Mixing
Inequality regarding sum of squared probabilities
I'm working on a problem set for a course on Machine Learning and one the problems asks me to prove a given inequality. As an aid for that, the problem gives me the hint to use the following result, which I can understand but I am not able to prove:
Show that $Sigma_i a_i^2 geq a_1^2 + frac{(1-a_1)^2}{C-1}$ for any $C$ real numbers such that $a_1 geq a_2 geq ... geq a_C geq 0$ and $Sigma_i a_i = 1$
This is speacially useful in the exercise when the $a_i$ are probabilities. However, I could not prove the result. I tried using contradictions and even induction but without any success. Any ideas on how to prove? Thanks a lot in advance!
probability inequality induction sums-of-squares
asked Nov 20 '18 at 11:46
Raul Guarini
460211
add a comment |
I'm working on a problem set for a course on Machine Learning and one the problems asks me to prove a given inequality. As an aid for that, the problem gives me the hint to use the following result, which I can understand but I am not able to prove:
Show that $Sigma_i a_i^2 geq a_1^2 + frac{(1-a_1)^2}{C-1}$ for any $C$ real numbers such that $a_1 geq a_2 geq ... geq a_C geq 0$ and $Sigma_i a_i = 1$
This is speacially useful in the exercise when the $a_i$ are probabilities. However, I could not prove the result. I tried using contradictions and even induction but without any success. Any ideas on how to prove? Thanks a lot in advance!
probability inequality induction sums-of-squares
asked Nov 20 '18 at 11:46
Raul Guarini
460211
add a comment |
I'm working on a problem set for a course on Machine Learning and one the problems asks me to prove a given inequality. As an aid for that, the problem gives me the hint to use the following result, which I can understand but I am not able to prove:
Show that $Sigma_i a_i^2 geq a_1^2 + frac{(1-a_1)^2}{C-1}$ for any $C$ real numbers such that $a_1 geq a_2 geq ... geq a_C geq 0$ and $Sigma_i a_i = 1$
This is speacially useful in the exercise when the $a_i$ are probabilities. However, I could not prove the result. I tried using contradictions and even induction but without any success. Any ideas on how to prove? Thanks a lot in advance!
probability inequality induction sums-of-squares
asked Nov 20 '18 at 11:46
Raul Guarini
460211
I'm working on a problem set for a course on Machine Learning and one the problems asks me to prove a given inequality. As an aid for that, the problem gives me the hint to use the following result, which I can understand but I am not able to prove:
Show that $Sigma_i a_i^2 geq a_1^2 + frac{(1-a_1)^2}{C-1}$ for any $C$ real numbers such that $a_1 geq a_2 geq ... geq a_C geq 0$ and $Sigma_i a_i = 1$
This is speacially useful in the exercise when the $a_i$ are probabilities. However, I could not prove the result. I tried using contradictions and even induction but without any success. Any ideas on how to prove? Thanks a lot in advance!
probability inequality induction sums-of-squares
probability inequality induction sums-of-squares
asked Nov 20 '18 at 11:46
Raul Guarini
460211
asked Nov 20 '18 at 11:46
Raul Guarini
460211
asked Nov 20 '18 at 11:46
Raul Guarini
460211
asked Nov 20 '18 at 11:46
Raul Guarini
460211
asked Nov 20 '18 at 11:46
Raul Guarini
460211
460211
add a comment |
add a comment |
1 Answer
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By Cauchy-Schwarz,
$$(1-a_1)^2 = left( sum_{j=2}^C a_jright)^2 le sum_{j=2}^C a_j^2 cdot (C-1)$$
answered Nov 20 '18 at 12:06
Richard Martin
1,63618
Thanks for your answer! I think this solves the problem.
– Raul Guarini
Nov 20 '18 at 12:26
add a comment |
Your Answer
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
By Cauchy-Schwarz,
$$(1-a_1)^2 = left( sum_{j=2}^C a_jright)^2 le sum_{j=2}^C a_j^2 cdot (C-1)$$
answered Nov 20 '18 at 12:06
Richard Martin
1,63618
Thanks for your answer! I think this solves the problem.
– Raul Guarini
Nov 20 '18 at 12:26
add a comment |
By Cauchy-Schwarz,
$$(1-a_1)^2 = left( sum_{j=2}^C a_jright)^2 le sum_{j=2}^C a_j^2 cdot (C-1)$$
answered Nov 20 '18 at 12:06
Richard Martin
1,63618
Thanks for your answer! I think this solves the problem.
– Raul Guarini
Nov 20 '18 at 12:26
add a comment |
By Cauchy-Schwarz,
$$(1-a_1)^2 = left( sum_{j=2}^C a_jright)^2 le sum_{j=2}^C a_j^2 cdot (C-1)$$
answered Nov 20 '18 at 12:06
Richard Martin
1,63618
By Cauchy-Schwarz,
$$(1-a_1)^2 = left( sum_{j=2}^C a_jright)^2 le sum_{j=2}^C a_j^2 cdot (C-1)$$
answered Nov 20 '18 at 12:06
Richard Martin
1,63618
answered Nov 20 '18 at 12:06
Richard Martin
1,63618
answered Nov 20 '18 at 12:06
Richard Martin
1,63618
answered Nov 20 '18 at 12:06
Richard Martin
1,63618
1,63618
Thanks for your answer! I think this solves the problem.
– Raul Guarini
Nov 20 '18 at 12:26
add a comment |
Thanks for your answer! I think this solves the problem.
– Raul Guarini
Nov 20 '18 at 12:26
Thanks for your answer! I think this solves the problem.
– Raul Guarini
Nov 20 '18 at 12:26
Thanks for your answer! I think this solves the problem.
– Raul Guarini
Nov 20 '18 at 12:26
add a comment |
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