I forgot a simple regression. What I'm doing wrong?

[Gareth Keenan]

So, this came up while studying for exam 6, but won't be on the exam itself, so I thought I'd through it out here.

The situation is at 24 and 36 months, there are 500 and 700 closed claims respectively.

Also at 24 and 36 months, there are 1100 and 1600 unadjusted paid claims respectively.

It turns out, that the 24 month closed claims are actually supposed to be 540 closed claims, and not 500. So, I need to adjust my 24 month paid claims to recognize this difference according to the method I was using.

Now, the easy method is to just linearly interpelate between 1100 and 1600, but I wanted to test my regression skills.

My material indicates I can fit a curve to where Y is paid claims, X is closed claim counts, and a and b are the parameters you solve for. This situation, is should be pretty easy, because there are only two coordinates in a (x,y) format. The ones for 24 months(500,1100) and (700,1600)

So, you natural log the whole thing, to reduce to

When I solve for the parameters a and b, am I only supposed to use the natural logs of the y's, but not the x's? I recall not. From there b is the covariance of the dependent and independent variable divided by the variance of the independent variable, and the y intercept adjustment, in this case ln(a) is used solved for using the average values of the dependent and independent variable, and the b you just solved for.

Also, the parameters were supposed to have been a = 431 and b = 0.001873

Regards,

Gareth Keenan

[BassFreq]

OK, Gareth - Imagine a warehouse, where a little midget fellow is driving a forklift. He can't see over the top, he's got great big platform shoes on so he can reach the pedals, cos of his little legs. I mean...

Let me start again - Imagine z=ln(y) and w=ln(a). So now we just restate the equation -
ln(y) = ln(a) + bx
turns into
z=w+bx.
Now regress z as a function of x.

I get b=0.00187, and w=6.06633. But that w doesn't help directly. We want a value for a. So, we can take our definition of w and determine a.
w=ln(a)
6.06633=ln(a)
e^6.06633=a
a=431.0957

[MGN]

Why do you recall not using the ln() of just the y and not the x? You've got the equation down and the rest is standard one-variable linear regression. So ya, transform the y's, do the regression and exponentiate the ln(a) and you get the parameters of best fit for the exponential regression.