# 2.1 Machine learning lecture 2 course notes  (Page 6/6)

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$\begin{array}{ccc}\hfill {\Phi }_{35000|y=1}& =& \frac{{\sum }_{i=1}^{m}1\left\{{x}_{35000}^{\left(i\right)}=1\wedge {y}^{\left(i\right)}=1\right\}}{{\sum }_{i=1}^{m}1\left\{{y}^{\left(i\right)}=1\right\}}=0\hfill \\ \hfill {\Phi }_{35000|y=0}& =& \frac{{\sum }_{i=1}^{m}1\left\{{x}_{35000}^{\left(i\right)}=1\wedge {y}^{\left(i\right)}=0\right\}}{{\sum }_{i=1}^{m}1\left\{{y}^{\left(i\right)}=0\right\}}=0\hfill \end{array}$

I.e., because it has never seen “nips” before in either spam or non-spam training examples, it thinks the probability of seeing it ineither type of email is zero. Hence, when trying to decide if one of these messages containing “nips” is spam, it calculates theclass posterior probabilities, and obtains

$\begin{array}{ccc}\hfill p\left(y=1|x\right)& =& \frac{{\prod }_{i=1}^{n}p\left({x}_{i}|y=1\right)p\left(y=1\right)}{{\prod }_{i=1}^{n}p\left({x}_{i}|y=1\right)p\left(y=1\right)+{\prod }_{i=1}^{n}p\left({x}_{i}|y=0\right)p\left(y=0\right)}\hfill \\ & =& \frac{0}{0}.\hfill \end{array}$

This is because each of the terms “ ${\prod }_{i=1}^{n}p\left({x}_{i}|y\right)$ ” includes a term $p\left({x}_{35000}|y\right)=0$ that is multiplied into it. Hence, our algorithm obtains $0/0$ , and doesn't know how to make a prediction.

Stating the problem more broadly, it is statistically a bad idea to estimate the probability of some event to be zero just because you haven't seen it before in your finite training set. Take the problem of estimatingthe mean of a multinomial random variable $z$ taking values in $\left\{1,...,k\right\}$ . We can parameterize our multinomial with ${\Phi }_{i}=p\left(z=i\right)$ . Given a set of $m$ independent observations $\left\{{z}^{\left(1\right)},...,{z}^{\left(m\right)}\right\}$ , the maximum likelihood estimates are given by

${\Phi }_{j}=\frac{{\sum }_{i=1}^{m}1\left\{{z}^{\left(i\right)}=j\right\}}{m}.$

As we saw previously, if we were to use these maximum likelihood estimates, then some of the ${\Phi }_{j}$ 's might end up as zero, which was a problem.To avoid this, we can use Laplace smoothing , which replaces the above estimate with

${\Phi }_{j}=\frac{{\sum }_{i=1}^{m}1\left\{{z}^{\left(i\right)}=j\right\}+1}{m+k}.$

Here, we've added 1 to the numerator, and $k$ to the denominator. Note that ${\sum }_{j=1}^{k}{\Phi }_{j}=1$ still holds (check this yourself!), which is a desirable property since the ${\Phi }_{j}$ 's are estimates for probabilities that we know must sum to 1. Also, ${\Phi }_{j}\ne 0$ for all values of $j$ , solving our problem of probabilities being estimated as zero. Under certain (arguably quite strong) conditions, it can be shown that the Laplace smoothing actually gives the optimal estimator of the ${\Phi }_{j}$ 's.

Returning to our Naive Bayes classifier, with Laplace smoothing, we therefore obtain the following estimates of the parameters:

$\begin{array}{ccc}\hfill {\Phi }_{j|y=1}& =& \frac{{\sum }_{i=1}^{m}1\left\{{x}_{j}^{\left(i\right)}=1\wedge {y}^{\left(i\right)}=1\right\}+1}{{\sum }_{i=1}^{m}1\left\{{y}^{\left(i\right)}=1\right\}+2}\hfill \\ \hfill {\Phi }_{j|y=0}& =& \frac{{\sum }_{i=1}^{m}1\left\{{x}_{j}^{\left(i\right)}=1\wedge {y}^{\left(i\right)}=0\right\}+1}{{\sum }_{i=1}^{m}1\left\{{y}^{\left(i\right)}=0\right\}+2}\hfill \end{array}$

(In practice, it usually doesn't matter much whether we apply Laplace smoothing to ${\Phi }_{y}$ or not, since we will typically have a fair fraction each of spam and non-spam messages, so ${\Phi }_{y}$ will be a reasonable estimate of $p\left(y=1\right)$ and will be quite far from 0 anyway.)

## Event models for text classification

To close off our discussion of generative learning algorithms, let's talk about one more model that is specifically for text classification. While Naive Bayes as we've presentedit will work well for many classification problems, for text classification, there is a related model that does even better.

In the specific context of text classification, Naive Bayes as presented uses the what's called the multi-variate Bernoulli event model . In this model, we assumed that the way an email is generated is that first it is randomly determined (according to the classpriors $p\left(y\right)$ ) whether a spammer or non-spammer will send you your next message. Then, the person sending the email runs through the dictionary, deciding whether to include each word $i$ in that email independently and according to the probabilities $p\left({x}_{i}=1|y\right)={\Phi }_{i|y}$ . Thus, the probability of a message was given by $p\left(y\right){\prod }_{i=1}^{n}p\left({x}_{i}|y\right)$ .

Here's a different model, called the multinomial event model . To describe this model, we will use a different notation and set of features for representing emails. We let ${x}_{i}$ denote the identity of the $i$ -th word in the email. Thus, ${x}_{i}$ is now an integer taking values in $\left\{1,...,|V|\right\}$ , where $|V|$ is the size of our vocabulary (dictionary). An email of $n$ words is now represented by a vector $\left({x}_{1},{x}_{2},...,{x}_{n}\right)$ of length $n$ ; note that $n$ can vary for different documents. For instance, if an email starts with “A NIPS ...,” then ${x}_{1}=1$ (“a” is the first word in the dictionary), and ${x}_{2}=35000$ (if “nips” is the 35000th word in the dictionary).

In the multinomial event model, we assume that the way an email is generated is via a random process in which spam/non-spam is first determined (according to $p\left(y\right)$ ) as before. Then, the sender of the email writes the email by first generating ${x}_{1}$ from some multinomial distribution over words ( $p\left({x}_{1}|y\right)$ ). Next, the second word ${x}_{2}$ is chosen independently of ${x}_{1}$ but from the same multinomial distribution, and similarly for ${x}_{3}$ , ${x}_{4}$ , and so on, until all $n$ words of the email have been generated. Thus, the overall probability of a message is given by $p\left(y\right){\prod }_{i=1}^{n}p\left({x}_{i}|y\right)$ . Note that this formula looks like the one we had earlier for the probability of a message under themulti-variate Bernoulli event model, but that the terms in the formula now mean very different things. In particular ${x}_{i}|y$ is now a multinomial, rather than a Bernoulli distribution.

The parameters for our new model are ${\Phi }_{y}=p\left(y\right)$ as before, ${\Phi }_{k|y=1}=p\left({x}_{j}=k|y=1\right)$ (for any $j$ ) and ${\Phi }_{i|y=0}=p\left({x}_{j}=k|y=0\right)$ . Note that we have assumed that $p\left({x}_{j}|y\right)$ is the same for all values of $j$ (i.e., that the distribution according to which a word is generated does not depend on its position $j$ within the email).

If we are given a training set $\left\{\left({x}^{\left(i\right)},{y}^{\left(i\right)}\right);i=1,...,m\right\}$ where ${x}^{\left(i\right)}=\left({x}_{1}^{\left(i\right)},{x}_{2}^{\left(i\right)},...,{x}_{{n}_{i}}^{\left(i\right)}\right)$ (here, ${n}_{i}$ is the number of words in the $i$ -training example), the likelihood of the data is given by

$\begin{array}{ccc}\hfill \mathcal{L}\left(\Phi ,{\Phi }_{k|y=0},{\Phi }_{k|y=1}\right)& =& \prod _{i=1}^{m}p\left({x}^{\left(i\right)},{y}^{\left(i\right)}\right)\hfill \\ & =& \prod _{i=1}^{m}\left(\prod _{j=1}^{{n}_{i}},p,\left({x}_{j}^{\left(i\right)}|y;{\Phi }_{k|y=0},{\Phi }_{k|y=1}\right)\right)p\left({y}^{\left(i\right)};{\Phi }_{y}\right).\hfill \end{array}$

Maximizing this yields the maximum likelihood estimates of the parameters:

$\begin{array}{ccc}\hfill {\Phi }_{k|y=1}& =& \frac{{\sum }_{i=1}^{m}{\sum }_{j=1}^{{n}_{i}}1\left\{{x}_{j}^{\left(i\right)}=k\wedge {y}^{\left(i\right)}=1\right\}}{{\sum }_{i=1}^{m}1\left\{{y}^{\left(i\right)}=1\right\}{n}_{i}}\hfill \\ \hfill {\Phi }_{k|y=0}& =& \frac{{\sum }_{i=1}^{m}{\sum }_{j=1}^{{n}_{i}}1\left\{{x}_{j}^{\left(i\right)}=k\wedge {y}^{\left(i\right)}=0\right\}}{{\sum }_{i=1}^{m}1\left\{{y}^{\left(i\right)}=0\right\}{n}_{i}}\hfill \\ \hfill {\Phi }_{y}& =& \frac{{\sum }_{i=1}^{m}1\left\{{y}^{\left(i\right)}=1\right\}}{m}.\hfill \end{array}$

If we were to apply Laplace smoothing (which needed in practice for good performance) when estimating ${\Phi }_{k|y=0}$ and ${\Phi }_{k|y=1}$ , we add 1 to the numerators and $|V|$ to the denominators, and obtain:

$\begin{array}{ccc}\hfill {\Phi }_{k|y=1}& =& \frac{{\sum }_{i=1}^{m}{\sum }_{j=1}^{{n}_{i}}1\left\{{x}_{j}^{\left(i\right)}=k\wedge {y}^{\left(i\right)}=1\right\}+1}{{\sum }_{i=1}^{m}1\left\{{y}^{\left(i\right)}=1\right\}{n}_{i}+|V|}\hfill \\ \hfill {\Phi }_{k|y=0}& =& \frac{{\sum }_{i=1}^{m}{\sum }_{j=1}^{{n}_{i}}1\left\{{x}_{j}^{\left(i\right)}=k\wedge {y}^{\left(i\right)}=0\right\}+1}{{\sum }_{i=1}^{m}1\left\{{y}^{\left(i\right)}=0\right\}{n}_{i}+|V|}.\hfill \end{array}$

While not necessarily the very best classification algorithm, the Naive Bayes classifier often works surprisingly well. It is often also a very good “first thing to try,”given its simplicity and ease of implementation.

#### Questions & Answers

where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
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da
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Bhagvanji
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Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
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ya I also want to know the raman spectra
Bhagvanji
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please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
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Professor
I think
Professor
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Alexandre
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Rafiq
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Damian
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LITNING
scanning tunneling microscope
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Santosh
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Rafiq
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Mahi
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Rafiq
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Anam
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Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
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write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
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Renato
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Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
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