Notation and major confusion with the definition of the probability simplex












0












$begingroup$


So it starts like this:



Given a discrete set $N$, the probability simplex over $N$, denoted $∆(N)$ is defined to be:



$$Δ(N) = left{ x in mathbb{R}^{|N|} ;:; x_{i} geq 0 text{ for all } i, text{ and } sum_{i=1}^{|N|} x_i = 1right}$$



I guess I kind of understand what it tries to convey, but in practice something doesn't add up.



For example, suppose, that we have $x= {x_1,x_2}$ with dimentionality $|N| = 3$ according to the specification above.



Say:



that $x_1 = {mathbb{R}_1, mathbb{R}_2, mathbb{R}_3}$
and
$x_2 = {mathbb{R}_4, mathbb{R}_5, mathbb{R}_6}$.



How does this summation $sum_{i=1}^{|N|} x_i = 1$ work?






linear-algebra probability probability-theory probability-distributions summation







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    So it starts like this:



    Given a discrete set $N$, the probability simplex over $N$, denoted $∆(N)$ is defined to be:



    $$Δ(N) = left{ x in mathbb{R}^{|N|} ;:; x_{i} geq 0 text{ for all } i, text{ and } sum_{i=1}^{|N|} x_i = 1right}$$



    I guess I kind of understand what it tries to convey, but in practice something doesn't add up.



    For example, suppose, that we have $x= {x_1,x_2}$ with dimentionality $|N| = 3$ according to the specification above.



    Say:



    that $x_1 = {mathbb{R}_1, mathbb{R}_2, mathbb{R}_3}$
    and
    $x_2 = {mathbb{R}_4, mathbb{R}_5, mathbb{R}_6}$.



    How does this summation $sum_{i=1}^{|N|} x_i = 1$ work?






    linear-algebra probability probability-theory probability-distributions summation







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      So it starts like this:



      Given a discrete set $N$, the probability simplex over $N$, denoted $∆(N)$ is defined to be:



      $$Δ(N) = left{ x in mathbb{R}^{|N|} ;:; x_{i} geq 0 text{ for all } i, text{ and } sum_{i=1}^{|N|} x_i = 1right}$$



      I guess I kind of understand what it tries to convey, but in practice something doesn't add up.



      For example, suppose, that we have $x= {x_1,x_2}$ with dimentionality $|N| = 3$ according to the specification above.



      Say:



      that $x_1 = {mathbb{R}_1, mathbb{R}_2, mathbb{R}_3}$
      and
      $x_2 = {mathbb{R}_4, mathbb{R}_5, mathbb{R}_6}$.



      How does this summation $sum_{i=1}^{|N|} x_i = 1$ work?






      linear-algebra probability probability-theory probability-distributions summation







      share|cite|improve this question











      $endgroup$




      So it starts like this:



      Given a discrete set $N$, the probability simplex over $N$, denoted $∆(N)$ is defined to be:



      $$Δ(N) = left{ x in mathbb{R}^{|N|} ;:; x_{i} geq 0 text{ for all } i, text{ and } sum_{i=1}^{|N|} x_i = 1right}$$



      I guess I kind of understand what it tries to convey, but in practice something doesn't add up.



      For example, suppose, that we have $x= {x_1,x_2}$ with dimentionality $|N| = 3$ according to the specification above.



      Say:



      that $x_1 = {mathbb{R}_1, mathbb{R}_2, mathbb{R}_3}$
      and
      $x_2 = {mathbb{R}_4, mathbb{R}_5, mathbb{R}_6}$.



      How does this summation $sum_{i=1}^{|N|} x_i = 1$ work?






      linear-algebra probability probability-theory probability-distributions summation




      linear-algebra probability probability-theory probability-distributions summation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 20 at 14:33









      Blue

      48.5k870154




      48.5k870154










      asked Jan 20 at 14:18









      bdpsbdps

      82




      82






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          I think, this is a kind of question that should be answered via examples.



          For any set $N_2={a,b}$ with 2 elements the following vectors $x_1, x_2, x_3$ belong to $Delta(N_2)$:



          $$
          x_1 =
          begin{bmatrix}
          0.2 \ 0.8
          end{bmatrix}, quad
          x_2 =
          begin{bmatrix}
          0.4 \ 0.6
          end{bmatrix}, quad
          x_3 =
          begin{bmatrix}
          0 \ 1
          end{bmatrix}.
          $$
          For each of them we have $x_i in Delta(N_2)$. Of course there are many more such vectors.



          For any set $N_3={a,b,c}$ with 3 elements the following vectors $x_1, x_2, x_3$ belong to $Delta(N_3)$:



          $$
          x_1 =
          begin{bmatrix}
          0.2 \ 0.3 \ 0.5
          end{bmatrix}, quad
          x_2 =
          begin{bmatrix}
          0.4 \ 0.4 \ 0.2
          end{bmatrix}, quad
          x_3 =
          begin{bmatrix}
          0 \ 1 \ 0
          end{bmatrix}.
          $$
          For each of them we have $x_i in Delta(N_3)$. Again, there are many more elements of $Delta(N_3)$.



          Note that the set $N$ is actually of no matter for constructing $Delta(N)$, it is only its size $|N|$ that matters.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Oh thanks a lot for the input. Then, I guess that any probability distribution can be explained in terms of probability simplex, or say it could be the result of sampling from some $|N|-$dimensional/size simplex distribution.
            $endgroup$
            – bdps
            Jan 20 at 15:06





















          1












          $begingroup$

          Well, what is meant under $triangle(N)$ is the following: if your vector $x in mathbf{R}^{|N|}$ has the following properties:




          1. $x_i ge 0$ for all $i$


          2. $sum_{i=1}^{|N|} x_i = 1$



          Then, your vector is from simplex $triangle(N)$. The sum is taken over all coordinates of the vector.



          Clearly, if one of these conditions is violated then your vector doesn't belong to simplex $triangle(N)$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            So if I get it well, every sample $x_1,x_2...x_k$ is made up of a vector of $N$ elements/categories, whose real values sum to 1, then, each sample has its own probability distribution?
            $endgroup$
            – bdps
            Jan 20 at 14:49








          • 1




            $begingroup$
            I didn’t quite get your question. Sampling from distribution $triangle(N)$ means that you’ll get a vector from $mathbf{R}^{|N|}$ with two conditions given above.
            $endgroup$
            – pointguard0
            Jan 20 at 14:55











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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          I think, this is a kind of question that should be answered via examples.



          For any set $N_2={a,b}$ with 2 elements the following vectors $x_1, x_2, x_3$ belong to $Delta(N_2)$:



          $$
          x_1 =
          begin{bmatrix}
          0.2 \ 0.8
          end{bmatrix}, quad
          x_2 =
          begin{bmatrix}
          0.4 \ 0.6
          end{bmatrix}, quad
          x_3 =
          begin{bmatrix}
          0 \ 1
          end{bmatrix}.
          $$
          For each of them we have $x_i in Delta(N_2)$. Of course there are many more such vectors.



          For any set $N_3={a,b,c}$ with 3 elements the following vectors $x_1, x_2, x_3$ belong to $Delta(N_3)$:



          $$
          x_1 =
          begin{bmatrix}
          0.2 \ 0.3 \ 0.5
          end{bmatrix}, quad
          x_2 =
          begin{bmatrix}
          0.4 \ 0.4 \ 0.2
          end{bmatrix}, quad
          x_3 =
          begin{bmatrix}
          0 \ 1 \ 0
          end{bmatrix}.
          $$
          For each of them we have $x_i in Delta(N_3)$. Again, there are many more elements of $Delta(N_3)$.



          Note that the set $N$ is actually of no matter for constructing $Delta(N)$, it is only its size $|N|$ that matters.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Oh thanks a lot for the input. Then, I guess that any probability distribution can be explained in terms of probability simplex, or say it could be the result of sampling from some $|N|-$dimensional/size simplex distribution.
            $endgroup$
            – bdps
            Jan 20 at 15:06


















          1












          $begingroup$

          I think, this is a kind of question that should be answered via examples.



          For any set $N_2={a,b}$ with 2 elements the following vectors $x_1, x_2, x_3$ belong to $Delta(N_2)$:



          $$
          x_1 =
          begin{bmatrix}
          0.2 \ 0.8
          end{bmatrix}, quad
          x_2 =
          begin{bmatrix}
          0.4 \ 0.6
          end{bmatrix}, quad
          x_3 =
          begin{bmatrix}
          0 \ 1
          end{bmatrix}.
          $$
          For each of them we have $x_i in Delta(N_2)$. Of course there are many more such vectors.



          For any set $N_3={a,b,c}$ with 3 elements the following vectors $x_1, x_2, x_3$ belong to $Delta(N_3)$:



          $$
          x_1 =
          begin{bmatrix}
          0.2 \ 0.3 \ 0.5
          end{bmatrix}, quad
          x_2 =
          begin{bmatrix}
          0.4 \ 0.4 \ 0.2
          end{bmatrix}, quad
          x_3 =
          begin{bmatrix}
          0 \ 1 \ 0
          end{bmatrix}.
          $$
          For each of them we have $x_i in Delta(N_3)$. Again, there are many more elements of $Delta(N_3)$.



          Note that the set $N$ is actually of no matter for constructing $Delta(N)$, it is only its size $|N|$ that matters.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Oh thanks a lot for the input. Then, I guess that any probability distribution can be explained in terms of probability simplex, or say it could be the result of sampling from some $|N|-$dimensional/size simplex distribution.
            $endgroup$
            – bdps
            Jan 20 at 15:06
















          1












          1








          1





          $begingroup$

          I think, this is a kind of question that should be answered via examples.



          For any set $N_2={a,b}$ with 2 elements the following vectors $x_1, x_2, x_3$ belong to $Delta(N_2)$:



          $$
          x_1 =
          begin{bmatrix}
          0.2 \ 0.8
          end{bmatrix}, quad
          x_2 =
          begin{bmatrix}
          0.4 \ 0.6
          end{bmatrix}, quad
          x_3 =
          begin{bmatrix}
          0 \ 1
          end{bmatrix}.
          $$
          For each of them we have $x_i in Delta(N_2)$. Of course there are many more such vectors.



          For any set $N_3={a,b,c}$ with 3 elements the following vectors $x_1, x_2, x_3$ belong to $Delta(N_3)$:



          $$
          x_1 =
          begin{bmatrix}
          0.2 \ 0.3 \ 0.5
          end{bmatrix}, quad
          x_2 =
          begin{bmatrix}
          0.4 \ 0.4 \ 0.2
          end{bmatrix}, quad
          x_3 =
          begin{bmatrix}
          0 \ 1 \ 0
          end{bmatrix}.
          $$
          For each of them we have $x_i in Delta(N_3)$. Again, there are many more elements of $Delta(N_3)$.



          Note that the set $N$ is actually of no matter for constructing $Delta(N)$, it is only its size $|N|$ that matters.






          share|cite|improve this answer









          $endgroup$



          I think, this is a kind of question that should be answered via examples.



          For any set $N_2={a,b}$ with 2 elements the following vectors $x_1, x_2, x_3$ belong to $Delta(N_2)$:



          $$
          x_1 =
          begin{bmatrix}
          0.2 \ 0.8
          end{bmatrix}, quad
          x_2 =
          begin{bmatrix}
          0.4 \ 0.6
          end{bmatrix}, quad
          x_3 =
          begin{bmatrix}
          0 \ 1
          end{bmatrix}.
          $$
          For each of them we have $x_i in Delta(N_2)$. Of course there are many more such vectors.



          For any set $N_3={a,b,c}$ with 3 elements the following vectors $x_1, x_2, x_3$ belong to $Delta(N_3)$:



          $$
          x_1 =
          begin{bmatrix}
          0.2 \ 0.3 \ 0.5
          end{bmatrix}, quad
          x_2 =
          begin{bmatrix}
          0.4 \ 0.4 \ 0.2
          end{bmatrix}, quad
          x_3 =
          begin{bmatrix}
          0 \ 1 \ 0
          end{bmatrix}.
          $$
          For each of them we have $x_i in Delta(N_3)$. Again, there are many more elements of $Delta(N_3)$.



          Note that the set $N$ is actually of no matter for constructing $Delta(N)$, it is only its size $|N|$ that matters.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 20 at 14:51









          ZeeklessZeekless

          577111




          577111








          • 1




            $begingroup$
            Oh thanks a lot for the input. Then, I guess that any probability distribution can be explained in terms of probability simplex, or say it could be the result of sampling from some $|N|-$dimensional/size simplex distribution.
            $endgroup$
            – bdps
            Jan 20 at 15:06
















          • 1




            $begingroup$
            Oh thanks a lot for the input. Then, I guess that any probability distribution can be explained in terms of probability simplex, or say it could be the result of sampling from some $|N|-$dimensional/size simplex distribution.
            $endgroup$
            – bdps
            Jan 20 at 15:06










          1




          1




          $begingroup$
          Oh thanks a lot for the input. Then, I guess that any probability distribution can be explained in terms of probability simplex, or say it could be the result of sampling from some $|N|-$dimensional/size simplex distribution.
          $endgroup$
          – bdps
          Jan 20 at 15:06






          $begingroup$
          Oh thanks a lot for the input. Then, I guess that any probability distribution can be explained in terms of probability simplex, or say it could be the result of sampling from some $|N|-$dimensional/size simplex distribution.
          $endgroup$
          – bdps
          Jan 20 at 15:06













          1












          $begingroup$

          Well, what is meant under $triangle(N)$ is the following: if your vector $x in mathbf{R}^{|N|}$ has the following properties:




          1. $x_i ge 0$ for all $i$


          2. $sum_{i=1}^{|N|} x_i = 1$



          Then, your vector is from simplex $triangle(N)$. The sum is taken over all coordinates of the vector.



          Clearly, if one of these conditions is violated then your vector doesn't belong to simplex $triangle(N)$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            So if I get it well, every sample $x_1,x_2...x_k$ is made up of a vector of $N$ elements/categories, whose real values sum to 1, then, each sample has its own probability distribution?
            $endgroup$
            – bdps
            Jan 20 at 14:49








          • 1




            $begingroup$
            I didn’t quite get your question. Sampling from distribution $triangle(N)$ means that you’ll get a vector from $mathbf{R}^{|N|}$ with two conditions given above.
            $endgroup$
            – pointguard0
            Jan 20 at 14:55
















          1












          $begingroup$

          Well, what is meant under $triangle(N)$ is the following: if your vector $x in mathbf{R}^{|N|}$ has the following properties:




          1. $x_i ge 0$ for all $i$


          2. $sum_{i=1}^{|N|} x_i = 1$



          Then, your vector is from simplex $triangle(N)$. The sum is taken over all coordinates of the vector.



          Clearly, if one of these conditions is violated then your vector doesn't belong to simplex $triangle(N)$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            So if I get it well, every sample $x_1,x_2...x_k$ is made up of a vector of $N$ elements/categories, whose real values sum to 1, then, each sample has its own probability distribution?
            $endgroup$
            – bdps
            Jan 20 at 14:49








          • 1




            $begingroup$
            I didn’t quite get your question. Sampling from distribution $triangle(N)$ means that you’ll get a vector from $mathbf{R}^{|N|}$ with two conditions given above.
            $endgroup$
            – pointguard0
            Jan 20 at 14:55














          1












          1








          1





          $begingroup$

          Well, what is meant under $triangle(N)$ is the following: if your vector $x in mathbf{R}^{|N|}$ has the following properties:




          1. $x_i ge 0$ for all $i$


          2. $sum_{i=1}^{|N|} x_i = 1$



          Then, your vector is from simplex $triangle(N)$. The sum is taken over all coordinates of the vector.



          Clearly, if one of these conditions is violated then your vector doesn't belong to simplex $triangle(N)$.






          share|cite|improve this answer









          $endgroup$



          Well, what is meant under $triangle(N)$ is the following: if your vector $x in mathbf{R}^{|N|}$ has the following properties:




          1. $x_i ge 0$ for all $i$


          2. $sum_{i=1}^{|N|} x_i = 1$



          Then, your vector is from simplex $triangle(N)$. The sum is taken over all coordinates of the vector.



          Clearly, if one of these conditions is violated then your vector doesn't belong to simplex $triangle(N)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 20 at 14:33









          pointguard0pointguard0

          1,4801021




          1,4801021








          • 1




            $begingroup$
            So if I get it well, every sample $x_1,x_2...x_k$ is made up of a vector of $N$ elements/categories, whose real values sum to 1, then, each sample has its own probability distribution?
            $endgroup$
            – bdps
            Jan 20 at 14:49








          • 1




            $begingroup$
            I didn’t quite get your question. Sampling from distribution $triangle(N)$ means that you’ll get a vector from $mathbf{R}^{|N|}$ with two conditions given above.
            $endgroup$
            – pointguard0
            Jan 20 at 14:55














          • 1




            $begingroup$
            So if I get it well, every sample $x_1,x_2...x_k$ is made up of a vector of $N$ elements/categories, whose real values sum to 1, then, each sample has its own probability distribution?
            $endgroup$
            – bdps
            Jan 20 at 14:49








          • 1




            $begingroup$
            I didn’t quite get your question. Sampling from distribution $triangle(N)$ means that you’ll get a vector from $mathbf{R}^{|N|}$ with two conditions given above.
            $endgroup$
            – pointguard0
            Jan 20 at 14:55








          1




          1




          $begingroup$
          So if I get it well, every sample $x_1,x_2...x_k$ is made up of a vector of $N$ elements/categories, whose real values sum to 1, then, each sample has its own probability distribution?
          $endgroup$
          – bdps
          Jan 20 at 14:49






          $begingroup$
          So if I get it well, every sample $x_1,x_2...x_k$ is made up of a vector of $N$ elements/categories, whose real values sum to 1, then, each sample has its own probability distribution?
          $endgroup$
          – bdps
          Jan 20 at 14:49






          1




          1




          $begingroup$
          I didn’t quite get your question. Sampling from distribution $triangle(N)$ means that you’ll get a vector from $mathbf{R}^{|N|}$ with two conditions given above.
          $endgroup$
          – pointguard0
          Jan 20 at 14:55




          $begingroup$
          I didn’t quite get your question. Sampling from distribution $triangle(N)$ means that you’ll get a vector from $mathbf{R}^{|N|}$ with two conditions given above.
          $endgroup$
          – pointguard0
          Jan 20 at 14:55


















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