# Getting linear regression of huge numbers

I'm trying to get a linear regression slope and intercept for a large set of huge numbers.

I'm doing this on a computer, but I keep getting overflow errors (attempting to calculate a number too large for a standard data type). I figured I'd ask this here, since it's primarily a math question.

How can I "normalize" the input set, so that I don't have an overflow? Or perhaps, there is another method for calculating the slope and intercept that wouldn't result in multiplication of all the X values, summation of X*Y, etc.

(y, x)
2103.00 @ 1233687329.20
2104.00 @ 1233687329.50
2103.00 @ 1233687329.20
2104.00 @ 1233687329.50
2105.00 @ 1233687329.80
2106.00 @ 1233687330.10
2107.00 @ 1233687330.40
2108.00 @ 1233687330.70
2109.00 @ 1233687331.00
2110.00 @ 1233687331.30
2111.00 @ 1233687331.60
2112.00 @ 1233687331.90
2113.00 @ 1233687332.20
2114.00 @ 1233687332.50
2115.00 @ 1233687332.80
2116.00 @ 1233687333.10
2117.00 @ 1233687333.40
2118.00 @ 1233687333.70
2119.00 @ 1233687334.00


For example, trying to get the slope / intercept for this data set in Excel or Numbers will just result in an error.

Is there a way to normalize the set prior to doing the regression (and after to get the right answer), or perhaps a less intensive way of getting the regression?

Update: Normalizing by subtracting from Y doesn't work.

x  (x-5)    y
1   -4  1
2   -3  2
3   -2  4


slope works fine: 1.5

intercept non-"normalized": -0.66666

intercept "normalized": 6.83333 <-- problem, can't just add 5 to intercept to get the value

-
To get the non-normalised intercept from the normalised, you need to subtract 5*slope = 7.5. – TonyK Sep 5 '11 at 11:14

Subtract 1233687329 from each of your x values (in other words, do the change of variables $t = x - 1233687329$). Then you can change back to $x$ if you wish, although for most purposes $t$ would be a more sensible variable to use.
It seems silly to talk about a $y$-intercept in a case like this, since you'd be extrapolating a long way from all of the $x$ values in your data. Microscopic errors, or just rounding, of the $y$ values will throw off your estimated intercept a long way. The least squares line passes through the point whose coordinates are the average $x$ value and the average $y$ value. Just let the slope and the average $y$ value be the things you're trying to estimate. – Michael Hardy Sep 5 '11 at 1:51
If the equation in the $(t,y)$ variables is $y = m t + b$, and $t = x - c$, then $y = m (x - c) + b = m x + (b - m c)$. So the $y$ intercept when using the $(x,y)$ variables is $b - m c$. – Robert Israel Sep 5 '11 at 7:31