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May 9, 2017 at 18:17 vote accept Abhishek Halder
May 9, 2017 at 15:31 answer added martin cripps timeline score: 4
May 9, 2017 at 13:01 comment added Federico Poloni @martincripps Want to post it as an answer?
May 9, 2017 at 8:52 comment added Abhishek Halder @martin: very nice. looks trivial now :-) Thanks.
May 9, 2017 at 8:41 comment added martin cripps I wrote it as a system of equations $\frac{\nu_i}{1-\alpha_i}=\sum_j\frac{\alpha_i\nu_j}{1-\alpha_j}$ (here the sum runs over all $j$. Then changed variable $x_i=\nu_i/(1-\alpha_i)$ and observed that $x_i$ has to be proportional to $\alpha_i$.
May 9, 2017 at 8:38 comment added Abhishek Halder @ martin: I see. How do you prove it?
May 9, 2017 at 8:25 comment added martin cripps Try $\nu_i=\alpha_i(1-\alpha_i)k$ where $k$ is chosen to normalise.
May 9, 2017 at 8:04 history asked Abhishek Halder CC BY-SA 3.0