For a statistical learning problem (classification), I have the data set $\{ (x_i,y_i) \}_{i=1}^n$ with $x_i \in \mathbb{R}^2$ being the input data and $y_i \in \{0,1\}$ the possible classes.

The data is used to compute the log-likelihood for the data, in that equation I have to compute the logistic sigmoid function

$$\sigma(x_i) = \frac{1}{e^{-x_i} + 1}$$

My problem is:

The input data of $x$ is a matrix $X \in \mathbb{R}^{n \times 2}$, now I am confused how I can compute the $\sigma(x_i)$ for a certain value, since one value of the matrix is a tuple, respectively a vector, one row of this matrix.

Any hints on how to approach this problem and compute my $\sigma(x_i)$?

My matrix looks like that:

$$\begin{pmatrix} 1.55545 & -1.00055\\ -1.24155 & 1.58778\\ 1.28068 & -1.0224\\ \vdots & \vdots\\ -1.68505 & 0.290898\\ 1.73686 & 0.793386\\ \end{pmatrix}$$

Hence $x_1 = (1.55545, -1.00055)$, but what is then:

$$\sigma(1.55545, -1.00055) = \frac{1}{e^{????} + 1}$$

The only thing I have found is the Vector exponential which claims that it can be computed by:

$$exp(v) = 1 \cosh(|v|) + \frac{v}{|v|} \sinh(|v|)$$

  • $\begingroup$ Are you sure the matrix structure of your input data is relevant to the problem? Maybe you can just interpret the matrix as an ordinary vector. $\endgroup$ Apr 29, 2012 at 10:15
  • $\begingroup$ @Raskolnikov I cannot imagine how and dropping one column of the matrix seems not the right way to do it. $\endgroup$
    – Mahoni
    Apr 30, 2012 at 6:26
  • $\begingroup$ An $n\times 2$-matrix is just a structured $2n$-vector. What's the problem? $\endgroup$ Apr 30, 2012 at 7:39
  • $\begingroup$ @Raskolnikov The problem is, for all the input values $x_i$ I will compute with $\sigma(x)$ a vector $p$ which is used in further computations and the dimension has to be $n$, otherwise there will be dimension mismatch. $\endgroup$
    – Mahoni
    Apr 30, 2012 at 8:57
  • 2
    $\begingroup$ I think you are forgetting an essential step in your logistic model. Which is that you first have to do a linear regression $y_i\sim x_i$ and it is then for the $y_i$ that you compute $\pi_i\sim\sigma(y_i)$. You are just confused about how logistic regression works. Take a look here before proceeding. $\endgroup$ Apr 30, 2012 at 9:08

1 Answer 1


The $x_i$ of the input data is not the input of the logistic sigmoid function $\sigma(x)$, that $x$ there is only an arbitrary chosen variable name. The actual input of the $\sigma$ function is a scalar function $f$.

In this case here the function is $f(y,x)$, more specific: $$f(1,x) = \phi(x)^T \cdot \beta$$

The $\phi(x)^T$ is just a transposed version of the input vector with additional features. The result is a scalar and therefore $$\frac{1}{e^{f(1,x)} + 1}$$ can be easily computed.


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