# 2.3 Machine learning lecture 4 course notes  (Page 4/5)

 Page 4 / 5
$\stackrel{^}{\epsilon }\left({h}_{i}\right)=\frac{1}{m}\sum _{j=1}^{m}{Z}_{j}.$

Thus, $\stackrel{^}{\epsilon }\left({h}_{i}\right)$ is exactly the mean of the $m$ random variables ${Z}_{j}$ that are drawn iid from a Bernoulli distribution with mean $\epsilon \left({h}_{i}\right)$ . Hence, we can apply the Hoeffding inequality, and obtain

$P\left(|\epsilon \left({h}_{i}\right)-\stackrel{^}{\epsilon }\left({h}_{i}\right)|>\gamma \right)\le 2exp\left(-2{\gamma }^{2}m\right).$

This shows that, for our particular ${h}_{i}$ , training error will be close to generalization error with high probability, assuming $m$ is large. But we don't just want to guarantee that $\epsilon \left({h}_{i}\right)$ will be close to $\stackrel{^}{\epsilon }\left({h}_{i}\right)$ (with high probability) for just only one particular ${h}_{i}$ . We want to prove that this will be true for simultaneously for all $h\in \mathcal{H}$ . To do so, let ${A}_{i}$ denote the event that $|\epsilon \left({h}_{i}\right)-\stackrel{^}{\epsilon }\left({h}_{i}\right)|>\gamma$ . We've already show that, for any particular ${A}_{i}$ , it holds true that $P\left({A}_{i}\right)\le 2exp\left(-2{\gamma }^{2}m\right)$ . Thus, using the union bound, we have that

$\begin{array}{ccc}\hfill P\left(\exists \phantom{\rule{0.166667em}{0ex}}h\in \mathcal{H}.|\epsilon \left({h}_{i}\right)-\stackrel{^}{\epsilon }\left({h}_{i}\right)|>\gamma \right)& =& P\left({A}_{1}\cup \cdots \cup {A}_{k}\right)\hfill \\ & \le & \sum _{i=1}^{k}P\left({A}_{i}\right)\hfill \\ & \le & \sum _{i=1}^{k}2exp\left(-2{\gamma }^{2}m\right)\hfill \\ & =& 2kexp\left(-2{\gamma }^{2}m\right)\hfill \end{array}$

If we subtract both sides from 1, we find that

$\begin{array}{ccc}\hfill P\left(¬\exists \phantom{\rule{0.166667em}{0ex}}h\in \mathcal{H}.|\epsilon \left({h}_{i}\right)-\stackrel{^}{\epsilon }\left({h}_{i}\right)|>\gamma \right)& =& P\left(\forall h\in \mathcal{H}.|\epsilon \left({h}_{i}\right)-\stackrel{^}{\epsilon }\left({h}_{i}\right)|\le \gamma \right)\hfill \\ & \ge & 1-2kexp\left(-2{\gamma }^{2}m\right)\hfill \end{array}$

(The “ $¬$ ” symbol means “not.”) So, with probability at least $1-2kexp\left(-2{\gamma }^{2}m\right)$ , we have that $\epsilon \left(h\right)$ will be within $\gamma$ of $\stackrel{^}{\epsilon }\left(h\right)$ for all $h\in \mathcal{H}$ . This is called a uniform convergence result, because this is a bound that holds simultaneously for all (as opposed to just one) $h\in \mathcal{H}$ .

In the discussion above, what we did was, for particular values of $m$ and $\gamma$ , give a bound on the probability that for some $h\in \mathcal{H}$ , $|\epsilon \left(h\right)-\stackrel{^}{\epsilon }\left(h\right)|>\gamma$ . There are three quantities of interest here: $m$ , $\gamma$ , and the probability of error; we can bound either one in terms of the other two.

For instance, we can ask the following question: Given $\gamma$ and some $\delta >0$ , how large must $m$ be before we can guarantee that with probability at least $1-\delta$ , training error will be within $\gamma$ of generalization error? By setting $\delta =2kexp\left(-2{\gamma }^{2}m\right)$ and solving for $m$ , [you should convince yourself this is the right thing to do!], we find that if

$m\ge \frac{1}{2{\gamma }^{2}}log\frac{2k}{\delta },$

then with probability at least $1-\delta$ , we have that $|\epsilon \left(h\right)-\stackrel{^}{\epsilon }\left(h\right)|\le \gamma$ for all $h\in \mathcal{H}$ . (Equivalently, this shows that the probability that $|\epsilon \left(h\right)-\stackrel{^}{\epsilon }\left(h\right)|>\gamma$ for some $h\in \mathcal{H}$ is at most $\delta$ .) This bound tells us how many training examples we need in order makea guarantee. The training set size $m$ that a certain method or algorithm requires in order to achieve a certain level of performance is also calledthe algorithm's sample complexity .

The key property of the bound above is that the number of training examples needed to make this guarantee is only logarithmic in $k$ , the number of hypotheses in $\mathcal{H}$ . This will be important later.

Similarly, we can also hold $m$ and $\delta$ fixed and solve for $\gamma$ in the previous equation, and show [again, convince yourself that this is right!]that with probability $1-\delta$ , we have that for all $h\in \mathcal{H}$ ,

$|\stackrel{^}{\epsilon }\left(h\right)-\epsilon \left(h\right)|\le \sqrt{\frac{1}{2m}log\frac{2k}{\delta }}.$

Now, let's assume that uniform convergence holds, i.e., that $|\epsilon \left(h\right)-\stackrel{^}{\epsilon }\left(h\right)|\le \gamma$ for all $h\in \mathcal{H}$ . What can we prove about the generalization of our learning algorithm that picked $\stackrel{^}{h}=arg{min}_{h\in \mathcal{H}}\stackrel{^}{\epsilon }\left(h\right)$ ?

Define ${h}^{*}=arg{min}_{h\in \mathcal{H}}\epsilon \left(h\right)$ to be the best possible hypothesis in $\mathcal{H}$ . Note that ${h}^{*}$ is the best that we could possibly do given that we are using $\mathcal{H}$ , so it makes sense to compare our performance to that of ${h}^{*}$ . We have:

$\begin{array}{ccc}\hfill \epsilon \left(\stackrel{^}{h}\right)& \le & \stackrel{^}{\epsilon }\left(\stackrel{^}{h}\right)+\gamma \hfill \\ & \le & \stackrel{^}{\epsilon }\left({h}^{*}\right)+\gamma \hfill \\ & \le & \epsilon \left({h}^{*}\right)+2\gamma \hfill \end{array}$

The first line used the fact that $|\epsilon \left(\stackrel{^}{h}\right)-\stackrel{^}{\epsilon }\left(\stackrel{^}{h}\right)|\le \gamma$ (by our uniform convergence assumption). The second used the fact that $\stackrel{^}{h}$ was chosen to minimize $\stackrel{^}{\epsilon }\left(h\right)$ , and hence $\stackrel{^}{\epsilon }\left(\stackrel{^}{h}\right)\le \stackrel{^}{\epsilon }\left(h\right)$ for all $h$ , and in particular $\stackrel{^}{\epsilon }\left(\stackrel{^}{h}\right)\le \stackrel{^}{\epsilon }\left({h}^{*}\right)$ . The third line used the uniform convergence assumption again, to show that $\stackrel{^}{\epsilon }\left({h}^{*}\right)\le \epsilon \left({h}^{*}\right)+\gamma$ . So, what we've shown is the following: If uniform convergence occurs,then the generalization error of $\stackrel{^}{h}$ is at most $2\gamma$ worse than the best possible hypothesis in $\mathcal{H}$ !

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