# Why is the exponential loss used in this case?

I am reading a paper "Tracking-by-Segmentation With Online Gradient Boosting Decision Tree". In Section 2.1, the paper says I cannot understand the exponential loss function. In my opinion, the loss function should get the smallest value when $$y_i=f(x_i)$$. But the loss function in the image obtains a smaller value if $$(-y_i f(x_i))$$ becomes smaller.

• Hello, welcome to Artificial Intelligence Stack Exchange. If possible, please try to quote the data provided in text rather than image. Jul 25 at 11:32
• @hanugm Thanks for your suggestions! Jul 25 at 12:27

The loss is

$$\mathcal{L}=\sum_{i=1}^{N} \ell\left(y_{i}, f\left(\mathbf{x}_{i}\right)\right) \equiv \sum_{i=1}^{N} \exp \left(-y_{i} f\left(\mathbf{x}_{i}\right)\right),$$

which can also be written as follows

$$\mathcal{L} = \sum_{i=1}^{N} e^{-y_{i} f\left(\mathbf{x}_{i}\right)} \tag{1}\label{1}$$

The important thing to note here is the $$-$$ in the exponent, which allows us to write \ref{1} as follows (see this)

$$\mathcal{L} = \sum_{i=1}^{N} \frac{1}{e^{y_{i} f\left(\mathbf{x}_{i}\right)}} \tag{2}\label{2}$$

So, the loss becomes smaller the higher $$y_{i}$$ and $$f\left(\mathbf{x}_{i}\right)$$ are. If, for example, $$y_{i}$$ is negative and $$f\left(\mathbf{x}_{i}\right)$$ positive (or vice-versa), then \ref{2} would be higher. If both are positive or negative, then the loss will be smaller, as we will be summing fractions of the form $$\frac{1}{e^k} < 1$$ for some positive constant $$k$$: the higher the $$k$$, the smaller the loss.

So, if you use this loss, it seems that you want that

1. $$y_{i}$$ and $$f\left(\mathbf{x}_{i}\right)$$ have the same sign (either both negative or both positive)
2. $$f\left(\mathbf{x}_{i}\right)$$ is as high as possible (note that you cannot change $$y_{i}$$), which could even be much higher (in terms of magnitude) than $$y_{i}$$ (not sure why they would want this)

I don't know why they chose this loss function because I didn't read the paper (yet), but it seems to me that this is how you should interpret this loss function.