Hw5 Q8 E_out

marek · #1 05-05-2013, 06:55 PM

I am struggling to understand how to calculate $E_{out}$ in this question. I have two competing theories, which I will describe below. Any help is greatly appreciated.

Once the algorithm terminates, I have $w^{(t)}$ . I now generate a new set of data points $\{X_i\}_{i=1}^M$ . Using my original target function to generate the corresponding

.

Case 1. Just use the same cross entropy error calculation but on this new data set.

$E_{out} = \frac{1}{M} \sum_{i=1}^M \ln (1+e^{-Y_i w^\top X_i})$

Case 2. Directly calculate the expected output of our hypothesis function and compare to

.

with probability $\theta (w^\top X_i) = \frac{1}{1+e^{-w^\top X_i}}$

Ultimately this gives us the probability that our hypothesis aligns with Y:

$P(Y_i | X_i) = \theta(Y_i w^\top X_i)$

In the lectures/book, we would multiply these probabilities to get the "likelihood" that the data was generated by this hypothesis. However, it seems that averaging over these should give the expected error in this sample.

$E_{out} = \frac{1}{M} \sum_{i=1}^{M} (1-P(Y_i | X_i))$

It feels as though the first approach is the correct one, but I struggle because the second approach makes intuitive sense since that is how I historically I would have calculated $E_{out}$ . To make matters worse, the two approaches very closely approximate different answers in the question!

yaser · #2 05-05-2013, 10:07 PM

Quote:

Originally Posted by marek

I am struggling to understand how to calculate $E_{out}$ in this question. I have two competing theories, which I will describe below. Any help is greatly appreciated.

Once the algorithm terminates, I have $w^{(t)}$ . I now generate a new set of data points $\{X_i\}_{i=1}^M$ . Using my original target function to generate the corresponding

.

Case 1. Just use the same cross entropy error calculation but on this new data set.

$E_{out} = \frac{1}{M} \sum_{i=1}^M \ln (1+e^{-Y_i w^\top X_i})$

The above approach is correct. The problem specifies the cross entropy error measure, so $E_{\rm out} = {\rm E} [ \ln (1+e^{-y{\bf w}^\top {\bf x}})]$ , where the expectation is w.r.t. both ${\bf x},y$ . The above formula estimates that through a random sample.

marek · #3 05-05-2013, 10:12 PM

Quote:

Originally Posted by yaser

The above approach is correct. The problem specifies the cross entropy error measure, so $E_{\rm out} = {\rm E} [ \ln (1+e^{-y{\bf w}^\top {\bf x}})]$ , where the expectation is w.r.t. both ${\bf x},y$ . The above formula estimates that through a random sample.

I suspected as much. I'll try to figure out why my other approach is wrong tomorrow. I think I've burned out on it today and am probably not seeing something obvious. Thanks for your help!

arcticblue · #4 05-06-2013, 03:37 AM

I am also a little unsure about exactly how this equation works:
$E_{out} = \frac{1}{M} \sum_{i=1}^M \ln (1+e^{-Y_i w^\top X_i})$

Obviously the more negative ${-Y_i w^\top X_i}$ is the closer E_out is to zero which is good. So is w supposed to be normalized? I presume so because otherwise I could just scale w and then E_out becomes very small. And if it is normalized then the values I'm getting for E_in and E_out are both much greater than any of the options. (Maybe it's meant to be like that, if so it's quite unnerving.)

yaser · #5 05-06-2013, 11:14 AM

Quote:

Originally Posted by arcticblue

I am also a little unsure about exactly how this equation works:
$E_{out} = \frac{1}{M} \sum_{i=1}^M \ln (1+e^{-Y_i w^\top X_i})$

Obviously the more negative ${-Y_i w^\top X_i}$ is the closer E_out is to zero which is good. So is w supposed to be normalized? I presume so because otherwise I could just scale w and then E_out becomes very small. And if it is normalized then the values I'm getting for E_in and E_out are both much greater than any of the options. (Maybe it's meant to be like that, if so it's quite unnerving.)

No normalization. The value of ${\bf w}$ is determined iteratively by the specific algorithm given in the lecture. If ${\bf w}$ 'agrees' with all the training examples, then indeed the algorithm will try to scale it up to get the value of the logistic function closer to a hard threshold. When you evaluate the quoted formula on a test set, ${\bf w}$ is frozen and no scaling or any other change in it is allowed.

Michael Reach · #6 05-06-2013, 03:39 PM

Quote:

If 'agrees' with all the training examples, then indeed the algorithm will try to scale it up to get the value of the logistic function closer to a hard threshold.

Thank you for this comment - I finally have some idea what I'm seeing in the homework. This point is worth stressing: the scale of w determines the sharpness of the threshold.

arcticblue · #7 05-06-2013, 05:22 PM

Thank you for explaining that normalization is not required. Your explanation now makes a bit more sense why I see the weights continue to increase in value the more iterations that I run.

hsolo · #8 07-16-2013, 06:27 AM

One very minor point which may help error prone folks such as myself:
In Linear methods we always have a d+1 dimensional weight vector by adding an extra pseudo-coordinate of 1 for each training and test point. I completely overlooked this and was struggling with the error never getting as small as the expected answer. Spent a long time rechecking code etc.

As soon as I fixed this, things fell into place. Likely I will never forget this :-)

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