# Confusing summation of $i$ not $j$, but $j$ is not defined

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}

• Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures. – José Carlos Santos Jun 12 '18 at 9:46
• Thank you José. I was on stack overflow earlier and this wasn't possible! I will change it! – 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.
• There is no confusion. The first sum is over $i$ and the second one is over all pairs as I described above. – Michal Adamaszek Jun 12 '18 at 10:24