bias and variance - definition of g bar

[magdon]

All the numbers you mention below are approximately correct. You can now explicitly compute bias(x) and var(x) in terms of x, mean(a), mean(b), var(a) and var(b) (mean(b)=0):





Bias is the average of bias(x) over x; var is the average of var(x) over x. Set . One can show that





Note: you can also compute the bias and variance via simulation.

Originally Posted by the cyclist

I am struggling to replicate the variance of H_1 of Ex. 2.8 in the text. I was able to get the bias correct (and both bias and variance for H_0), as well as getting the related quiz problem correct, so this is really puzzling me.

I'm trying to narrow down where my mistake might be. Can someone please verify whether or not the correct average hypothesis is

g_bar(x) = a_mean * x + b_mean

where

a_mean ~= 0.776

and

b_mean ~= 0.

I plot that, and it does look like the figure in the book.

Also, when I take the standard deviation (over the data sets) of the coefficients a and b, I get

std(a) ~= 1.52
std(b) ~= 0.96

Do those look right? I am truly puzzled here!

[munchkin]

I'm having doubts about the variance value in example 2.8 since it indicates that the root mean square deviation of the test data from the sinusoid line is 1.3= sqrt(1.69). So the magnitude of the average (a*x+b) difference from (a_mean*x+b_mean) evaluated at a given point on the sinusoid is bigger than the root mean square value (.7071) of the sinusoid that generated the data point in the first place? I'm inclined to doubt that.

The mean square deviation between the slope of each generated line and a_mean is larger than 1.69 so at this point I have no idea where that variance number came from.

[yaser]

Originally Posted by munchkin

I'm having doubts about the variance value in example 2.8 since it indicates that the root mean square deviation of the test data from the sinusoid line is 1.3= sqrt(1.69). So the magnitude of the average (a*x+b) difference from (a_mean*x+b_mean) evaluated at a given point on the sinusoid is bigger than the root mean square value (.7071) of the sinusoid that generated the data point in the first place? I'm inclined to doubt that.

The average is taken over the entire domain so that includes points where the line (which fits the two training points on the sinusoid) diverges significantly from the sinusoid and from the average line. The figures on page 65 illustrate that.

[munchkin]

Yes, I can see that on the charts for example 2.8 but those outlying points do not exert an effect (at a given x) for the averaged g(D)[x] calculation I am using. So I am wrong on both counts!

A careful rereading of page 63 has led me to try averaging over the calculated data set g's at an arbitrary (generic?) point x and using that to calculate the variance of g(D)[x]. This seems to be a step in the right direction since the calculated variance is now a function of that arbitrary x point and has a minimum around x=0 just like the chart in the example 2.8 but based on the values at the extremes and in the middle I can't see how my average variance over the domain [-1,1] would be as low as 1.69. We shall see.

Thanks so much for your helpful comments, they are really appreciated and this is a great class even if I am a little dense in absorbing some of the material. Have a great day.

[the cyclist]

Finally got it. Thanks to magdon for confirming one part of my calculation, so that I did not need to waste time poring over it. Thanks also to yaser for a tip, in another thread, that helped me a lot. It turns out that I was incorrectly reusing the sample dataset to calculate (via simulation) the variance. Instead, I needed to generate a fresh dataset for that.

It's funny how sometimes making mistakes at first leads to a much more solid understanding later!

[munchkin]

I've got it too! Repeatedly evaluate var[ g(D)[x] ] over the entire data set with x ranging from -1 to 1 and average those values to get 1.69 ! ! ! Feeling a sense of real accomplishment here.