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#3
01-31-2013, 12:37 AM
 scottedwards2000 Junior Member Join Date: Jan 2013 Posts: 9
Re: How is linearity of PLA obvious?

Thanks Dr. Magdon-Ismail, that was very helpful. I'm glad it wasn't obvious! Just to ensure I understand it correctly, the way I'm conceptualizing it is slightly differently, but I hope is still correct. I see the function being in 3-d space, with y (the output) as the vertical axis, and and forming the "horizontal plane." If you plot all the points, such that you would get a tilted plane, that intersects the y axis at . The boundary line would then be all values of and which results in y=0.

If you imagine looking straight down on this tilted plane (so it does not appear tilted anymore), the points where y=0 would form a straight line.

Is this a correct way of conceptualizing it?

Also, since this process reminded me of multiple regression, it occurred to me that you could get a curved boundary if you included interaction terms in the equation; however this would require another weight/parameter, and I doubt the PLA algorithm would still work, since it depends on the angle between straight lines.

Finally, just to confirm another suspicion, since there are 3 unknowns in your explication above, it is impossible to derive the 3 weights from just being given the boundary line (e.g. two points in the space) - correct?