goal and specify what you want to minimize and then calculate the constant based on that requirement.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .