*ANSWER* Questions 4-6 of Hwk 4 (bias and variance)

[gmathew]

I solved this problem very carefully many times. But I am not getting the answers given in the homework solution for 5 and 6. These are the answers I am getting.

I am getting ghat = 1.4272x
bias = 0.5413
variance = 0.4725

Is anybody else getting similar answers? Can the intructor please verify the answers given in the homework solution?

thanks

[Iverson]

I got :
y = 1.4305x
bias : 0.2804
variance : 0.2381

My results for the bias and the variance are quite different from yours. I'm not sure they are correct though.

[danielfm]

I got:

gbar = 1.415863
bias = 0.261949
variance = 0.231310

Of course the results of each run vary, but if I round the numbers to 1 decimal place I always get bias = 0.3 and variance = 0.2.

[cassio]

Originally Posted by gmathew

I solved this problem very carefully many times. But I am not getting the answers given in the homework solution for 5 and 6. These are the answers I am getting.

I am getting ghat = 1.4272x
bias = 0.5413
variance = 0.4725

Is anybody else getting similar answers? Can the intructor please verify the answers given in the homework solution?

thanks

look this topic: http://book.caltech.edu/bookforum/showthread.php?t=424
there is a good discussion about these problems

[ramin]

Maybe don't forget that the expectation "integral" has a p(x) term in it, and p(x) is uniform in the interval [-1,1]

[dtyrp]

Bump for interest in question 6! And I can't access the above link that cassio posted a few years ago

I'm able to get gbar and calculate the bias correctly but calculating the variance still has me stumped.

Here's my thought process:

To calculate variance, compute the integral of (g(x) - gbar(x))^2*p(x) from x = -1 to x = 1. Do this for each hypothesis g in your your hypothesis set (I generated 1000 different g's). p(x) is the uniform distribution that produces your x-axis data points. Now the expected value of a uniform random variable from -1 to 1 is 1/(1- -1) or 1/2. So the integral you calculate for each hypothesis g is really (g(x) - gbar(x))^2/2 from -1 to 1.

Take the average of these integrals to get your variance.

I'm not getting the answer so clearly my logic is wrong. Can anybody point me in the right direction?

[dtyrp]

Wow major misunderstanding/typo. The pdf of a uniform random variable from -1 to 1 is 1/2 (i.e. 1/1--1). The expected value is 1/2*(a+b) or 1/2*0, or just zero. Anyway, so adding the uniform pdf p(x) into the integral just halves the value of each integral. Still not understanding what is wrong with my logic, though.