Perceptron Update $1$

$1$ point possible $($graded$)$

Now consider the Perceptron algorithm with Offset. Whenever there is a "mistake" $($or equivalently, whenever

$y^{(i)}(*x^{(i)}+{}_0)0$ i$.$e$.$ When the label $y^i$ and $h(x)$ do not match$),$ perceptron updates

$$ with $+y^{(i)}{x}^{(i)}$

and

${}_0$ with ${}_0+y^{(i)}$

More formally, the Perceptron Algorithm with Offset is defined as follows:

Perceptron $(\{(x^{(i)},y^{(i)}),i=1,$dots,$n\},T)$ :

initialize $=0(vector)$; ${}_0=0(scalar)$

for $t=1,$dots,$Tdo$

for $i=1,$dots,$ndo$

$if{y}^{(i)}(*x^{(i)}+{}_0)0$ then

update $=+y^{(i)}{x}^{(i)}$

update ${}_0={}_0+y^{(i)}$

In the next set of problems, we will try to understand why such an update is a reasonable one.

When a mistake is spotted, do the updated values of $$ and ${}_0$ provide a better prediction? In other words, is

$y^{(i)}((+y^{(i)}{x}^{(i)})*x^{(i)}+{}_0+y^{(i)})$

always greater than or equal to

$y^{(i)}(*x^{(i)}+{}_0)$

Yes, because $+y^{(i)}{x}^{(i)}$ is always larger than $$

Yes, because $(y^{(i)}{)}^2||x^{(i)}|{|}^2+(y^{(i)}{)}^{2}0$

No$,$ because $(y^{(i)}{)}^2||x^{(i)}|{|}^2-(y^i{)}^{2}0$

No$,$ because $+y^{(i)}{x}^{(i)}$ is always larger than $$