# Machine learning lecture 1 course notes  (Page 13/13)

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## Softmax regression

Let's look at one more example of a GLM. Consider a classification problem in which the response variable $y$ can take on any one of $k$ values, so $y\in \left\{1,2,...,k\right\}$ . For example, rather than classifying email into the two classes spam or not-spam—which would have been a binaryclassification problem—we might want to classify it into three classes, such as spam, personal mail, and work-related mail. The response variable isstill discrete, but can now take on more than two values. We will thus model it as distributed according to a multinomial distribution.

Let's derive a GLM for modelling this type of multinomial data. To do so, we will begin by expressing the multinomial as an exponential family distribution.

To parameterize a multinomial over $k$ possible outcomes, one could use $k$ parameters ${\Phi }_{1},...,{\Phi }_{k}$ specifying the probability of each of the outcomes. However, these parameters would be redundant, or more formally, they would not beindependent (since knowing any $k-1$ of the ${\Phi }_{i}$ 's uniquely determines the last one, as they must satisfy ${\sum }_{i=1}^{k}{\Phi }_{i}$ = 1). So, we will instead parameterize the multinomial with only $k-1$ parameters, ${\Phi }_{1},...,{\Phi }_{k-1}$ , where ${\Phi }_{i}=p\left(y=i;\Phi \right)$ , and $p\left(y=k;\Phi \right)=1-{\sum }_{i=1}^{k-1}{\Phi }_{i}$ . For notational convenience, we will also let ${\Phi }_{k}=1-{\sum }_{i=1}^{k-1}{\Phi }_{i}$ , but we should keep in mind that this is not a parameter, and that it is fully specified by ${\Phi }_{1},...,{\Phi }_{k-1}$ .

To express the multinomial as an exponential family distribution, we will define $T\left(y\right)\in {\mathbb{R}}^{k-1}$ as follows:

$T\left(1\right)=\left[\begin{array}{c}1\\ 0\\ 0\\ ⋮\\ 0\end{array}\right],\phantom{\rule{0.277778em}{0ex}}T\left(2\right)=\left[\begin{array}{c}0\\ 1\\ 0\\ ⋮\\ 0\end{array}\right],\phantom{\rule{0.277778em}{0ex}}T\left(3\right)=\left[\begin{array}{c}0\\ 0\\ 1\\ ⋮\\ 0\end{array}\right],\phantom{\rule{0.277778em}{0ex}}\cdots ,T\left(k-1\right)=\left[\begin{array}{c}0\\ 0\\ 0\\ ⋮\\ 1\end{array}\right],\phantom{\rule{0.277778em}{0ex}}T\left(k\right)=\left[\begin{array}{c}0\\ 0\\ 0\\ ⋮\\ 0\end{array}\right],\phantom{\rule{0.277778em}{0ex}}$

Unlike our previous examples, here we do not have $T\left(y\right)=y$ ; also, $T\left(y\right)$ is now a $k-1$ dimensional vector, rather than a real number. We will write ${\left(T\left(y\right)\right)}_{i}$ to denote the $i$ -th element of the vector $T\left(y\right)$ .

We introduce one more very useful piece of notation. An indicator function $1\left\{·\right\}$ takes on a value of 1 if its argument is true, and 0 otherwise ( $1\left\{\mathrm{True}\right\}=1$ , $1\left\{\mathrm{False}\right\}=0$ ). For example, $1\left\{2=3\right\}=0$ , and $1\left\{3=5-2\right\}=1$ . So, we can also write the relationship between $T\left(y\right)$ and $y$ as ${\left(T\left(y\right)\right)}_{i}=1\left\{y=i\right\}$ . (Before you continue reading, please make sure you understand why this is true!) Further, wehave that $\mathrm{E}\left[{\left(T\left(y\right)\right)}_{i}\right]=P\left(y=i\right)={\Phi }_{i}$ .

We are now ready to show that the multinomial is a member of the exponential family. We have:

$\begin{array}{ccc}\hfill p\left(y;\Phi \right)& =& {\Phi }_{1}^{1\left\{y=1\right\}}{\Phi }_{2}^{1\left\{y=2\right\}}\cdots {\Phi }_{k}^{1\left\{y=k\right\}}\hfill \\ & =& {\Phi }_{1}^{1\left\{y=1\right\}}{\Phi }_{2}^{1\left\{y=2\right\}}\cdots {\Phi }_{k}^{1-{\sum }_{i=1}^{k-1}1\left\{y=i\right\}}\hfill \\ & =& {\Phi }_{1}^{{\left(T\left(y\right)\right)}_{1}}{\Phi }_{2}^{{\left(T\left(y\right)\right)}_{2}}\cdots {\Phi }_{k}^{1-{\sum }_{i=1}^{k-1}{\left(T\left(y\right)\right)}_{i}}\hfill \\ & =& exp\left({\left(T\left(y\right)\right)}_{1}log\left({\Phi }_{1}\right)+{\left(T\left(y\right)\right)}_{2}log\left({\Phi }_{2}\right)+\hfill \\ & & \phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\cdots +\left(1-{\sum }_{i=1}^{k-1}{\left(T\left(y\right)\right)}_{i}\right)log\left({\Phi }_{k}\right)\right)\hfill \\ & =& exp\left({\left(T\left(y\right)\right)}_{1}log\left({\Phi }_{1}/{\Phi }_{k}\right)+{\left(T\left(y\right)\right)}_{2}log\left({\Phi }_{2}/{\Phi }_{k}\right)+\hfill \\ & & \phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\cdots +{\left(T\left(y\right)\right)}_{k-1}log\left({\Phi }_{k-1}/{\Phi }_{k}\right)+log\left({\Phi }_{k}\right)\right)\hfill \\ & =& b\left(y\right)exp\left({\eta }^{T}T\left(y\right)-a\left(\eta \right)\right)\hfill \end{array}$

where

$\begin{array}{ccc}\hfill \eta & =& \left[\begin{array}{c}log\left({\Phi }_{1}/{\Phi }_{k}\right)\\ log\left({\Phi }_{2}/{\Phi }_{k}\right)\\ ⋮\\ log\left({\Phi }_{k-1}/{\Phi }_{k}\right)\end{array}\right],\hfill \\ \hfill a\left(\eta \right)& =& -log\left({\Phi }_{k}\right)\hfill \\ \hfill b\left(y\right)& =& 1.\hfill \end{array}$

This completes our formulation of the multinomial as an exponential family distribution.

The link function is given (for $i=1,...,k$ ) by

${\eta }_{i}=log\frac{{\Phi }_{i}}{{\Phi }_{k}}.$

For convenience, we have also defined ${\eta }_{k}=log\left({\Phi }_{k}/{\Phi }_{k}\right)=0$ . To invert the link function and derive the response function, we therefore havethat

$\begin{array}{ccc}\hfill {e}^{{\eta }_{i}}& =& \frac{{\Phi }_{i}}{{\Phi }_{k}}\hfill \\ \hfill {\Phi }_{k}{e}^{{\eta }_{i}}& =& {\Phi }_{i}\hfill \\ \hfill {\Phi }_{k}\sum _{i=1}^{k}{e}^{{\eta }_{i}}& =& \sum _{i=1}^{k}{\Phi }_{i}=1\hfill \end{array}$

This implies that ${\Phi }_{k}=1/{\sum }_{i=1}^{k}{e}^{{\eta }_{i}}$ , which can be substituted back into Equation  [link] to give the response function

${\Phi }_{i}=\frac{{e}^{{\eta }_{i}}}{{\sum }_{j=1}^{k}{e}^{{\eta }_{j}}}$

This function mapping from the $\eta$ 's to the $\Phi$ 's is called the softmax function.

To complete our model, we use Assumption 3, given earlier, that the ${\eta }_{i}$ 's are linearly related to the $x$ 's. So, have ${\eta }_{i}={\theta }_{i}^{T}x$ (for $i=1,...,k-1$ ), where ${\theta }_{1},...,{\theta }_{k-1}\in {\mathbb{R}}^{n+1}$ are the parameters of our model. For notational convenience, we can also define ${\theta }_{k}=0$ , so that ${\eta }_{k}={\theta }_{k}^{T}x=0$ , as given previously. Hence, our model assumes that the conditional distribution of $y$ given $x$ is given by

$\begin{array}{ccc}\hfill p\left(y=i|x;\theta \right)& =& {\Phi }_{i}\hfill \\ & =& \frac{{e}^{{\eta }_{i}}}{{\sum }_{j=1}^{k}{e}^{{\eta }_{j}}}\hfill \\ & =& \frac{{e}^{{\theta }_{i}^{T}x}}{{\sum }_{j=1}^{k}{e}^{{\theta }_{j}^{T}x}}\hfill \end{array}$

This model, which applies to classification problems where $y\in \left\{1,...,k\right\}$ , is called softmax regression . It is a generalization of logistic regression.

Our hypothesis will output

$\begin{array}{ccc}\hfill {h}_{\theta }\left(x\right)& =& \mathrm{E}\left[T\left(y\right)|x;\theta \right]\hfill \\ & =& \mathrm{E}\left[\left(\begin{array}{c}1\left\{y,=,1\right\}\\ 1\left\{y,=,2\right\}\\ ⋮\\ 1\left\{y,=,k,-,1\right\}\end{array}|,\phantom{\rule{0.166667em}{0ex}},x,;,\theta \right]\hfill \\ & =& \left[\begin{array}{c}{\Phi }_{1}\\ {\Phi }_{2}\\ ⋮\\ {\Phi }_{k-1}\end{array}\right]\hfill \\ & =& \left[\begin{array}{c}\frac{exp{\theta }_{1}^{T}x}{{\sum }_{j=1}^{k}exp{\theta }_{j}^{T}x}\\ \frac{exp{\theta }_{2}^{T}x}{{\sum }_{j=1}^{k}exp{\theta }_{j}^{T}x}\\ ⋮\\ \frac{exp{\theta }_{k-1}^{T}x}{{\sum }_{j=1}^{k}exp{\theta }_{j}^{T}x}\end{array}\right].\hfill \end{array}$

In other words, our hypothesis will output the estimated probability that $p\left(y=i|x;\theta \right)$ , for every value of $i=1,...,k$ . (Even though ${h}_{\theta }\left(x\right)$ as defined above is only $k-1$ dimensional, clearly $p\left(y=k|x;\theta \right)$ can be obtained as $1-{\sum }_{i=1}^{k-1}{\Phi }_{i}$ .)

Lastly, let's discuss parameter fitting. Similar to our original derivation of ordinary least squares and logistic regression, if we have a training set of $m$ examples $\left\{\left({x}^{\left(i\right)},{y}^{\left(i\right)}\right);i=1,...,m\right\}$ and would like to learn the parameters ${\theta }_{i}$ of this model, we would begin by writing down the log-likelihood

$\begin{array}{ccc}\hfill \ell \left(\theta \right)& =& \sum _{i=1}^{m}logp\left({y}^{\left(i\right)}|{x}^{\left(i\right)};\theta \right)\hfill \\ & =& \sum _{i=1}^{m}log\prod _{l=1}^{k}{\left(\frac{{e}^{{\theta }_{l}^{T}{x}^{\left(i\right)}}}{{\sum }_{j=1}^{k}{e}^{{\theta }_{j}^{T}{x}^{\left(i\right)}}}\right)}^{1\left\{{y}^{\left(i\right)}=l\right\}}\hfill \end{array}$

To obtain the second line above, we used the definition for $p\left(y|x;\theta \right)$ given in Equation  [link] . We can now obtain the maximum likelihood estimate of the parameters by maximizing $\ell \left(\theta \right)$ in terms of $\theta$ , using a method such as gradient ascent or Newton's method.

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Source:  OpenStax, Machine learning. OpenStax CNX. Oct 14, 2013 Download for free at http://cnx.org/content/col11500/1.4
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