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I know about multinoulli distribution, but I found a different explanation in a book that I have been reading and I didn't quite get it. It says:

The multinoulli, or categorical, distribution is a distribution over a single discrete variable with $k$ different states, where k is finite. The multinoulli distribution is parameterized by a vector $p ∈[0,1]^𝑘-1$, where $p_i$ gives the probability of the $i$-th state. The final, $k$-th state's probability is given by $1−(1^𝑇)\cdot 𝐩$. Note that we must constrain (1^𝑇)𝐩 ≤ 1.

I didn't understand how this represents the final state like this $1−(1^𝑇)\cdot 𝐩$ and anything about the vector.

If someone can provide detailed explanation it will be a real help.

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  • $\begingroup$ $1^T \mathbf{p}$ is just a "fancy" compact way to write $\sum_{i=1}^{k-1} \mathbf{p}_i$. (this is the dot product of the all-ones vector with the vector $\mathbf{p}$) $\endgroup$ – Clement C. Feb 11 at 6:47
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here is a clarifying example using $k = 3$. Suppose that your experiment consists of drawing one ball. There are balls with three diverent colors: blue (b), green (g) and red (r). This corresponds to a multinoulli distribution with $k = 3$. We just have to specify the probabilities for the different outcomes. So we set $$ p_b = \Bbb P (\text{Ball is blue}) = \frac 16 \quad \text{and} \quad p_g = \Bbb P (\text{Ball is green}) = \frac 13. $$
These two probabilities are enough to determine $p_r = \Bbb P (\text{Ball is red})$: $$ p_r = 1 - (p_g + p_b) = \frac 12. $$ Or in more fancy notation if we set $\mathbf{p} = (p_b, p_g)$ and $1^T = (1, 1)^T$ then $$ p_r = 1 - 1^T \cdot \mathbf{p}. $$

Personally, I would not use the last notation.

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