Return to Answer

Assuming $p_r>0$, your problem has at most a solution, and it admits a solution provided the $p_i$'s are not too dispersed, in a sense to be made precise below.

Consider the polynomial $f(t):=\frac{1-(1-t)^n}{t}$. It defines a strictly decreasing homeo $[0,1]\rightarrow [1,n]$ (because it is the incremental ratio of a concave function), and your system writes $p_i f(t_i)=p_{i+1} f(t_{i+1}),$ for $i=1,\dots, r-1$. So it has at most one solution, given by $t_i=f^{-1}(\lambda/p_i)$, where $\lambda$ solves $$ p_1\le\lambda \le n p_r $$ $$\sum_{i=0}^rf^{-1}(\lambda/p_i)=1.$$ Since the LHS in the latter equation is strictly decreasing, a necessary and sufficient condition for existence is then expressed on the $p_i$'s, requiring
$$ p_1\le n p_r $$ and $$\sum_{i=0}^rf^{-1}(np_r /p_i)\le 1 .$$

Assuming $p_r>0$, your problem has at most a solution with all $t_i>0$, and it admits a solution provided the $p_i$'s are not too dispersed, in a sense to be made precise below.

Consider the polynomial $f(t):=\frac{1-(1-t)^n}{t}$. It defines a strictly decreasing homeo $[0,1]\rightarrow [1,n]$ (because it is the incremental ratio of a concave function), and your system writes $p_i f(t_i)=p_{i+1} f(t_{i+1}),$ for $i=1,\dots, r-1$. So it has at most one strictly positive solution, given by $t_i=f^{-1}(\lambda/p_i)$, where $\lambda$ solves $$ p_1\le\lambda \le n p_r $$ $$\sum_{i=0}^rf^{-1}(\lambda/p_i)=1.$$ Since the LHS in the latter equation is strictly decreasing, a sufficient condition for existence is then expressed on the $p_i$'s, requiring
$$ p_1\le n p_r $$ and $$\sum_{i=0}^rf^{-1}(np_r /p_i)\le 1 .$$

Assuming $p_1>0$, your problem has at most a solution, and it admits a solution provided the $p_i$'s are not too dispersed, in a sense to be made precise below.

Consider the polynomial $f(t):=\frac{1-(1-t)^n}{t}$. It defines a strictly decreasing homeo $[0,1]\rightarrow [1,n]$ (because it is the incremental ratio of a concave function), and your system writes $p_i f(t_i)=p_{i+1} f(t_{i+1}),$ for $i=1,\dots, r-1$. So it has at most one solution, given by $t_i=f^{-1}(\lambda/p_i)$, where $\lambda$ solves $$ p_1\le\lambda \le n p_r $$ $$\sum_{i=0}^rf^{-1}(\lambda/p_i)=1.$$ Since the LHS in the latter equation is strictly decreasing, a necessary and sufficient condition for existence is then expressed on the $p_i$'s, requiring
$$ p_1\le n p_r $$ and $$\sum_{i=0}^rf^{-1}(np_r /p_i)\le 1 .$$

Assuming $p_r>0$, your problem has at most a solution, and it admits a solution provided the $p_i$'s are not too dispersed, in a sense to be made precise below.

Consider the polynomial $f(t):=\frac{1-(1-t)^n}{t}$. It defines a strictly decreasing homeo $[0,1]\rightarrow [1,n]$ (because it is the incremental ratio of a concave function), and your system writes $p_i f(t_i)=p_{i+1} f(t_{i+1}),$ for $i=1,\dots, r-1$. So it has at most one solution, given by $t_i=f^{-1}(\lambda/p_i)$, where $\lambda$ solves $$ p_1\le\lambda \le n p_r $$ $$\sum_{i=0}^rf^{-1}(\lambda/p_i)=1.$$ Since the LHS in the latter equation is strictly decreasing, a necessary and sufficient condition for existence is then expressed on the $p_i$'s, requiring
$$ p_1\le n p_r $$ and $$\sum_{i=0}^rf^{-1}(np_r /p_i)\le 1 .$$

Assuming $p_1>0$, your problem has at most a solution, and it admits a solution provided the $p_i$'s are not too dispersed, in a sense to be made precise below.

Consider the polynomial $f(t):=\frac{1-(1-t)^n}{t}$. It defines a strictly decreasing homeo $[0,1]\rightarrow [1,n]$ (because it is the incremental ratio of a concave function), and your system writes $p_i f(t_i)=p_{i+1} f(t_{i+1}),$ for $i=1,\dots, r-1$. So it has at most one solution, given by $t_i=f^{-1}(\lambda/p_i)$, where $\lambda$ solves $$ p_1\le\lambda \le n p_r $$ $$\sum_{i=0}^rf^{-1}(\lambda/p_i)=1.$$ Since the LHS in the latter equation is strictly decreasing, a necessary and sufficient condition for existence is then expressed on the $p_i$'s, requiring
$$ p_1\le\lambda \le n p_r $$ and $$\sum_{i=0}^rf^{-1}(np_r /p_i)\le 1 .$$

Assuming $p_1>0$, your problem has at most a solution, and it admits a solution provided the $p_i$'s are not too dispersed, in a sense to be made precise below.

Consider the polynomial $f(t):=\frac{1-(1-t)^n}{t}$. It defines a strictly decreasing homeo $[0,1]\rightarrow [1,n]$ (because it is the incremental ratio of a concave function), and your system writes $p_i f(t_i)=p_{i+1} f(t_{i+1}),$ for $i=1,\dots, r-1$. So it has at most one solution, given by $t_i=f^{-1}(\lambda/p_i)$, where $\lambda$ solves $$ p_1\le\lambda \le n p_r $$ $$\sum_{i=0}^rf^{-1}(\lambda/p_i)=1.$$ Since the LHS in the latter equation is strictly decreasing, a necessary and sufficient condition for existence is then expressed on the $p_i$'s, requiring
$$ p_1\le n p_r $$ and $$\sum_{i=0}^rf^{-1}(np_r /p_i)\le 1 .$$