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In section 3.3 of the paper, they state that they use the cross entropy. Then they define the probability for a label to be a false positive as $\theta_0$ and a false negative as $\theta_1$.

They use it to somehow modify the loss function but never actually state what this new loss is.

I was expecting something like

$f_\theta(\hat{m_i}, \tilde{m_i}) = \tilde{m_i} *g_{\theta_0}(\hat{m_i}) + (1-\tilde{m_i})*g_{\theta_1}(1-\hat{m_i})$

but $g$ is not given.

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Section 3.3 simply gives the equation for the negative log-likelihood. They say that it takes the form of cross-entropy (because it just looks like a cross-entropy equation, perhaps?), but mathematically it seems to come from the fact that they define the model in equation 2 to follow a Bernoulli distribution, which can be 0 or 1 with probabilities p or q respectively:

$$ \Pr(X=1)=p=1-\Pr(X=0)=1-q $$

The probability mass function of the Bernoulli looks like this:

$$ f(x;p)=px+(1-p)(1-x)\!\quad {\text{for }}x\in \{0,1\} $$

and the maximum likelihood for the Bernoulli looks like this:

$$ L(p) = \prod_{i+1}^n p^{x_1} (1 - p)^{1 - x_1} $$

Taking the log (to get the log-likelihood that they mention), gives something of the form we see in equation 3:

$$ \log{L(p)} = \log{p}\sum_{i=1}^n x_i + \log{(1-p)}\sum_{i=1}^n (1-x_i) $$

Then you can takes the sum out the front as it is over the image patches:

$$ \sum_{i}^{w^2_m} (\tilde{m}_i ln\hat{m}_i + (1-\tilde{m}_i)ln(1- \hat{m}_i)) $$ So here This sums over all the pixel-wise output units in each image patch (map: m), hence why the summation goes to $w_m^2$; the width of the image patch squared. $\tilde{m}$ is a realisation (possible outcome) of $\hat{m}$, just as $x$ is a realisation of distribution $p$ in the above equations.

This looks like the binary cross-entropy loss function, which is why I think they say:

For the model given in Equation 2 the negative log likelihood takes the form of a cross entropy between the patch $\tilde{m}$ derived from the given map and the predicted patch $\hat{m}$

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  • $\begingroup$ Thank you for your answer. Yes this is the "base" loss function they use. But in section 4, they introduce the thetas parameters, chances of false positive and false negative in the labels themselves. They use those theta to modify the loss function. We can see in their graph (Figure 2), that when m̃ is 0, if theta_1 is 0 then the loss is -m̂ and when theta_1 is 0.05, the loss is whatever the equation of the green curve is. So I'm expecting a loss function parametrized by theta, isn't it? $\endgroup$ – Borbag Jul 22 '18 at 6:43
  • $\begingroup$ @Borbag - I'm not sure that you would expect functions of the introduced $\theta$-values ($g_{\theta_{0}}$, as you have it in your post), as they are fixed model parameters. I think it might be the case, that they two fixed $\theta$-values are used as a mask on all predictions, to map these two specific cases of FP or FN to their chosen heuristic values. In this case, the loss function itself would not be altered, just am extra step before the summing over all pixels. I agree that more information here from the authors would have been helpful! $\endgroup$ – n1k31t4 Jul 22 '18 at 11:00
  • $\begingroup$ If this is true, then what is represented in Figure 2? What is the equation of the curve? I really think this is the equation of my function g, when theta_1 = 0.05. $\endgroup$ – Borbag Jul 22 '18 at 16:55
  • $\begingroup$ thank you for your help, a colleague of mine found the answer $\endgroup$ – Borbag Jul 23 '18 at 12:16
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$m_i$ is the true unobserved label for pixel $i$ (0 for background, 1 for building/road/whatever the model is segmenting)
$\tilde{m}_i$ is the observed label for pixel $i$
$\hat{m}_i$ is the prediction for pixel $i$.

$\theta_0$ and $\theta_1$ are the probability of false positives and false negatives in the labels.

\begin{equation} \label{thetas} \begin{split} \theta_0 &= p(\tilde{m}_i = 1 | m_i = 0) \\ \theta_1 &= p(\tilde{m}_i = 0 | m_i = 1) \end{split} \end{equation}

We don't try to minimize the difference between label and prediction anymore ($\epsilon = \tilde{m}_i - \hat{m}_i$) but the difference between the probability that the true unobserved label is 1 and the prediction (for an input $s$: $\epsilon = p(m_i = 1 | \tilde{m_i}, s) - \hat{m}_i$).

Bayes law gives us:

\begin{equation} p(m_i = 1 | \tilde{m}_i) - \hat{m}_i = \frac{p(\tilde{m}_i | m_i = 1) * p(m_i=1)}{p(\tilde{m}_i)} -\hat{m}_i \end{equation}

and since $m_i$ can only be $0$ or $1$,

\begin{equation} p(\tilde{m}_i) = p(\tilde{m}_i | m_i=1) * p(m_i=1) + p(\tilde{m}_i | m_i=0) * p(m_i=0) \end{equation}

The definitions of $\theta_0$ and $\theta_1$ give us using the Bernoulli law those two distributions:

\begin{equation} \begin{split} p(\tilde{m}_i | m_i=0) & = \theta_0^{\tilde{m}_i} * (1-\theta_0)^{(1-\tilde{m}_i)} \\ p(\tilde{m}_i | m_i=1) & = \theta_1^{(1-\tilde{m}_i)} * (1-\theta_1)^{\tilde{m}_i} \end{split} \end{equation}

since $p(m_i =1) = 1 - p(m_i = 0)$ and $ p(m_i=1 ) = \hat{m}_i$ we get if $\tilde{m}_i = 0$

\begin{equation} \epsilon = \frac{\theta_1 * \hat{m}_i}{\theta_1 * \hat{m}_i + (1- \theta_0) * (1-\hat{m}_i))} - \hat{m}_i \end{equation}

and if $\tilde{m}_i = 1$

\begin{equation} \epsilon = \frac{(1-\theta_1) * \hat{m}_i}{(1-\theta_1) * \hat{m}_i + \theta_0 * (1-\hat{m}_i)} -\hat{m}_i \end{equation}

that is what is plotted in Figure 2.

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