I am trying to write out the following log-likelihood: $$\mathcal L(\vec{x}, \vec{y}) = \sum_{i} \left[ k_{i}^{out} (\boldsymbol{A}^*) \ln x_i + k_i^{in} (\boldsymbol{A}^*) \ln y_i\right] - \sum_{i \neq j} \ln(1 + x_i y_i). \tag{1}$$

I read it as, in the first term, the sum over all $i$, which is logical. But then, in the second sum, it says, a sum over all $i$ but not $j$. Which in principle is fine, but how does one determine $j$ in that instance? It's also not that the second sum, should be included in the first sum, because both sums have a subscript $i$ Is this a mistake in the publication, or is this still solvable?

In case the second sum should be a double sum, is the equation below, also a double sum? (this is from the exact same paper, but I don't think a double sum would be logical here)

\begin{align} & \sum_{j \neq i} \dfrac{x_i^* y_j^*}{1 + x_I^* y_j^*} = k_i^{out}(\boldsymbol{A}^*); \qquad \forall i. \tag{2} \\ & \sum_{j \neq i} \dfrac{x_j^* y_i^*}{1 + x_j^* y_i^*} = k_i^{in}(\boldsymbol{A}^*); \qquad \forall i. \tag{3} \end{align}

  • 1
    $\begingroup$ Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures. $\endgroup$ – José Carlos Santos Jun 12 '18 at 9:46
  • $\begingroup$ Thank you José. I was on stack overflow earlier and this wasn't possible! I will change it! $\endgroup$ – Thomas Mc Donald Jun 12 '18 at 10:45

The notation means sum over all pairs $(i,j)$ such that $i\neq j$. What exactly "all pairs" means should follow from the context.

  • $\begingroup$ I am sorry I included the wrong picture! I have edited the new version $\endgroup$ – Thomas Mc Donald Jun 12 '18 at 10:20
  • $\begingroup$ now you see why it is so confusion right? $\endgroup$ – Thomas Mc Donald Jun 12 '18 at 10:22
  • $\begingroup$ There is no confusion. The first sum is over $i$ and the second one is over all pairs as I described above. $\endgroup$ – Michal Adamaszek Jun 12 '18 at 10:24
  • $\begingroup$ So is this a short notation of a double sum? $\endgroup$ – Thomas Mc Donald Jun 12 '18 at 10:24
  • $\begingroup$ Yes it is a double sum. $\endgroup$ – Michal Adamaszek Jun 12 '18 at 10:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.