Re: Proof that neural nets work

Three week results:

Ok here are the three week results.

Memorial day June 16 pred pred. change June 16 actual actual
change

QQQQ 39.5 40.5 +1 38.4
-1.24
MSFT 23.7 25.7 +2 22.1
-1.6
JNPR 15.7 17.7 +2 16.7
+1.05
CSCO 20.3 21.3 +1 20.0
- 0.3
AQNT 26.0 25.0 -1 22.2
- 3.8

Now for the correlation coefficient. The average of the vector
of pred changes is 1, so subtracting 1 from all entries gives

(0,1,1,0,-2)

The magnitude of this vector is the square root of six which is 2.45.

The average of the actual changes is -1.2 so adding 1.2 to all entries
of that gives

(-.13, -.4, 2.25, .9, -2.6)

The magnitude of this vector is

..0169+ .16+5.06+.81+6.76=12.8 and the square root of this is
3.57

The dot product of the vectors is twice 2.6 plus 2.25 minus .4 or 7.05.

The product of magnitudes is 2.45*3.57=8.75

The correlation coef is then r=7.05/8.75= 0.8

Now if we believe it is valid to apply the t distribution (and we have
discussed this)
we evaluate r times the square root of (5-2)/(1-r^2) in the t
distribution.
This number is .8 times the square root of 3/.96 or 1.41.

In a t distribution with 5-2 = three degrees of freedom we find the
probability of
obtaining such a good result ( a correlation coefficient of .8) by
random chance
is between 12% and 13% or as they write p<0.13

I havve discussed above the notion of normalizing the numbers so the
different standard deviations wouldn't affect this interpretation of
the correlation coefficient. This raises the question of at what time
do we measure the standard deviation used for the normalization; and
W. Reid has pointed out, we get even better correlation coefficients if
we normalize by price (so take actual and predicted percentage
changes).

Note that all the actual prices were lower than predicted, as the whole
tech market has gone down. This did not affect the correlation
coefficient, which does not see a change affecting all shares equally.

So the final results for the memorial day predictions were that the 2
week prediction had a positive correlation coefficient r around .5,
and the three week prediction had a positive correlation coefficient of
..8

It is not possible to obtain a better correlation coef than 1, of
course.

During the week-end I will make predictions for the next few weeks.

This is not, by the way, to prove how good I am at doing this. the
particular collection of shares which we are using was suggested by
Luke B. and we are just verifying that the relationship which he said
is there, actually does exist.

By the way, anyone following along with a copy of GoldenGem, I have
changed the prog so it finds indices like ^DJI now, and the numbers are
scaled so that the main ticks are the largest interval below +/- two
and a half standard deviations that differs from the average by one
significant digit (so the display looks nice).

.

Relevant Pages

  • Re: Proof that neural nets work
    ... E = the predicted variance among the predicted percentage changes. ... A/E will be larger than 1, and it will be lower if volatility ... The correlation coefficient r will be between -1 and 1, ... predictions made on memorial day we got r=.54 and r=.87 for this ...
    (misc.invest.stocks)
  • Re: Accuracy of correlation coefficient?
    ... I am assuming that you mean that x and y each have different standard deviations. ... where rho is the population correlation coefficient. ... A faster asymptotic result is the variance-stabilizing transformation, aka Fisher's z. ...
    (sci.stat.math)
  • Re: The Correlation Coefficient
    ... I am not usre why you are computing these conditional quantities. ... The usual definition of correlation coefficient is the covariance normalized by the product of the standard deviations. ...
    (sci.math)