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Old 12-18-2015, 04:26 AM
henry2015 henry2015 is offline
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Default Problem 1.3 c

||w(t)||^2
= ||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?

Thanks!
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Old 12-19-2015, 08:18 AM
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htlin htlin is offline
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Default 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.
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Old 06-10-2016, 03:19 AM
henry2015 henry2015 is offline
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Default Re: Problem 1.3 c

Hi,

I just revisited this problem again.

If I substitute w(t-1) with
\begin{bmatrix}w{_0}\\w{_1}\\w{_2}\end{bmatrix}

x(t-1) with
\begin{bmatrix}1\\x{_1}\\x{_2}\end{bmatrix}

I could get
||w(t)||{^2}
= ||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.

Last edited by henry2015; 06-10-2016 at 03:38 AM. Reason: syntax error
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