# Decision Boundary Single feature

Is there any decision boundary when there is one feature in case of logistic regression?
a) Is the line passing through the threshold point, or y = 0.5, a decision boundary?

When there are multiple features, then the decision boundary concept is clear to me, as Prof. Andrew taught in the lecture. For example, if x1+x2>=3 is obtained, we will take x1 as the x axis and x2 as the y axis, then just draw the equation line, which is the decision boundary.
But I am really confused: when there is a single feature, what is the decision boundary? for example:
g(z)>=0.5
z>=0
wx+b>=0
suppose I get by placing w and b values: x>=-2,
b) So will this equation be the decision boundary in the plot g(z) vs. x~single input?

Yes, though it would be more of a decision line, rather than a boundary. This is because with only one one feature, you can only make a 2D plot where βxβ is on the horizontal axis, and the y and prediction values are on the vertical axis.

With y and the predictions on the vertical axis, the boundary would be a horizontal line at 0.5.

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Hi, I edited the post by adding options a and b.
So a and b both are possible? or only a or only b?

This will be easier if I draw up a sketch. It may take some time.

I may do it.

I think both are possible, that you said in 1st comment.
Thanks !
IF I STILL doing wrong then correct me.

When x = 0, g(z) = 0.5738, but when x = -0.3, g(z) should be 0.5, so it would be better if you redraw B such that the steepest slope is at x = -0.3.

You see why g(z) = 0.5 always has the steepest slope?

In your A, you have drawn the boundary horizontally, but drawing it vertically is also fine, and the same applies to B too.

If you discuss the boundary with respect to x, that will be a vertical line in your B; if it is with respect to z, that will be a vertical line in your A; if with respect to g(z), it will be a horizontal line in either A or B.

All boundaries above are lines because we are drawing them on a 2D plane. If we have only the feature axis (without g(z)), then it is going to be a boundary point. So, one feature β boundary point; two features β boundary line; three features β boundary plane.

Your graphs are inconsistent, because the graph for A has Z on the horizontal axis, and your graph for B has X on the horizontal axis.

The challenge with visualizing this boundary is that even though there is only one feature x, there are still two learned parameters: w and b.

So if you plot x on the horizontal axis, and y on the vertical axis, you have to assume some constant values for w and b. But the exact w and b value will shift the result of the sigmoid (vertically). And unless you have completed training, w and b are both variables.

So this really is a complex system to plot in general.

This is very difficult to visualize, because you either have a complicated boundary plane in 3D, or you have to use markers for the βyβ values, but you cannot use markers for the g(z) values, because they are real numbers (not just labels).

Tomβs comment reminded me of this, although it looks totally fine to me you labeled your y axis in B as g(z), another way is to label it as g(wx+b) so that we can see that it is a function of x.

However, as I said, you might want to shift your sigmoid curve on B a bit so that the steepest slope is at x = -0.3.

Yes, I did that, actually. g(z)= f(x)