Did anybody do any interesting submissions they want to share?

I submitted $p_i = 0.18584427052136$ for all $i$ giving a public score of  0.486849.

If anybody has submitted all zeros, then we can calculate the mean of $a_i$ for the sample.

Thanks, you just saved me a submission. You're right about 0.522226 being the score for constant 0. I get the same log mean+1.

EDIT:

I calculate the score for the mean should be 0.486435. Can anyone confirm?

I submitting the constant value 0.18584427052136 and got a score of 0.486849

Hmm, isn't it

> print(sqrt(0.522226^2 - 0.189941^2))
0.4864591

Or maybe I am just sleepy again but it does agree with two submissions on the leaderboard.

> print(sqrt(0.522226^2 - 0.189941^2))
0.4864591

Yep. The way I calculated it was by fitting a parabola in Excel, so it had rounding errors, but your way is better.

Predicting that EVERYONE in the dataset will be hospitalized for about .19 days.

Kaggle is: http://www.heritagehealthprize.com/c/hhp/Leaderboard

Hmm, I get 0.188965 using your figures (0.522226 RMSE for constant zero submission, 0.486849 for constant 0.18584427 submission).

Too much maths: my head will hurt :-)  Anyhow, my calculations and reasoning are:

Let \\(A_i = \log(a_i + 1)\\).  For a submission of \\(p_i = 0\\) the score is \\(\epsilon = \sqrt{ \mean(A_{i}^2) }\\)and we have the value \\(0.522226\\) for this.

For a constant submission \\(p_i = p\\) and setting \\(P = \log(p + 1)\\) we have \\(\epsilon = \sqrt{P^2 + \mean(A_{i}^2) - 2 P \mean(A_i)}\\).  For \\(p = 0.18584427052136\\) we got a score of \\(\epsilon = 0.486849\\).  Solving for \\(\mean(A_i)\\) gives

$$
\mean(A_i) = \frac{P^2 + \mean(A_i^2) - \epsilon^2}{2 P}
$$

Using \\(\mean(A_i^2) = (0.522226)^2\\) and the \\(\epsilon\\) of \\(0.486849\\) gives \\(\mean(A_i) = 0.189941\\).

In R, I do the calculation as

P E MAI2 (P^2 + MAI2 - E^2)/(2*P)
## [1] 0.1899415

I still can't see my mistake, but that certainly doesn't mean there isn't one!

Erik Jackson,

Your mistake is that you probably that you used 0.18584427052136 in your formula instead of log(1.18584427052136)

Allan Engelhardt seems to be right and we get log(a+1)=0.189941 and it mean a=0.209179 and it means that the best submision of constant should be 0.209179 days for everybody(I hope that I have no errors).

Allan,Are you sure that you got 0.1863221 and 0.178212 in Y2 and Y3?

The first number is ok but the second number is for me 0.178228 so it may be possible that I have an error in reading the data of Y3.

P E MAI2 (P^2 + MAI2 - E^2)/(2*P)
## [1] 0.1899415

Hmm, Jeff Moser edited that after I posted so it now makes no sense at all.  Let me try again and see if Jeff can keep his fingers off the edit button this time....:

P <- log1p(0.18584427052136)
E <- 0.486849
MAI2 <- (0.522226)^2
(P^2 + MAI2 - E^2)/(2*P)

Sorry about that.. your post has shown me that I need to tweak how inline MathJax is rendered. That's what I was experimenting with. Currently only displaymode math works (i.e. math surrounded by double dollar signs on each side). Displaymode puts equations on a separate line which looks odd. I'd like to get inline math to work as well (i.e. with single dollar sign delimeters). The problem is that some programming languages use dollar signs and confuse MathJax.

Just wanted to give an update on what I was doing.

Uri was absolutely right on my error.  I now get 0.189941 like Allan.

More significantly for me, what Uri's correction made me realize was that I had been mistakenly producing predictions in the log domain rather than the real domain.  In other words, I have been submitting files with predictions for log(D+1) rather than predictions for D.  It certainly helped me to fix that problem, although not as much as you might have thought - my leaderboard score improved by only 0.001.

@Jeff Moser: No problem, thanks.  And don't worry about \mean; I have

#+LATEX_HEADER: \usepackage[fleqn]{amsmath}
#+LATEX_HEADER: \DeclareMathOperator \mean {mean}

in my Org-Mode headers.

More details on making beautiful math posts in these forums can be found at http://www.kaggle.com/forums/t/581/tips-for-beautiful-math-posts

For small numbers \\(\log(1+x) \approx x\\) so your score woudn’t change much.  For example, \\(\log(1+0.189941) = 0.1739037\\).

A simple linear model on Sex and AgeAtFirstClaim gives a public score of 0.478118

http://www.heritagehealthprize.com/c/hhp/forums/t/648/cases-missing-sex-and-age-code-in-release-3/4217#post4217