Question on word probability for hierarchical softmax used in natural language processing
I have figured out my confusion.

The indicator function is $\lbrack x \rbrack$ is set to output a $-1$ or a $1$ based on the argument being false or true, respectively. 

When the authors say "let $ch(n)$ be an arbitrary fixed child of node $n$", they mean arbitrary in the sense of left child or right child. But in order for our function to work, we must decide on this arbitrary decision. So, let us assume $ch(n)$ ouputs the left child of unit $n$.

Then as we go down our tree, if $\lbrack n(w,j+1) = ch(n(w,j))\rbrack = 1 $ it means the next node on our path is to the left of the parent node $n(w,j)$ and if $\lbrack n(w,j+1) = ch(n(w,j))\rbrack = -1$ it means that the next node is to the right of $n(w,j)$.

As a result, since we defined $ch(n)$ to indicate the left child, all left probabilities are calculated as $\sigma(v_{n(w,j)}'^T\cdot v_{w_I})$ since $\lbrack n(w,j+1) = ch(n(w,j))\rbrack = 1 $. Now we know that the probability on the right must then be   $1 -\sigma(v_{n(w,j)}'^T\cdot v_{w_I})$. This is where I was confused since the function outputs $\sigma(-v_{n(w,j)}'^T\cdot v_{w_I})$. But let $a=v_{n(w,j)}'^T\cdot v_{w_I}$ and we can verify that $\sigma(-a)=1-\sigma(a)$.

We have $\sigma(a) = \frac{1}{1+e^{-a}}$, now observe that

$\begin{align*}
1 - \sigma(a) &amp;= \frac{1+e^{-a}}{1+e^{-a}}-\frac{1}{1+e^{-a}}\\
&amp;=\frac{e^{-a}}{1+e^{-a}}\\
&amp;=\frac{1}{e^a}\times \frac{1}{1+e^{-a}}\\
&amp;=\frac{1}{1+e^a}\\
&amp;=\frac{1}{1+e^{-(-a)}}\\
&amp;=\sigma(-a).
\end{align*}$

Hence the function $$p(w\vert w_I) =\prod_{j=1}^{L(w)-1}\sigma(\lbrack n(w,j+1)=ch(n(w,j))\rbrack \cdot v_{n(w,j)}'^T v_{w_I})$$

simply multiplies probabilities along the branches corresponding to the path from the root node to the proper leaf node $w$.