Lower bound of the entropy over the probability distribution
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I am trying to show that:
$$inf_{z: 1^Tz=1} sum_{i=1}^m {z_i log z_i} = -log m$$
I thought about Jensen inequality or induction, but none of them provided me something.
calculus inequality upper-lower-bounds jensen-inequality
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up vote
1
down vote
favorite
I am trying to show that:
$$inf_{z: 1^Tz=1} sum_{i=1}^m {z_i log z_i} = -log m$$
I thought about Jensen inequality or induction, but none of them provided me something.
calculus inequality upper-lower-bounds jensen-inequality
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to show that:
$$inf_{z: 1^Tz=1} sum_{i=1}^m {z_i log z_i} = -log m$$
I thought about Jensen inequality or induction, but none of them provided me something.
calculus inequality upper-lower-bounds jensen-inequality
I am trying to show that:
$$inf_{z: 1^Tz=1} sum_{i=1}^m {z_i log z_i} = -log m$$
I thought about Jensen inequality or induction, but none of them provided me something.
calculus inequality upper-lower-bounds jensen-inequality
calculus inequality upper-lower-bounds jensen-inequality
edited Nov 28 at 2:28
Alex Ravsky
37.1k32078
37.1k32078
asked Nov 22 at 14:24
pl-94
1274
1274
add a comment |
add a comment |
1 Answer
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Jensen’s inequality works straightforwardly. Since for each $i$, $log z_i$ is involved in the given inequality, we assume that $z_i>0$. Consider a function $f(x)=xlog x$ for $x>0$. Since $f’’(x)=frac 1{x}>0$, the function $f$ is convex. Thus
$$frac{sum f(z_i)}mge f left(frac{sum z_i}mright)=fleft(frac 1mright)=frac 1mlogfrac 1m=-frac{log m}{m}.$$
and the minimum is attained when each $z_i$ equals $frac 1m$.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Jensen’s inequality works straightforwardly. Since for each $i$, $log z_i$ is involved in the given inequality, we assume that $z_i>0$. Consider a function $f(x)=xlog x$ for $x>0$. Since $f’’(x)=frac 1{x}>0$, the function $f$ is convex. Thus
$$frac{sum f(z_i)}mge f left(frac{sum z_i}mright)=fleft(frac 1mright)=frac 1mlogfrac 1m=-frac{log m}{m}.$$
and the minimum is attained when each $z_i$ equals $frac 1m$.
add a comment |
up vote
1
down vote
accepted
Jensen’s inequality works straightforwardly. Since for each $i$, $log z_i$ is involved in the given inequality, we assume that $z_i>0$. Consider a function $f(x)=xlog x$ for $x>0$. Since $f’’(x)=frac 1{x}>0$, the function $f$ is convex. Thus
$$frac{sum f(z_i)}mge f left(frac{sum z_i}mright)=fleft(frac 1mright)=frac 1mlogfrac 1m=-frac{log m}{m}.$$
and the minimum is attained when each $z_i$ equals $frac 1m$.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Jensen’s inequality works straightforwardly. Since for each $i$, $log z_i$ is involved in the given inequality, we assume that $z_i>0$. Consider a function $f(x)=xlog x$ for $x>0$. Since $f’’(x)=frac 1{x}>0$, the function $f$ is convex. Thus
$$frac{sum f(z_i)}mge f left(frac{sum z_i}mright)=fleft(frac 1mright)=frac 1mlogfrac 1m=-frac{log m}{m}.$$
and the minimum is attained when each $z_i$ equals $frac 1m$.
Jensen’s inequality works straightforwardly. Since for each $i$, $log z_i$ is involved in the given inequality, we assume that $z_i>0$. Consider a function $f(x)=xlog x$ for $x>0$. Since $f’’(x)=frac 1{x}>0$, the function $f$ is convex. Thus
$$frac{sum f(z_i)}mge f left(frac{sum z_i}mright)=fleft(frac 1mright)=frac 1mlogfrac 1m=-frac{log m}{m}.$$
and the minimum is attained when each $z_i$ equals $frac 1m$.
edited Nov 28 at 17:57
pl-94
1274
1274
answered Nov 28 at 2:28
Alex Ravsky
37.1k32078
37.1k32078
add a comment |
add a comment |
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