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-   -   Problem 1.3 c (http://book.caltech.edu/bookforum/showthread.php?t=4648)

henry2015 12-18-2015 04:26 AM

Problem 1.3 c
= ||w(t-1) + y(t-1)x(t-1)||^2 <--- this is from the PLA iteration
<=(||w(t-1)|| + ||y(t-1)x(t-1)||)^2 <--- a property: ||a + b|| <= ||a|| + ||b||
= ||w(t-1)||^2 + 2y(t-1)||w(t-1)||||x(t-1)|| + y(t-1)^2||x(t-1)||^2
= ||w(t-1)||^2 + 2y(t-1)||w(t-1)||||x(t-1)|| + ||x(t-1)||^2

Now, it seems like 2y(t-1)||w(t-1)||||x(t-1)|| is somehow <= 0.

||w(t-1)||||x(t-1)|| is >= 0 tho.

hence, it seems like 2y(t-1) is somehow <= 0.

It seems like I am on the wrong track as I am not using the hint mentioned in the question at all.

Any pointer?


htlin 12-19-2015 08:18 AM

Re: Problem 1.3 c
This inequality ||a + b|| <= ||a|| + ||b|| is quite loose in general and you may want to consider not using it. Hope this helps.

henry2015 06-10-2016 03:19 AM

Re: Problem 1.3 c

I just revisited this problem again.

If I substitute w(t-1) with

x(t-1) with

I could get
= ||w(t-1) + y(t-1)x(t-1)||{^2}
= ||w(t-1)||{^2} + y(t-1){^2}||x(t-1)||{^2} + 2y(t-1)w{^T}(t-1)x(t-1)

and then use the hint to get the answer.

However, the dimension of x and w can be more than 3. I just wonder whether there is a more generic proof.

Thanks in advance.

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