# Linear and Quadratic Discriminant Analysis for ML / statistics newbies

(Note: This post assumes that the reader has knowledge of basic statistics and terms used in machine learning. Inspite of that, I have provided links whereever necessary 🙂 ).

Linear Discriminant Analysis(LDA) and Quadratic Discriminant Analysis(QDA) are types of Bayesian classifiers.

A Bayesian classifier, in mathematical terms, does the following-

$C^{bayesian}(x) = argmax_{r \in \{1, 2, ..., k\}} Pr(Y = r | X = x)$

What does this mean? To put it in the form of steps, heres what happens-

1. The classifier is given an input($X$) that is the feature vector $x$. $x$ consists of $p$ different predictors, one for each input dimension.

2. Then, with respect to each possible output class($Y$) $r$ from $1, 2, ..., k$, the classifier computes $Pr(Y = r | X = x)$ – this is the probability that the actual output is $r$, given $x$ as the input.

3. The class (i.e. that value of $r$) that returns the maximum value for the above mentioned probability, is given as the output of the classifier – $C^{bayesian}(x)$.

In essence, a Bayesian classifier tries to minimize the probability of misclassification (which would be equal to $1 - Pr(Y = r | X = x)$).

But then the most natural question is- How is this probability computed? It is precisely at this point where all Bayesian classifiers differ. Each one of them uses a different technique to understand the data, and compute these crucial quantities. In fact, even a simple nearest-neighbors classifier can be used as a Bayesian classifier, with $Pr(Y = r | X = x)$ being the fraction of the nearest neighbors to $x$  that belong to class $r$. LDA and QDA, on the other hand, use a more mathematical approach.

Anyone acquainted with basic statistics knows Bayes theorem. Given two possible events $A$ and $B$, it states-

$Pr(A|B) = \frac{Pr(A)Pr(B|A)}{Pr(B)}$

Applying this to $Pr(Y = r | X = x)$, we would get

$Pr(Y = r | X = x) = \frac{Pr(Y = r)Pr(X = x | Y = r)}{Pr(X = x)}$.

Now, look at the first part in the above numerator- $Pr(Y = r)$. This is essentially the probability of the class of a random input being $r$ – prior to even looking at what the input is! Thats why its called the priori probability. During the training stage itself, it is computed as the fraction of the training samples that belonged to class $r$. As you might have understood by now, this number is independent of the what $x$ is, and reflects the trends seen in the training set.

Now lets consider the second part – $Pr(X = x | Y = r)$. This number is conceptually equivalent to answering the question- Given that a random data point(not necessarily in the training data) was being selected from class $r$, what would be the likelihood that it looked like $x$? To quantify this ‘likelihood’, LDA and QDA use a Multivariate Gaussian Distribution model for each class. Mathematically put, the probability density function for a multivariate Gaussian distribution is-

$f_{\boldsymbol\Sigma, \boldsymbol\mu}(x) = \frac{1}{\sqrt{(2\pi)^k|\boldsymbol\Sigma|}}\exp(-\frac{1}{2}({ x}-{\boldsymbol\mu})^\mathrm{T}{\boldsymbol\Sigma}^{-1}({x}-{\boldsymbol\mu}))$.

You definitely don’t need to remember this huge-ass formula. To put it simply, heres the intuition behind it- A probability distribution model is a way for an algorithm to understand how data points are distributed in a $p$-dimensional space. In the case of the multivariate Gaussian(also called Normal) distribution, the two parameters $\boldsymbol\mu$ (the $p$-dimensional mean vector) and $\boldsymbol\Sigma$ (the $p$x$p$-dimensional covariance matrix) are used to ‘summarize’ how the data ‘looks’ in the conceptual space. These are obviously learnt during training.

While $\boldsymbol\mu$ denotes the approximate average position, $\boldsymbol\Sigma$ summarizes the ‘shape’ of the distribution of data points. (Imagine drawing a random cluster of points on a graph paper. Now draw an ellipse-like shape over that region where most of the points are concentrated. The ‘central’ point of the ellipse will be a 2-dimensional vector $\boldsymbol\mu$, and the ‘shape’ would be approximated by a 2×2 matrix given by $\boldsymbol\Sigma$.) The corresponding probability density function(pdf) for a random $x$ in space, is proportional to how likely it is that a point at that location would be selected randomly from the distribution. The pdf is to probability, what pressure is to force. Pressure isn’t exactly force itself, but it shows how intense the latter is at a particular location- and the higher the pressure, the greater is the likelihood that the force will show its effect at that point.

Thus, LDA and QDA use the aforementioned $f(x)$ to quantify $Pr(X = x | Y = r)$. What about the denominator then – $Pr(X = x)$? It is nothing but the sum of the values of $Pr(Y = r)Pr(X = x | Y = r)$ over all possible $r$s. This is given by the Total Probability Theorem.

But then where do LDA and QDA differ? They differ in only one subtle point- While LDA assumes the same $\boldsymbol\Sigma$ for all classes, QDA computes a different $\boldsymbol\Sigma$ for each class. Essentially, LDA computes a separate $\boldsymbol\mu$ for each class(using training points that belonged to it), but $\boldsymbol\Sigma$ is computed using the entire training data. And this common covariance matrix is used in the computations corresponding to every class. QDA, on the other hand, computes a separate $\boldsymbol\Sigma$ and $\boldsymbol\mu$ for each possible output class.

We will now tackle the two most important question that might be plaguing your mind-

1. Why is LDA ‘linear’, and QDA ‘quadratic’?

The answer lies in the probability function. Consider the simplest case of one input variable only. The answer can be generalized to multiple dimensions using matrix algebra.

$Pr(Y = r | X = x) = \frac{Pr(Y = r)Pr(X = x | Y = r)}{Pr(X = x)}$

We are maximizing over all possible $r$s right? Intuitively, that should tell you that the denominator in the above equation, doesn’t really matter in comparisons- since its the same for all values of $r$. So we are now down to finding that $r$ which gives the maximum value for

$Pr(Y = r)Pr(X = x | Y = r)$

Let $Pr(Y = r) = \beta_{r}$ (to reduce my LaTeX efforts). Using the formula for 1-D Normal distribution density function, we have

$Pr(X = x | Y = r) = \frac{1}{\sigma_r\sqrt{2\pi}}e^\frac{(x - \mu_r)^2}{\sigma_r^2}$

Multiplying the two and taking a natural logarithm of the resultant expression, we are now down to finding that $r$ that gives the maximum value for

$log(\beta_{r}/\sigma_r) - \frac{x^2}{2\sigma_r^2} - \frac{\mu_r^2}{2\sigma_r^2} + \frac{x\mu_r}{\sigma_r^2}$

Observe the above quadratic(in x) expression carefully. Once again, I have omitted those terms that are constants or which remain the same irrespective of $r$(NOT $x$), which is the focus of the whole process. Or have I? Consider the second term. IF I am talking of the LDA classifier, would $\sigma_r$ be dependent on $r$ at all? It wouldn’t, since LDA assumes the same variance/covariance for all classes. As a result, the second term could no longer be something worth taking into the comparison process. Incidentally, that very term is what makes the above expression quadratic. Otherwise, its linear!

And thats (more or less) why LDA is ‘linear’ and QDA is ‘quadratic’ 🙂 .

2. Why would we ever choose LDA? Isn’t QDA conceptually more flexible?

QDA isn’t always better. In a certain way(too mathematically convoluted for this post), LDA does learn a linear boundary between the points belonging to different classes. QDA, on the other hand, learns a quadratic one. If the actual boundaries are indeed linear, QDA may have a higher model bias. – especially if the available data isn’t dense enough. Another(and easier to imagine) scenario is when you have a very limited set of training points. In such a case, if you divide your already sparse dataset into the constituent classes, the covariance matrix computed for each might be extremely inaccurate. In such cases, its better to simplify the entire process and use a common covariance matrix thats computed from the entire dataset we use while training.

## 6 thoughts on “Linear and Quadratic Discriminant Analysis for ML / statistics newbies”

1. Erik says:

“If the actual boundaries are indeed linear, QDA would obviously have a higher model bias.”

Is not the problem higher model variance? I.e., tendency to overfit. QDA has more flexibility. A quadratic decision boundary can be linear.

1. Aahh…I agree. Though for a quadratic boundary to be linear there needs to be a sufficient number of data points. I will edit the post to say ‘may have a higher bias’. Thanks for pointing this out! 🙂

2. Erik says:

That is not the point. A quadratic boundary will always fit the training set better, i.e., _always have lower bias_. But it will have higher variance, especially if the right model is a linear one, in which case the extra flexibility will only model noise.

1. I agree on the variance part. But given a small dataset, would QDA necessarily have lower bias even with respect to a test dataset? Or is bias only defined with respect to the training set?

3. Erik says:

Don’t think you can talk about bias really with respect to unseen data. What does it mean to underfit something you have not seen? Good generalization on a test set, hopefully, happens when you are balancing the bias-variance trade-off.

4. @Sachin: I am a newbie, learning ML now. And the way you have explained Linear vs quadratic Discriminant analysis is just too clear. Thanks for making this so easy to comprehend. I suggest you post more, if your time persists. Also, if possible, would love to connect with you.