# Fisher LDA - What is the difference between a discriminant function and a linear decision boundary?

I am studying Fisher LDA, the case where there are K=2 classes of data.

It is my understanding that Fisher LDA looks for the 1-dimensional space onto which the data should be projected in order to maximize the ratio of the between class variance to the within class variance.

If this direction is a vector $w$, then how do I compute the linear decision boundary?

Many online resources write this as $y(x)=w^{t}x+w_{0}$.

If $w$, then isn't $w^{t}x$ the projection of a vector $x$ onto that line? And if so, what is $w_{0}$ and how can one solve for it?

I suppose my main question is this: is the decision boundary a function orthogonal to the direction vector that separates data points before projection, or is the decision boundary a point on a line through the vector of maximum separation $w$ that separates the projected points (and if so what is $w_{0}$?

• Since what discriminates in LDA is the extracted discriminant variate, which is single in the 2-class case - the boundary, strictly speaking, is a point on the discriminant line: it is the point of zero discriminant score. However, you are in right to extend the point onto the parental p-dim. space of the p analyzed variables; that will form a p-1-dim plane perpendicular to the discriminant line (and coming through that zero point on it). "Perpendicular" because nothing discriminative is being added extra to the single discriminant function. Jun 10 '18 at 16:23

The dimensions of the decision boundary match the number of decision models you have. The reason $K-1$ models is common is that the $K^{th}$ model is redundant as it is the samples that have not been positively assigned by the previous $K-1$ models. One exception is if you want to have an invalid result where the data does not conform to the expected distribution of any group then you would need $K$ models and samples that fall outside this will have an invalid label. In such situations you would then use $K$ models. So for simplicity I will use the number of models $m$ below, which will typically be $K$ or $K-1$. $v$ will be the number of variables in a sample vector, $o$ the number of observations in the data.

$w^tx$ is indeed a projection of your function w into a vector x, and as it is the inner product of the two, its dimensions depends on the outer dimensions of each matrix/vector. For this to work the dimensions of $w^t$ must be $mxv$ and for $x$ the dimensions are $vxo$.

If $w^t$ is a vector, and $x$ is a vector the product will be a scalar, a single number i.e. if $m>1,o=1$ then you get a $mx1$ vector where each element is the scalar for each model. If $m=1,o>1$ you get an $1xo$ vector where each element is the score for model 1 for each observation. If $m>1,o>1$ then you get a $mxo$ matrix of scores for each model and each observation.

If $o>1$ i.e. x is a matrix, you would to replicate $w0$ to the outer dimension of $x$ which is the number of samples $o$, i.e. you apply the offset to each sample.

If $m>1$ i.e. w is a matrix, you would calculate $w0$ for each model $x$ which is the number of models $m$, i.e. you apply the individual offsets to each model.

$w_0$ is the offset you choose to tune false positives vs false negatives, e.g.through a ROC curve.

• Thank for that link, I am finding it very useful. I am under the impression that for $K=2$ classes measured in $m=2$ dimensions (as the graph in the link shows), that the hyperplane (a line which lives in 2-space) is normal to $w$, which is the weight vector computed with Fisher's ratio? If so it should make sense to me that setting $g(x) = 0$ would classify vectors whose distance from $g(x) \geq 0$ elements of one class and vectors whose distance from $g(x) \leq 0$ to be elements of the other.
– TYBG
Jun 9 '18 at 20:25
• see slide 4 'It corresponds to a (D-1)- dimensional hyperplane in a D-dimensional input space ' Jun 11 '18 at 7:58
• I've updated my answer and hopefully it walks through it a bit better. The graphs in the linked article show $K=2$ and $m = 1$, not $m = 2$ in the early slides and keeps things simple by always assuming $m = K-1$. I've tried to avoid this assumption and hopefully it doesn't make things more confusing. Jun 11 '18 at 8:29