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  #1  
Old 08-07-2012, 04:57 AM
itooam itooam is offline
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Default Q4) h(x) = ax

This question is similar to that in the lectures i.e.,

in the lecture H1 equals

h(x) = ax + b

Is this question different to the lecture in the respect we shouldn't add "b" (i.e., X0 the bias/intercept) when applying? Or should I treat the same?

My confusion is because in many papers etc a bias/intercept is assumed even if not specified i.e., h(x) = ax could be considered the same as h(x) = ax + b
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Old 08-07-2012, 05:24 AM
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yaser yaser is offline
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Default Re: Q4) h(x) = ax

Quote:
Originally Posted by itooam View Post
This question is similar to that in the lectures i.e.,

in the lecture H1 equals

h(x) = ax + b

Is this question different to the lecture in the respect we shouldn't add "b" (i.e., X0 the bias/intercept) when applying? Or should I treat the same?

My confusion is because in many papers etc a bias/intercept is assumed even if not specified i.e., h(x) = ax could be considered the same as h(x) = ax + b
There is no bias/intercept in this problem, only the slope (one parameter which is a).
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Old 08-07-2012, 05:36 AM
itooam itooam is offline
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Default Re: Q4) h(x) = ax

Thanks for comfirmation, much appreciated
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Old 01-31-2013, 11:16 AM
geekoftheweek geekoftheweek is offline
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Default Re: Q4) h(x) = ax

Is there a best way to minimize the mean-squared error? I am doing gradient descent with a very low learning rate (0.00001) and my solution is diverging! not converging. Is it not feasible to do gradient descent with two points when approximating a sine?
Thanks
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Old 01-31-2013, 12:09 PM
geekoftheweek geekoftheweek is offline
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Default Re: Q4) h(x) = ax

Never mind, I got my solution to converge, though I do not trust my answer. Oh well.
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Old 01-31-2013, 04:34 PM
sanbt sanbt is offline
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Default Re: Q4) h(x) = ax

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Originally Posted by geekoftheweek View Post
Never mind, I got my solution to converge, though I do not trust my answer. Oh well.
You can use linear regression to calculate each hypothesis.
(since linear regression is basically analytical formula for minimizing mean square error).

Also, you can confirm if your g_bar from simulation makes sense by calculate it directly. (calculate expectation of the hypothesis from each (x1,x2) over [-1,1] x [-1,1] ). This involves two integrals but you can plug in the expression to wolfram or mathematica.
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Old 02-01-2013, 07:49 AM
melipone melipone is offline
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Default Re: Q4) h(x) = ax

I thought it would simply be (y1/x1 + y2/x2)/2 to find an a that minimizes the mean square error on two points, no?
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Old 02-01-2013, 11:36 AM
Anne Paulson Anne Paulson is offline
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Default Re: Q4) h(x) = ax

So, in this procedure we:

Pick two points;
Find the best slope a for those two points, the one that minimizes the squared error for those two points;
Do this N times and average all the as

Rather than:

Pick two points;
Calculate the squared error for those two points as a function of a;
Do this N times, then find the a that minimizes the sum of all of the squared errors, as we do with linear regression

Are we doing the first thing here or the second thing? Either way there's a simple analytic solution, but I'm not sure which procedure we're doing.
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Old 02-01-2013, 12:19 PM
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yaser yaser is offline
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Default Re: Q4) h(x) = ax

Quote:
Originally Posted by Anne Paulson View Post
So, in this procedure we:

Pick two points;
Find the best slope a for those two points, the one that minimizes the squared error for those two points;
Do this N times and average all the as

Rather than:

Pick two points;
Calculate the squared error for those two points as a function of a;
Do this N times, then find the a that minimizes the sum of all of the squared errors, as we do with linear regression

Are we doing the first thing here or the second thing? Either way there's a simple analytic solution, but I'm not sure which procedure we're doing.
The first method estimates a for the average hypothesis {\bar g} (which takes into consideration only two points at a time). The second method estimates a for the best approximation of the target function (which takes into consideration all the points in the input space {\cal X} at once).
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Old 02-01-2013, 12:28 PM
Anne Paulson Anne Paulson is offline
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Default Re: Q4) h(x) = ax

OK, and then the average value of \bar{g} *is* the expected value of \bar{g}.
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