# How to plot logistic regression decision boundary?

I am running logistic regression on a small dataset which looks like this: After implementing gradient descent and the cost function, I am getting a 100% accuracy in the prediction stage, However I want to be sure that everything is in order so I am trying to plot the decision boundary line which separates the two datasets.

Below I present plots showing the cost function and theta parameters. As can be seen, currently I am printing the decision boundary line incorrectly. Extracting data

clear all; close all; clc;

alpha = 0.01;
num_iters = 1000;

%% Plotting data
x1 = linspace(0,3,50);
mqtrue = 5;
cqtrue = 30;
dat1 = mqtrue*x1+5*randn(1,50);

x2 = linspace(7,10,50);
dat2 = mqtrue*x2 + (cqtrue + 5*randn(1,50));

x = [x1 x2]'; % X

subplot(2,2,1);
dat = [dat1 dat2]'; % Y

scatter(x1, dat1); hold on;
scatter(x2, dat2, '*'); hold on;
classdata = (dat>40);


%  Setup the data matrix appropriately, and add ones for the intercept term
[m, n] = size(x);

% Add intercept term to x and X_test
x = [ones(m, 1) x];

% Initialize fitting parameters
theta = zeros(n + 1, 1);
%initial_theta = [0.2; 0.2];

J_history = zeros(num_iters, 1);

plot_x = [min(x(:,2))-2,  max(x(:,2))+2]

for iter = 1:num_iters
% Compute and display initial cost and gradient
[cost, grad] = logistic_costFunction(theta, x, classdata);
theta = theta - alpha * grad;
J_history(iter) = cost;

fprintf('Iteration #%d - Cost = %d... \r\n',iter, cost);

subplot(2,2,2);
hold on; grid on;
plot(iter, J_history(iter), '.r');  title(sprintf('Plot of cost against number of iterations. Cost is %g',J_history(iter)));
xlabel('Iterations')
ylabel('MSE')
drawnow

subplot(2,2,3);
grid on;
plot3(theta(1), theta(2), J_history(iter),'o')
title(sprintf('Tita0 = %g, Tita1=%g', theta(1), theta(2)))
xlabel('Tita0')
ylabel('Tita1')
zlabel('Cost')
hold on;
drawnow

subplot(2,2,1);
grid on;
% Calculate the decision boundary line
plot_y = theta(2).*plot_x + theta(1);  % <--- Boundary line
% Plot, and adjust axes for better viewing
plot(plot_x, plot_y)
hold on;
drawnow

end

fprintf('Cost at initial theta (zeros): %f\n', cost);
fprintf('Gradient at initial theta (zeros): \n');


The above code is implementing gradient descent correctly (I think) but I am still unable to show the boundary line plot. Any suggestions would be appreciated.

logistic_costFunction.m

function [J, grad] = logistic_costFunction(theta, X, y)

% Initialize some useful values
m = length(y); % number of training examples

h = sigmoid(X * theta);
J = -(1 / m) * sum( (y .* log(h)) + ((1 - y) .* log(1 - h)) );

for i = 1 : size(theta, 1)
grad(i) = (1 / m) * sum( (h - y) .* X(:, i) );
end

end


EDIT:

As per the below answer by @Esmailian, now I have something like this:

[m, n] = size(x);

x1_class = [ones(m, 1) x1' dat1'];
x2_class = [ones(m, 1) x2' dat2'];

x = [x1_class ; x2_class]


### Regarding the code

1. You should plot the decision boundary after training is finished, not inside the training loop, parameters are constantly changing there; unless you are tracking the change of decision boundary.

2. x1 (x2) is the first feature and dat1 (dat2) is the second feature for the first (second) class, so the extended feature space x for both classes should be the union of (1, x1, dat1) and (1, x2, dat2).

### Decision boundary

Assuming that data is $$\boldsymbol{x}=(x_1, x_2)$$ ((x, dat) or (plot_x, plot_y) in the code), and parameter is $$\boldsymbol{\theta}=(\theta_0, \theta_1,\theta_2)$$ ((theta(1), theta(2), theta(3)) in the code), here is the line that should be drawn as decision boundary: $$x_2 = -\frac{\theta_1}{\theta_2} x_1 - \frac{\theta_0}{\theta_2}$$ which can be drawn as a segment by connecting two points $$(0, - \frac{\theta_0}{\theta_2})$$ and $$(- \frac{\theta_0}{\theta_1}, 0)$$. However, if $$\theta_2=0$$, the line would be $$x_1=-\frac{\theta_0}{\theta_1}$$.

### Where this comes from?

Decision boundary of Logistic regression is the set of all points $$\boldsymbol{x}$$ that satisfy $${\Bbb P}(y=1|\boldsymbol{x})={\Bbb P}(y=0|\boldsymbol{x}) = \frac{1}{2}.$$ Given $${\Bbb P}(y=1|\boldsymbol{x})=\frac{1}{1+e^{-\boldsymbol{\theta}^t\boldsymbol{x_+}}}$$ where $$\boldsymbol{\theta}=(\theta_0, \theta_1,\cdots,\theta_d)$$, and $$\boldsymbol{x}$$ is extended to $$\boldsymbol{x_+}=(1, x_1, \cdots, x_d)$$ for the sake of readability to have$$\boldsymbol{\theta}^t\boldsymbol{x_+}=\theta_0 + \theta_1 x_1+\cdots+\theta_d x_d,$$ decision boundary can be derived as follows \begin{align*} &\frac{1}{1+e^{-\boldsymbol{\theta}^t\boldsymbol{x_+}}} = \frac{1}{2} \\ &\Rightarrow \boldsymbol{\theta}^t\boldsymbol{x_+} = 0\\ &\Rightarrow \theta_0 + \theta_1 x_1+\cdots+\theta_d x_d = 0 \end{align*} For two dimensional data $$\boldsymbol{x}=(x_1, x_2)$$ we have \begin{align*} & \theta_0 + \theta_1 x_1+\theta_2 x_2 = 0 \\ & \Rightarrow x_2 = -\frac{\theta_1}{\theta_2} x_1 - \frac{\theta_0}{\theta_2} \end{align*} which is the separation line that should be drawn in $$(x_1, x_2)$$ plane.

### Weighted decision boundary

If we want to weight the positive class ($$y = 1$$) more or less using $$w$$, here is the general decision boundary: $$w{\Bbb P}(y=1|\boldsymbol{x}) = {\Bbb P}(y=0|\boldsymbol{x}) = \frac{w}{w+1}$$

For example, $$w=2$$ means point $$\boldsymbol{x}$$ will be assigned to positive class if $${\Bbb P}(y=1|\boldsymbol{x}) > 0.33$$ (or equivalently if $${\Bbb P}(y=0|\boldsymbol{x}) < 0.66$$), which implies favoring the positive class (increasing the true positive rate).

Here is the line for this general case: \begin{align*} &\frac{1}{1+e^{-\boldsymbol{\theta}^t\boldsymbol{x_+}}} = \frac{1}{w+1} \\ &\Rightarrow e^{-\boldsymbol{\theta}^t\boldsymbol{x_+}} = w\\ &\Rightarrow \boldsymbol{\theta}^t\boldsymbol{x_+} = -\text{ln}w\\ &\Rightarrow \theta_0 + \theta_1 x_1+\cdots+\theta_d x_d = -\text{ln}w \end{align*}

• Thanks for your insight, I understand why I need three thetas and the weighted decision boundary you explain. I am still having trouble implementing this in code however. ABove you say: "Assuming that input x=(x1, x2) ... Isn't this what I present in the code above? – Rrz0 Apr 20 at 8:53
• Gradient descent is implemented correctly, but I am having issue with mismatching matrix dimensions when it comes to the input X and classdata. Shouldn't x  be nx3 ? [x1, x2, 1] – Rrz0 Apr 20 at 9:06
• @Rrz0 that is the extended 3D version as I explained, which is actually X+ = [1, x, dat], you do not need this for drawing the line. Please check my last update regarding 'plot_x'. – Esmailian Apr 20 at 9:11
• Thanks for the update. I edited the question to show the cost function.. When computing the sigmoid we get X * theta. I believe this results in a dimension error if X is not [n x 3] – Rrz0 Apr 20 at 9:15
• @Rrz0 I spotted the problem. Check out the update. – Esmailian Apr 20 at 9:29

Your decision boundary is a surface in 3D as your points are in 2D.

With Wolfram Language

Create the data sets.

mqtrue = 5;
cqtrue = 30;
With[{x = Subdivide[0, 3, 50]},
dat1 = Transpose@{x, mqtrue x + 5 RandomReal[1, Length@x]};
];
With[{x = Subdivide[7, 10, 50]},
dat2 = Transpose@{x, mqtrue x + cqtrue + 5 RandomReal[1, Length@x]};
];


View in 2D (ListPlot) and the 3D (ListPointPlot3D) regression space.

ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"] I Append the response variable to the data.

datPlot =
ListPointPlot3D[
] Perform a Logistic regression (LogitModelFit). You could use GeneralizedLinearModelFit with ExponentialFamily set to "Binomial" as well.

With[{dat = Join[dat1, dat2]},
model =
LogitModelFit[
{x, y}, {x, y}]
] From the FittedModel "Properties" we need "Function".

model["Properties"]


{AdjustedLikelihoodRatioIndex, DevianceTableDeviances, ParameterConfidenceIntervalTableEntries, AIC, DevianceTableEntries, ParameterConfidenceRegion, AnscombeResiduals, DevianceTableResidualDegreesOfFreedom, ParameterErrors, BasisFunctions, DevianceTableResidualDeviances, ParameterPValues, BestFit, EfronPseudoRSquared, ParameterTable, BestFitParameters, EstimatedDispersion, ParameterTableEntries, BIC, FitResiduals, ParameterZStatistics, CookDistances, Function, PearsonChiSquare, CorrelationMatrix, HatDiagonal, PearsonResiduals, CovarianceMatrix, LikelihoodRatioIndex, PredictedResponse, CoxSnellPseudoRSquared, LikelihoodRatioStatistic, Properties, CraggUhlerPseudoRSquared, LikelihoodResiduals, ResidualDeviance, Data, LinearPredictor, ResidualDegreesOfFreedom, DesignMatrix, LogLikelihood, Response, DevianceResiduals, NullDeviance, StandardizedDevianceResiduals, Deviances, NullDegreesOfFreedom, StandardizedPearsonResiduals, DevianceTable, ParameterConfidenceIntervals, WorkingResiduals, DevianceTableDegreesOfFreedom, ParameterConfidenceIntervalTable}

model["Function"] Use this for prediction

model["Function"][8, 54]

0.0196842


and plot the decision boundary surface in 3D along with the data (datPlot) using Show and Plot3D

modelPlot =
Show[
datPlot,
Plot3D[
model["Function"][x, y],
Evaluate[
Sequence @@
MapThread[Prepend, {MinMax /@ Transpose@Join[dat1, dat2], {x, y}}]],
Mesh -> None,
PlotStyle -> Opacity[.25, Green],
PlotPoints -> 30
]
] With ParametricPlot3D and Manipulate you can examine decision boundary curves for values of the variables. For example, keeping x fixed and letting y vary or vice versa.

Manipulate[
Show[
modelPlot,
ParametricPlot3D[
{x, u, model["Function"][x, u]}, {u, 0, 80}, PlotStyle -> Orange],
ParametricPlot3D[
{u, y, model["Function"][u, y]}, {u, 0, 10}, PlotStyle -> Purple],
PlotLabel ->
StringTemplate["model[1, 2] = 3"] @@ {x, y, model["Function"][x, y]}
],
{{x, 6, Style["x", Orange, Bold]}, 0, 10, Appearance -> "Labeled"},
{{y, 40, Style["y", Purple, Bold]}, 0, 80, Appearance -> "Labeled"}
] ## Update

You can also plot contours of the probability in 2D.

plot = ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"];

Manipulate[
db = y /. First@Quiet@Solve[model["Function"][x, y] == p, y];
Show[
plot,
Plot[db, {x, 0, 10}, PlotStyle -> Red]
],
{{p, .5}, 0, 1, Appearance -> "Labeled"}
] Hope this helps.

• Beautiful plots. Some important notes: Logistic regression is used by OP for "classification" in 2D space, therefore "decision boundary" should be drawn in the same dimension $d$ as feature space (2D here) and it is a straight 2D line (unlike the last plot), which is also not the same as those animated lines (it must be parallel to that waterfall). However, "output of logistic regression", i.e. $(\boldsymbol{x},P(y=1|\boldsymbol{x}))$, as you have beautifully illustrated, needs $d+1$ for visualization. – Esmailian Apr 19 at 19:38
• @Esmailian See update. – Edmund Apr 19 at 22:56
• Thank you for your very helpful insight and excellent visualizations. – Rrz0 Apr 20 at 8:54