#1




*ANSWER* Questions 46 of Hwk 4 (bias and variance)
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 
#2




Re: Questions 46 of Hwk 4 (bias and variance)
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. 
#3




Re: Questions 46 of Hwk 4 (bias and variance)
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. 
#4




Re: Questions 46 of Hwk 4 (bias and variance)
Quote:
there is a good discussion about these problems 
#5




Re: Questions 46 of Hwk 4 (bias and variance)
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]

#6




Re: *ANSWER* Questions 46 of Hwk 4 (bias and variance)
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 xaxis 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? 
#7




Re: *ANSWER* Questions 46 of Hwk 4 (bias and variance)
Wow major misunderstanding/typo. The pdf of a uniform random variable from 1 to 1 is 1/2 (i.e. 1/11). 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.

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