Optimal two variable linear regression calculation

Answer 1

This has now been verified as correct:

SELECT
  ((sum(t.YEAR) * sum(t.AMOUNT)) - (count(1) * sum(t.YEAR * t.AMOUNT))) /
  (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as SLOPE,

  ((sum( t.YEAR ) * sum( t.YEAR * t.AMOUNT )) -
  (sum( t.AMOUNT ) * sum(power(t.YEAR, 2)))) /
  (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as INTERCEPT,

  ((avg(t.AMOUNT * t.YEAR)) - avg(t.AMOUNT) * avg(t.YEAR)) /
  (stddev( t.AMOUNT ) * stddev( t.YEAR )) as CORRELATION
FROM (
  SELECT
    AVG(D.AMOUNT) as AMOUNT,
    Y.YEAR as YEAR
  FROM
    CITY C,
    STATION S,
    YEAR_REF Y,
    MONTH_REF M,
    DAILY D
  WHERE
    C.ID = 8590 AND

    SQRT(
      POW( C.LATITUDE - S.LATITUDE, 2 ) +
      POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) < 15 AND

    S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND

    Y.YEAR BETWEEN 1900 AND 2009 AND

    M.YEAR_REF_ID = Y.ID AND

    M.CATEGORY_ID = '001' AND

    M.ID = D.MONTH_REF_ID AND
    D.DAILY_FLAG_ID <> 'M'
  GROUP BY
    Y.YEAR
) t

See the image for details on slope, intercept, and (Pearson's) correlation.

Answer 2

Try to split up the function, you have miscalculated the parameters. Have a look here for reference.

I would do something like the following (please excuse the fact that I don't remember much about SQL syntax and temporary variables, so the code might actually be wrong):

SELECT

sum(t.YEAR) / count(1) AS avgX,

sum(t.AMOUNT) / count(1) AS avgY,

sum(t.AMOUNT*t.YEAR) / count(1) AS avgXY,

sum(power(t.YEAR, 2)) / count(1) AS avgXsq,

( avgXY - avgX * avgY ) / ( avgXsq - power(avgX, 2) )  as SLOPE,

avgY - SLOPE * avgX as INTERCEPT,