Linear discriminant analysis (LDA) and quadratic discriminant analysis (QDA) are supervised machine learning algorithms used for multinomial classification tasks. Both LDA and QDA estimate $P(y \;\vert\;\mathbf{x})$ relying on Bayes theorem, i.e., first calculating $P(\mathbf{x} \;\vert\; y)$, and eventually $P(y)$ too. For each class $k \in \{1, \ldots, K\}$, LDA assumes that features follow a Gaussian distribution $N(\boldsymbol{\mu_k}, \Sigma)$, while for QDA they are of the type $N(\boldsymbol{\mu_k}, \Sigma_k)$. Relying on a training set, both LDA and QDA fit the mean value of each class. Moreover, while LDA fits a unique covariance matrix assumed to be shared by all classes, QDA allows to fit one per class. As a consequence, while LDA may only draw linear decision boundaries, QDA is able to produce quadratic ones. Once the Gaussian distributions are known, and thus $P(\mathbf{x} \;\vert\; y)$ may be evaluated, assigning a label to new samples is straightforward.

## Gaussian Distributions

Since LDA and QDA assume that features of each class follow Gaussian distributions, before diving into these algorithms, it is actually convenient to first understand how to characterize a random variable $\mathit{X}$ that has either a univariate or multivariate Gaussian distribution.

### Case Univariate

In this case, $\mathit{X}$ has a one-dimensional Gaussian distribution $\mathit{X} \sim N(\mu, \sigma^2)$, i.e.,

$f_\mathit{X}(x \;\vert\; \mu, \sigma) = \frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{1}{2\sigma^2}(x-\mu)^2\right)$

where $\mu$ is the mean or expectation of $\mathit{X}$, i.e., $\mu = \mathbb{E}[\mathit{X}]$, and $\sigma^2$ is the variance of $\mathit{X}$ , i.e., $\sigma^2 =V[\mathit{X}]$.

To characterize $\mathit{x}$, according to the shape of $f_\mathit{X}(x \;\vert\; \mu, \sigma)$, we only need to determine $\mu$ and $\sigma$. To estimate their values (for $\sigma$ we actually focus on $\sigma^2$) with a training set of $m$ i.i.d. samples $X = \left\{x^{(1)}, \ldots, x^{(m)} \right\}$, we can compute the log-likelihood

\begin{align} l(X, \mu, \sigma^2) &= \sum_{i=1}^m\log\left(f_\mathit{X} (x^{(i)}\vert\; \mu, \sigma)\right)\\ &= -\frac{m}{2}\log(2\pi) - \frac{m}{2}\log(\sigma^2)-\frac{1}{2\sigma^2}\sum_{i=1}^m\left(x^{(i)}-\mu\right)^2 \end{align}

To find $\hat{\mu}$ and $\hat{\sigma^2}$, we can calculate the partial derivatives of $l(X, \mu, \sigma^2)$ with respect to $\mu$ and $\sigma^2$, and set them to zero. Applying some algebra, it can be shown that

\begin{align} \hat{\mu} &= \frac{1}{m}\sum_{i=1}^m x^{(i)}\\ \hat{\sigma^2} &= \frac{1}{m}\sum_{i=1}^m \left(x^{(i)} - \mu\right)^2 \end{align}

The way it is defined, $\hat{\mu}$ can be directly computed from the samples, and thus is known as the sample mean. On the other hand, the variance estimator $\hat{\sigma^2}$ has the problem that it depends on the “real” value of $\mu$, that is actually unknown. Hence, in general, $\hat{\sigma^2}$ is approximated by the sample variance, which only requires replacing $\mu$ by $\hat{\mu}$ in the previous expression, i.e.,

$\hat{\sigma^2} = \frac{1}{m}\sum_{i=1}^m \left(x^{(i)} - \hat{\mu}\right)^2$

Since $E[\hat{\sigma^2}] = \frac{m-1}{m}\sigma^2\neq \sigma^2$, then the sample variance is a biased estimator. We can define an unbiased estimator $\tilde{\sigma^2}$ as follows

\begin{align} \tilde{\sigma^2} &= \frac{m}{m-1}\hat{\sigma^2}\\ &= \frac{1}{m-1}\sum_{i=1}^m \left(x^{(i)} - \hat{\mu}\right)^2 \end{align}

Once $\hat{\mu}$ and $\tilde{\sigma^2}$ are computed, we can approximate the behavior of $\mathit{X}$ with that of a random variable with a normal distribution $N(\hat{\mu}, \tilde{\sigma^2})$.

### Case Multivariate

In these scenarios, $\mathit{X}$ has a n-dimensional Gaussian distribution $\mathit{X} \sim N(\boldsymbol{\mu}, \Sigma)$, i.e.,

$f_\mathit{X}(\mathbf{x} \;\vert\; \boldsymbol{\mu}, \Sigma) = \frac{1}{(2\pi)^{n/2}|\Sigma|^{1/2}}\,\exp\left(-\frac{1}{2}(\mathbf{x}-\boldsymbol{\mu})^\intercal~\Sigma^{-1}~(\mathbf{x}-\boldsymbol{\mu})\right)$

where $\mathbf{x} \in \mathbb{R}^{n\times1}$ is a vector that may take any values, $\boldsymbol{\mu} \in \mathbb{R}^{n\times1}$ is the mean or expectation of $\mathit{X}$, i.e., $E[\mathit{X}]=\boldsymbol{\mu}$, and $\Sigma \in \mathbb{R}^{n \times n}$ is the covariance matrix of $\mathit{X}$, i.e., $\Sigma = E[(\mathit{X}-\boldsymbol{\mu})(\mathit{X}-\boldsymbol{\mu})^\intercal]$.

For a training set of $m$ i.i.d. samples $X = \left\{\mathbf{x}^{(1)}, \ldots, \mathbf{x}^{(m)} \right\}$, following the same reasoning as for the univariate case, it is straightforward to show that

\begin{align} \hat{\boldsymbol{\mu}} &= \frac{1}{m}\sum_{i=1}^m \mathbf{x}^{(i)}\\ \hat{\Sigma} &= \frac{1}{m-1}\sum_{i=1}^m (\mathbf{x}^{(i)} - \hat{\boldsymbol{\mu}})(\mathbf{x}^{(i)} - \hat{\boldsymbol{\mu}})^\intercal \end{align}

Then we can emulate the behavior of $\mathit{X}$ as that of random variable with a normal distribution $N(\hat{\boldsymbol{\mu}}, \hat{\Sigma})$.

For each class $k \in \{1, \ldots, K\}$, features are modeled as random variables $\mathit{X_k} \sim N(\boldsymbol{\mu_k}, \Sigma_k)$. The objective is to determine the mean values and covariance matrixes. For this, we assume we have a training dataset of $m$ labeled samples $X = \left\{(\mathbf{x}^{(1)}, y^{(1)}) \ldots, (\mathbf{x}^{(m)}, y^{(m)})\right\}$, such that for each class $k$, there are $m_k$ i.i.d. samples, i.e., $m =\sum_{k=1}^K m_k$.

To solve this problem, QDA assumes that the $m_k$ samples of any class $k$ only give information about the parameters of that particular class, that is $\boldsymbol{\mu_k}$ and $\Sigma_k$. Relying on this hypothesis, it turns out that we can actually split the problem of estimating all the required parameters in $K$ problems, each aiming to estimate those parameters associated to a particular class.

For each given $k$, we can

• fit the prior probability $P(y=k)$ to the proportion of samples of that class in the dataset, i.e., $P(y = k) = \frac{m_k}{m}$, or assume they are all equiprobable, i.e., $\forall k, \, P(y = k) = \frac{1}{m}$.
• estimate the values of $\boldsymbol{\mu_k}$ and $\Sigma_k$ apply the same reasoning as we did to characterize multivariate Gaussian distributions, i.e., we only need to compute
\begin{align} \hat{\boldsymbol{\mu_k}} &= \frac{1}{m_k}\sum_{y^{(i)}=k}^m \mathbf{x}^{(i)}\\ \hat{\Sigma}_k &= \frac{1}{m_k-1}\sum_{y^{(i)}=k}^m (\mathbf{x}^{(i)} - \hat{\boldsymbol{\mu_k}})(\mathbf{x}^{(i)} - \hat{\boldsymbol{\mu_k}})^\intercal \end{align}

Finally, when classifying any new sample $\mathbf{x}$, we proceed to assign $\mathbf{x}$ to the class $k$ for which $f_{\mathit{X_k}}(\mathbf{x} \;\vert\; \boldsymbol{\mu_k}, \Sigma_k)P(y=k)$ is largest.

## Linear Discriminant Analysis

For each class $k$, features are modeled as random variables $\mathit{X_k} \sim N(\mu_k, \Sigma)$. Compared with QDA, less parameters need to be estimated, since $\Sigma$ is now modeled as being the same across all classes.

Considering the same dataset as for QDA, the prior probabilities and mean values of each class may be calculated as in the previous method. On the other hand, to estimate $\Sigma$, the complete dataset can be used to estimate its value, i.e., the $m$ samples. In particular, $\hat{\Sigma}$ can be obtained combining the covariance matrices estimated in LDA for each class $k$, as

\begin{align} \hat{\Sigma} &= \frac{\sum_{k=1}^K (m_k -1)\hat{\Sigma}_k}{\sum_{k=1}^K (m_k - 1)}\\ &= \frac{1}{m - K}\sum_{k=1}^K \sum_{y^{(i)}=k} (\mathbf{x}^{(i)} - \hat{\boldsymbol{\mu_k}})(\mathbf{x}^{(i)} - \hat{\boldsymbol{\mu_k}})^\intercal \end{align}

which is known as the pooled covariance matrix.

When classifying any new sample $\mathbf{x}$, we proceed to assign $\mathbf{x}$ to the class $k$ for which $f_{\mathit{X_k}}(\mathbf{x} \;\vert\; \boldsymbol{\mu_k}, \Sigma)P(y=k)$ is largest.