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→Data points: It is plainly incorrect to italicize digits in this context. |
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Using the fact that all ''K'' of the probabilities must sum to one, we find:
:<math>
:<math>\Pr(Y_i=K) \,=\, 1- \sum_{j=1}^{K-1} \Pr (Y_i = j) \,=\, 1 - \sum_{j=1}^{K-1}{\Pr(Y_i=K)}\;e^{\boldsymbol\beta_j \cdot \mathbf{X}_i} \;\;\Rightarrow\;\; \Pr(Y_i=K) \,=\, \frac{1}{1 + \sum_{j=1}^{K-1} e^{\boldsymbol\beta_j \cdot \mathbf{X}_i}}</math>.▼
\begin{align}
▲
= {} & \frac{1}{1 + \sum_{j=1}^{K-1} e^{\boldsymbol\beta_j \cdot \mathbf{X}_i}}.
\end{align}
</math>
We can use this to find the other probabilities:
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