Let x_1, ... x_n be real numbers and t_1, ... t_n be nonnegative numbers summing to 1.
A trivial consequence of Jensen's inequality is that
exp(t_1 x_1 + ... + t_n x_n) <= t_1 exp(x_1) + ... + t_n exp(x_n).
I claim that in the other direction, we have the following:
t_1 exp(x_1) + ... + t_n exp(x_n) <= exp( diam(x) + t_1 x_1 + ... + t_n x_n)
where diam(x) = max_i x_i - min_i x_i is the diameter of {x_1, ..., x_n}.
Any proof ideas? (Unlike my previous goof-ups, I can actually prove this one.)
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5 comments:
Easy one, as
diam + sum_i ( t_i x_i ) =
sum_i t_i ( diam + x_i ) <=
sum_i t_i max( x_i ) = max( x_i )
Therefore, the "converse" is true not only for exp but for any increasing function...
cjgb: your proof (and claim) fail for negative x. If the x's are all assumed to be positive and F is any increasing function, then trivially
t_1 F(x_1) + ... + t_n F(x_n)
<=
F(max x_i)
Sorry, there was a small typo in my "proof". However, I still claim that it holds.
I want to prove that
t_1 exp(x_1) + ... + t_n exp(x_n) <= exp( diam(x) + t_1 x_1 + ... + t_n x_n)
If F is increasing (and exp is), you have, as you agreed, that
t_1 F(x_1) + ... + t_n F(x_n)
<=
F(max x_i)
All I would need to prove is that
max x_i <= diam + sum_i ( t_i x_i )
and that is exactly what my previous "proof" says (only that I wrote "<=" where I meant ">=". In fact,
max( x_i ) = sum_i t_i max( x_i ) <= sum_i t_i ( diam + x_i ) = diam + sum_i ( t_i x_i )
regardless of the sign of the x_i as
max_i x_i <= x_i + diam
Indeed, that does it! Very nicely done.
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