Exercises

We use the iris_reduced dataframe and add the column prediction to it.

df = iris_reduced
df.prediction = σ.(X*w) .>= 0.5

Now we show all misclassified samples.

sort(df[df.label .!= df.prediction, :], [:PetalLength, :PetalWidth])

A quick look at the image shows that the point $(4.8,1.8)$ is misclassified, but the image shows it correctly. Let us show all such points.

df[(df.PetalLength .== 4.8) .& (df.PetalWidth .== 1.8), :]

As we can see, there are three samples with the same data. Two of them have label 1 and one label 0. Since the incorrectly classified sample was redrawn, it was not possible to see it.

First, we run the logistic regression as before, only with a different starting point

log_reg(X, y, [1;2;3])

3-element Vector{Float64}:
 NaN
 NaN
 NaN

This resulted in NaNs. When something fails, it may be a good idea to run a step-by-step analysis. In this case, we will run the first iteration of Newton's method

julia> w = [1;2;3];
julia> X_mult = [row*row' for row in eachrow(X)];
julia> y_hat = 1 ./(1 .+exp.(-X*w))100-element Vector{Float64}:
 0.9999724643088853
 0.9999724643088853
 0.9999815421067044
 0.9999322758503804
 0.9999750846110607
 0.9999589221322353
 0.9999815421067044
 0.9997515449181605
 0.9999628310628971
 0.9999387202603833
 ⋮
 0.9999969488837513
 0.999993209641302
 0.999998629042793
 0.9999988775548947
 0.9999972392350497
 0.9999924954984029
 0.9999949695696981
 0.999997739675702
 0.9999917062496261
julia> grad = X'*(y_hat.-y) / size(X,1)3-element Vector{Float64}:
 2.129852089784904
 0.66295432173409
 0.49996260866776915
julia> hess = y_hat.*(1 .-y_hat).*X_mult |> mean3×3 Matrix{Float64}:
 0.000598417  0.00018514  0.000147894
 0.00018514   5.80682e-5  4.56733e-5
 0.000147894  4.56733e-5  3.73868e-5
julia> w -= hess \ grad3-element Vector{Float64}:
 -10764.590886852531
  -1801.6675865928162
  31420.070268736363

Starting from the bottom, we can see that even though we started with relatively small $w$, the next iteration is four degrees of magnitude larger. This happened because the Hessian hess is much smaller than the gradient grad. This indicates that there is some kind of numerical instability. The prediction y_hat should lie in the interval $[0,1]$ but it seems that it is almost always close to 1. Let us verify this by showing the extrema of y_hat

extrema(y_hat)

(0.9997254218438986, 0.9999994956525918)

They are indeed too large.

Now we explain the reason. We know that the prediction equals to

\[\hat y_i = \sigma(w^\top x_i),\]

where $\sigma$ is the sigmoid function. Since the mimimum from $w^\top x_i$

minimum(X*[1;2;3])

8.2

is large, all $w^\top x_i$ are large. But plotting the sigmoid funtion

xs = -10:0.01:10
plot(xs, σ, label="", ylabel="Sigmoid function")

it is clear that all $w^\top x_i$ hit the part of the sigmoid which is flat. This means that the derivative is almost zero, and the Hessian is "even smaller" zero. Then the ratio of the gradient and Hessian is huge.

The gradient descent will probably run into the same difficulty. Since the gradient will be too small, it will take a huge number of iterations to escape the flat region of the sigmoid. This is a known problem of the sigmoid function. It is also the reason why it was replaced in neural networks by other activation functions.

Since $\hat y$ equals the probability of predicting $1$, we have

\[\hat y = \frac{1}{1+e^{-w^\top x}}\]

Then the cross-entropy loss reduces to

\[\begin{aligned} \operatorname{loss}(y,\hat y) &= - y\log \hat y - (1-y)\log(1-\hat y) \\ &= y\log(1+e^{-w^\top x}) - (1-y)\log(e^{-w^\top x}) + (1-y)\log(1+e^{-w^\top x}) \\ &= \log(1+e^{-w^\top x}) + (1-y)w^\top x. \end{aligned}\]

Then it remains to sum this term over all samples.

We derived that the Hessian of the objective function for logistic regression is

\[\nabla^2 L(w) = \frac 1n \sum_{i=1}^n\hat y_i(1-\hat y_i)x_i x_i^\top.\]

For any vector $a$, we have

\[a^\top x_i x_i^\top a = (x_i^\top a)^\top (x_i^\top a) = \|x_i^\top a\|^2 \ge 0,\]

which implies that $x_i x_i^\top$ is a positive semidefinite matrix (it is known as rank-1 matrix as its rank is always 1 if $x_i$ is a non-zero vector). Since $y_i(1-\hat y_i)\ge 0$, it follows that $\nabla^2 L(w)$ is a positive semidefinite matrix. If a Hessian of a function is positive semidefinite everywhere, the function is immediately convex. Since Newton's method found a stationary point, this points is a global minimum.