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
[Math Processing Error]
q_{23}=\quad
unanswered
{::}
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}
To find the distribution \mathbf{Q}, we need to calculate q_{12}=q_{13} and q_{23} in terms of y=e^{-a^2} and \delta =e^{-(2a)^2}.
First, let's calculate q_{12}=q_{13}:
From the given equation, we have:
q_{12} = \frac{\exp (-d_{12}^2)}{\sum _{k>l}\exp (-d_{kl}^2)}
Since d_{12} is the distance between \mathbf{y}^{(1)} and \mathbf{y}^{(2)} in 1 dimension, we can write:
d_{12} = |y^{(1)} - y^{(2)}|
Substituting this into the equation for q_{12}, we get:
q_{12} = \frac{\exp (-(y^{(1)} - y^{(2)})^2)}{\sum _{k>l}\exp (-(y^{(k)} - y^{(l)})^2)}
Similarly, for q_{13}, we have:
d_{13} = |y^{(1)} - y^{(3)}|
q_{13} = \frac{\exp (-(y^{(1)} - y^{(3)})^2)}{\sum _{k>l}\exp (-(y^{(k)} - y^{(l)})^2)}
Now, let's calculate q_{23}:
d_{23} = |y^{(2)} - y^{(3)}|
q_{23} = \frac{\exp (-(y^{(2)} - y^{(3)})^2)}{\sum _{k>l}\exp (-(y^{(k)} - y^{(l)})^2)}
So, the distribution \mathbf{Q} can be written as:
q_{12} = q_{13} = \frac{\exp (-(y^{(1)} - y^{(2)})^2)}{\sum _{k>l}\exp (-(y^{(k)} - y^{(l)})^2)}
q_{23} = \frac{\exp (-(y^{(2)} - y^{(3)})^2)}{\sum _{k>l}\exp (-(y^{(k)} - y^{(l)})^2)}
where y=e^{-a^2} and \delta =e^{-(2a)^2}.
To find the distribution \textbf{Q}, we need to calculate the values of q_{12}, q_{13}, and q_{23} in terms of y=e^{-a^2} and \delta =e^{-(2a)^2}, where a is the unknown distance between the image of the central node \textbf{y}^{(1)} and \textbf{y}^{(2)}.
Let's start by calculating q_{12}:
The Gaussian function for calculating q_{ij} is given by:
q_{ij}=\frac{\exp (-d_{ij}^2)}{\sum _{k>l}\exp (-d_{lk}^2)}
In this case, we have i = 1 and j = 2. The distance between the points \textbf{y}^{(1)} and \textbf{y}^{(2)} in 1 dimension is d_{12} = a.
Substituting these values into the Gaussian function, we get:
q_{12}=\frac{\exp (-a^2)}{\sum _{k>l}\exp (-d_{lk}^2)}
Since we have only three points, the sum in the denominator contains only one term, which is \exp (-d_{23}^2). The distance between the points \textbf{y}^{(2)} and \textbf{y}^{(3)} in 1 dimension is d_{23} = 2a.
Substituting \delta =e^{-(2a)^2} into \exp (-d_{23}^2), we get:
\exp (-d_{23}^2) = \exp (-(2a)^2) = \delta
Therefore, the distribution q_{12} can be written as:
q_{12} = \frac{\exp (-a^2)}{\delta}
Next, let's calculate q_{13}:
In this case, we have i = 1 and j = 3. The distance between the points \textbf{y}^{(1)} and \textbf{y}^{(3)} in 1 dimension is d_{13} = 2a.
Substituting these values into the Gaussian function, we get:
q_{13}=\frac{\exp (-d_{13}^2)}{\sum _{k>l}\exp (-d_{lk}^2)}
Again, the sum in the denominator contains only one term, which is \exp (-d_{23}^2) = \delta.
Substituting \delta =e^{-(2a)^2} into \exp (-d_{13}^2), we get:
\exp (-d_{13}^2) = \exp (-(2a)^2) = \delta
Therefore, the distribution q_{13} can be written as:
q_{13} = \frac{\exp (-2a^2)}{\delta}
Finally, let's calculate q_{23}:
In this case, we have i = 2 and j = 3. The distance between the points \textbf{y}^{(2)} and \textbf{y}^{(3)} in 1 dimension is d_{23} = 2a.
Substituting these values into the Gaussian function, we get:
q_{23}=\frac{\exp (-d_{23}^2)}{\sum _{k>l}\exp (-d_{lk}^2)} = \frac{\exp (-(2a)^2)}{\delta}
Therefore, the distribution q_{23} can be written as:
q_{23} = \frac{\exp (-(2a)^2)}{\delta}
So, in terms of y=e^{-a^2} and \delta =e^{-(2a)^2}, we have:
q_{12} = \frac{y}{\delta}
q_{13} = \frac{y^2}{\delta}
q_{23} = \frac{\delta}{\delta} = 1
These are the distributions \textbf{Q} in terms of y and \delta.