To find q_{12} and q_{13} in terms of y=e^{-a^2} and \delta =e^{-(2a)^2}, we need to calculate the distances between the points \mathbf{y}^{(i)} and \mathbf{y}^{(j)}. From the given configuration, it seems that the distances are as follows:
d_{12} = a
d_{13} = 2a
d_{23} = a
Now we can apply the equation for q_{ij}:
q_{12} = \frac{\exp(-d_{12}^2)}{\sum_{k > l}\exp(-d_{lk}^2)} = \frac{\exp(-a^2)}{\exp(-d_{13}^2) + \exp(-d_{23}^2)} = \frac{\exp(-a^2)}{e^{-(2a)^2} + e^{-a^2}}
Since q_{12} = q_{13}, we have:
q_{13} = \frac{\exp(-a^2)}{e^{-(2a)^2} + e^{-a^2}}
For q_{23}, we can directly apply the equation:
q_{23} = \frac{\exp(-d_{23}^2)}{\sum_{k > l}\exp(-d_{lk}^2)} = \frac{\exp(-a^2)}{\exp(-d_{12}^2) + \exp(-d_{13}^2)} = \frac{\exp(-a^2)}{\exp(-a^2) + \exp(-(2a)^2)} = \frac{\exp(-a^2)}{1 + \delta}
Therefore, the distribution \mathbf{Q} in terms of y=e^{-a^2} and \delta =e^{-(2a)^2} is:
q_{12} = q_{13} = \frac{\exp(-a^2)}{e^{-(2a)^2} + e^{-a^2}}
q_{23} = \frac{\exp(-a^2)}{1 + \delta}
Define \mathbf{Q} using Gaussian on each point in the target 1-dimensional space (just as we define \mathbf{P} in the original high dimensional space):
\displaystyle q_{ij}=\frac{\exp (-d_{ij}^2)}{\sum _{k>l}\exp (-d_{lk}^2)}\qquad (i<j)
where d_{ij} is the distance between the points \mathbf{y}^{(i)} and \mathbf{y}^{(j)} in 1-dimension.
Let the embedding into 1 dimension of the 3 given points in 2 dimensions look like the configuration below, with unknown distance a between the image of central node \mathbf{y}^{(1)} and \mathbf{y}^{(2)}:
Find the distribution \mathbf{Q}. That is, find q_{12}=q_{13} and q_{23} in terms of y=e^{-a^2} and \delta =e^{-(2a)^2}.
q_{12}=q_{13}\quad
unanswered
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q_{23}=\quad
unanswered
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