Total variation distance of probaiblity measures
$begingroup$
Let $E$ be a set and $mathcal Esubseteq 2^E$ with $emptysetinmathcal E$. If $mu:mathcal Etomathbb R$ with $mu(emptyset)=0$, then $$operatorname{Var}_mu(B):=supleft{sum_{i=1}^nleft|mu(B_i)right|:ninmathbb Ntext{ and }B_1,ldots,B_ninmathcal Btext{ are disjoint with }biguplus_{i=1}^nB_isubseteq Bright}$$ for $Bsubseteq E$. If $Einmathcal E$, let $$left|muright|:=operatorname{Var}_mu(E).$$
Assume $(E,mathcal E)$ is a measurable space and $mu,nu$ are probability measures on $(E,mathcal E)$. Is it possible to show that $$left|mu-nuright|=sup_{Binmathcal E}left|mu(B)-nu(B)right|tag1$$ or is there a counterexample?
I was able to show the claim assuming that $mu$ and $nu$ are both absolutely continuous with respect to a common reference measure. Is it possible to show $(1)$ in general?
probability-theory measure-theory total-variation
asked Dec 29 '18 at 11:57
0xbadf00d0xbadf00d
1,67041534
$endgroup$
add a comment |
$begingroup$
Let $E$ be a set and $mathcal Esubseteq 2^E$ with $emptysetinmathcal E$. If $mu:mathcal Etomathbb R$ with $mu(emptyset)=0$, then $$operatorname{Var}_mu(B):=supleft{sum_{i=1}^nleft|mu(B_i)right|:ninmathbb Ntext{ and }B_1,ldots,B_ninmathcal Btext{ are disjoint with }biguplus_{i=1}^nB_isubseteq Bright}$$ for $Bsubseteq E$. If $Einmathcal E$, let $$left|muright|:=operatorname{Var}_mu(E).$$
Assume $(E,mathcal E)$ is a measurable space and $mu,nu$ are probability measures on $(E,mathcal E)$. Is it possible to show that $$left|mu-nuright|=sup_{Binmathcal E}left|mu(B)-nu(B)right|tag1$$ or is there a counterexample?
I was able to show the claim assuming that $mu$ and $nu$ are both absolutely continuous with respect to a common reference measure. Is it possible to show $(1)$ in general?
probability-theory measure-theory total-variation
asked Dec 29 '18 at 11:57
0xbadf00d0xbadf00d
1,67041534
$endgroup$
$begingroup$
Your guess is wrong. The equation fails when $mu$ and $nu$ are singular.
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:15
$begingroup$
If you downvoted me please read my revised answer.
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:25
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@KaviRamaMurthy Why did you delete your answer?
$endgroup$
– 0xbadf00d
Feb 25 at 17:48
add a comment |
$begingroup$
Let $E$ be a set and $mathcal Esubseteq 2^E$ with $emptysetinmathcal E$. If $mu:mathcal Etomathbb R$ with $mu(emptyset)=0$, then $$operatorname{Var}_mu(B):=supleft{sum_{i=1}^nleft|mu(B_i)right|:ninmathbb Ntext{ and }B_1,ldots,B_ninmathcal Btext{ are disjoint with }biguplus_{i=1}^nB_isubseteq Bright}$$ for $Bsubseteq E$. If $Einmathcal E$, let $$left|muright|:=operatorname{Var}_mu(E).$$
Assume $(E,mathcal E)$ is a measurable space and $mu,nu$ are probability measures on $(E,mathcal E)$. Is it possible to show that $$left|mu-nuright|=sup_{Binmathcal E}left|mu(B)-nu(B)right|tag1$$ or is there a counterexample?
I was able to show the claim assuming that $mu$ and $nu$ are both absolutely continuous with respect to a common reference measure. Is it possible to show $(1)$ in general?
probability-theory measure-theory total-variation
asked Dec 29 '18 at 11:57
0xbadf00d0xbadf00d
1,67041534
$endgroup$
Let $E$ be a set and $mathcal Esubseteq 2^E$ with $emptysetinmathcal E$. If $mu:mathcal Etomathbb R$ with $mu(emptyset)=0$, then $$operatorname{Var}_mu(B):=supleft{sum_{i=1}^nleft|mu(B_i)right|:ninmathbb Ntext{ and }B_1,ldots,B_ninmathcal Btext{ are disjoint with }biguplus_{i=1}^nB_isubseteq Bright}$$ for $Bsubseteq E$. If $Einmathcal E$, let $$left|muright|:=operatorname{Var}_mu(E).$$
Assume $(E,mathcal E)$ is a measurable space and $mu,nu$ are probability measures on $(E,mathcal E)$. Is it possible to show that $$left|mu-nuright|=sup_{Binmathcal E}left|mu(B)-nu(B)right|tag1$$ or is there a counterexample?
I was able to show the claim assuming that $mu$ and $nu$ are both absolutely continuous with respect to a common reference measure. Is it possible to show $(1)$ in general?
probability-theory measure-theory total-variation
probability-theory measure-theory total-variation
asked Dec 29 '18 at 11:57
0xbadf00d0xbadf00d
1,67041534
asked Dec 29 '18 at 11:57
0xbadf00d0xbadf00d
1,67041534
asked Dec 29 '18 at 11:57
0xbadf00d0xbadf00d
1,67041534
asked Dec 29 '18 at 11:57
0xbadf00d0xbadf00d
1,67041534
asked Dec 29 '18 at 11:57
0xbadf00d0xbadf00d
1,67041534
1,67041534
$begingroup$
Your guess is wrong. The equation fails when $mu$ and $nu$ are singular.
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:15
$begingroup$
If you downvoted me please read my revised answer.
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:25
$begingroup$
@KaviRamaMurthy Why did you delete your answer?
$endgroup$
– 0xbadf00d
Feb 25 at 17:48
add a comment |
$begingroup$
Your guess is wrong. The equation fails when $mu$ and $nu$ are singular.
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:15
$begingroup$
If you downvoted me please read my revised answer.
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:25
$begingroup$
@KaviRamaMurthy Why did you delete your answer?
$endgroup$
– 0xbadf00d
Feb 25 at 17:48
$begingroup$
Your guess is wrong. The equation fails when $mu$ and $nu$ are singular.
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:15
$begingroup$
Your guess is wrong. The equation fails when $mu$ and $nu$ are singular.
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:15
$begingroup$
If you downvoted me please read my revised answer.
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:25
$begingroup$
If you downvoted me please read my revised answer.
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:25
$begingroup$
@KaviRamaMurthy Why did you delete your answer?
$endgroup$
– 0xbadf00d
Feb 25 at 17:48
$begingroup$
@KaviRamaMurthy Why did you delete your answer?
$endgroup$
– 0xbadf00d
Feb 25 at 17:48
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You are off by a factor of 2: $| mu-nu| =2sup_{Binmathcal E}|mu(B)-nu(B)|$. For this it's important that $mu $ and $nu$ are probabilities, so that $mu(E)-nu(E)=0$.
Later addition: Let $lambda=mu+nu$ and let $f$ and $g$ be the Radon-Nikodym densities of $mu$ and $nu$ with respect to $lambda$. It's not hard to check that $|mu-nu|=int_E|f-g|,dlambda$.
First, because $mu(E)=nu(E)=1$, if $Binmathcal E$ then,
$$
2|mu(B)-nu(B)|=|mu(B)-nu(B)|+|mu(B^c)-nu(B^c)|leoperatorname{Var}_{mu-nu}(E)=|mu-nu|.
$$
This shows that $|mu-nu|ge 2sup_{Binmathcal E}|mu(B)-nu(B)|$.
For the reverse inequality,
$$
eqalign{
|mu-nu|&=int_E|f-g|,dlambdacr
&=int_{{f>g}}(f-g),dlambda+int_{{f<g}}(g-f),dlambdacr
&=(mu-nu)({f>g})+(nu-mu)({f<g})cr
&=|(mu-nu)({f>g})|+|(mu-nu)({f<g})|cr
&le 2sup_{Binmathcal E}|mu(B)-nu(B)|.
}
$$
answered Dec 29 '18 at 18:30
John DawkinsJohn Dawkins
13.4k11017
$endgroup$
$begingroup$
Maybe I forgot the factor $2$, but how is $mu(E)-mu(E)=1-1=0$ answering my question? Could you provide a proof for the identity? (I didn't downvote your answer.)
$endgroup$
– 0xbadf00d
Dec 30 '18 at 0:16
$begingroup$
It answers your question in the sense that it points out that you have the wrong formula except in the trivial case when both sides of the = sign are 0. So the answer to your question, put plainly, is NO.
$endgroup$
– John Dawkins
Jan 4 at 23:04
$begingroup$
Yes, okay. I didn't ask explicitly how we can prove the correct formula if mine would turn out to be wrong. However, can you tell me how we can prove it?
$endgroup$
– 0xbadf00d
Feb 24 at 18:19
$begingroup$
I've added a proof to my original answer.
$endgroup$
– John Dawkins
Feb 25 at 20:06
add a comment |
$begingroup$
Your claim that the equality holds when the measures are absolutely continuous w.r.t. some measure is wrong. You can check that it fails even in simple cases like $mu (A)=m(Acap (0,1))$, $nu (A)= m(Acap (-1,0))$ where $m$ is Lebesgue measure. [Any two probability measures are absolutely continuous w.r.t some probability measure, e.g. $frac {mu +nu} 2$].
If $mu =delta_a$ and $nu =delta_b$ with $a neq b$ then $|mu-nu|=2$ and $|mu(B)-nu(B)| leq 1$ for all $B$.
[$|mu-nu| geq |(mu-nu)({a})|+|(mu-nu)({b})|=2$].
More generally if $P$ and $Q$ are any two probability measures with $Pperp Q$ the $|P-Q|=2$ and $|P(B)-Q(B)| leq 1$ for all $B$.
What is true in general is RHS $leq$ LHS $leq 2$RHS.
answered Dec 29 '18 at 12:09
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
Can someone tell me what is wrong with this answer?
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:13
$begingroup$
Can we even show that "$text{LHS}=2text{RHS}$"?
$endgroup$
– 0xbadf00d
Feb 24 at 19:27
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You are off by a factor of 2: $| mu-nu| =2sup_{Binmathcal E}|mu(B)-nu(B)|$. For this it's important that $mu $ and $nu$ are probabilities, so that $mu(E)-nu(E)=0$.
Later addition: Let $lambda=mu+nu$ and let $f$ and $g$ be the Radon-Nikodym densities of $mu$ and $nu$ with respect to $lambda$. It's not hard to check that $|mu-nu|=int_E|f-g|,dlambda$.
First, because $mu(E)=nu(E)=1$, if $Binmathcal E$ then,
$$
2|mu(B)-nu(B)|=|mu(B)-nu(B)|+|mu(B^c)-nu(B^c)|leoperatorname{Var}_{mu-nu}(E)=|mu-nu|.
$$
This shows that $|mu-nu|ge 2sup_{Binmathcal E}|mu(B)-nu(B)|$.
For the reverse inequality,
$$
eqalign{
|mu-nu|&=int_E|f-g|,dlambdacr
&=int_{{f>g}}(f-g),dlambda+int_{{f<g}}(g-f),dlambdacr
&=(mu-nu)({f>g})+(nu-mu)({f<g})cr
&=|(mu-nu)({f>g})|+|(mu-nu)({f<g})|cr
&le 2sup_{Binmathcal E}|mu(B)-nu(B)|.
}
$$
answered Dec 29 '18 at 18:30
John DawkinsJohn Dawkins
13.4k11017
$endgroup$
$begingroup$
Maybe I forgot the factor $2$, but how is $mu(E)-mu(E)=1-1=0$ answering my question? Could you provide a proof for the identity? (I didn't downvote your answer.)
$endgroup$
– 0xbadf00d
Dec 30 '18 at 0:16
$begingroup$
It answers your question in the sense that it points out that you have the wrong formula except in the trivial case when both sides of the = sign are 0. So the answer to your question, put plainly, is NO.
$endgroup$
– John Dawkins
Jan 4 at 23:04
$begingroup$
Yes, okay. I didn't ask explicitly how we can prove the correct formula if mine would turn out to be wrong. However, can you tell me how we can prove it?
$endgroup$
– 0xbadf00d
Feb 24 at 18:19
$begingroup$
I've added a proof to my original answer.
$endgroup$
– John Dawkins
Feb 25 at 20:06
add a comment |
$begingroup$
You are off by a factor of 2: $| mu-nu| =2sup_{Binmathcal E}|mu(B)-nu(B)|$. For this it's important that $mu $ and $nu$ are probabilities, so that $mu(E)-nu(E)=0$.
Later addition: Let $lambda=mu+nu$ and let $f$ and $g$ be the Radon-Nikodym densities of $mu$ and $nu$ with respect to $lambda$. It's not hard to check that $|mu-nu|=int_E|f-g|,dlambda$.
First, because $mu(E)=nu(E)=1$, if $Binmathcal E$ then,
$$
2|mu(B)-nu(B)|=|mu(B)-nu(B)|+|mu(B^c)-nu(B^c)|leoperatorname{Var}_{mu-nu}(E)=|mu-nu|.
$$
This shows that $|mu-nu|ge 2sup_{Binmathcal E}|mu(B)-nu(B)|$.
For the reverse inequality,
$$
eqalign{
|mu-nu|&=int_E|f-g|,dlambdacr
&=int_{{f>g}}(f-g),dlambda+int_{{f<g}}(g-f),dlambdacr
&=(mu-nu)({f>g})+(nu-mu)({f<g})cr
&=|(mu-nu)({f>g})|+|(mu-nu)({f<g})|cr
&le 2sup_{Binmathcal E}|mu(B)-nu(B)|.
}
$$
answered Dec 29 '18 at 18:30
John DawkinsJohn Dawkins
13.4k11017
$endgroup$
$begingroup$
Maybe I forgot the factor $2$, but how is $mu(E)-mu(E)=1-1=0$ answering my question? Could you provide a proof for the identity? (I didn't downvote your answer.)
$endgroup$
– 0xbadf00d
Dec 30 '18 at 0:16
$begingroup$
It answers your question in the sense that it points out that you have the wrong formula except in the trivial case when both sides of the = sign are 0. So the answer to your question, put plainly, is NO.
$endgroup$
– John Dawkins
Jan 4 at 23:04
$begingroup$
Yes, okay. I didn't ask explicitly how we can prove the correct formula if mine would turn out to be wrong. However, can you tell me how we can prove it?
$endgroup$
– 0xbadf00d
Feb 24 at 18:19
$begingroup$
I've added a proof to my original answer.
$endgroup$
– John Dawkins
Feb 25 at 20:06
add a comment |
$begingroup$
You are off by a factor of 2: $| mu-nu| =2sup_{Binmathcal E}|mu(B)-nu(B)|$. For this it's important that $mu $ and $nu$ are probabilities, so that $mu(E)-nu(E)=0$.
Later addition: Let $lambda=mu+nu$ and let $f$ and $g$ be the Radon-Nikodym densities of $mu$ and $nu$ with respect to $lambda$. It's not hard to check that $|mu-nu|=int_E|f-g|,dlambda$.
First, because $mu(E)=nu(E)=1$, if $Binmathcal E$ then,
$$
2|mu(B)-nu(B)|=|mu(B)-nu(B)|+|mu(B^c)-nu(B^c)|leoperatorname{Var}_{mu-nu}(E)=|mu-nu|.
$$
This shows that $|mu-nu|ge 2sup_{Binmathcal E}|mu(B)-nu(B)|$.
For the reverse inequality,
$$
eqalign{
|mu-nu|&=int_E|f-g|,dlambdacr
&=int_{{f>g}}(f-g),dlambda+int_{{f<g}}(g-f),dlambdacr
&=(mu-nu)({f>g})+(nu-mu)({f<g})cr
&=|(mu-nu)({f>g})|+|(mu-nu)({f<g})|cr
&le 2sup_{Binmathcal E}|mu(B)-nu(B)|.
}
$$
answered Dec 29 '18 at 18:30
John DawkinsJohn Dawkins
13.4k11017
$endgroup$
You are off by a factor of 2: $| mu-nu| =2sup_{Binmathcal E}|mu(B)-nu(B)|$. For this it's important that $mu $ and $nu$ are probabilities, so that $mu(E)-nu(E)=0$.
Later addition: Let $lambda=mu+nu$ and let $f$ and $g$ be the Radon-Nikodym densities of $mu$ and $nu$ with respect to $lambda$. It's not hard to check that $|mu-nu|=int_E|f-g|,dlambda$.
First, because $mu(E)=nu(E)=1$, if $Binmathcal E$ then,
$$
2|mu(B)-nu(B)|=|mu(B)-nu(B)|+|mu(B^c)-nu(B^c)|leoperatorname{Var}_{mu-nu}(E)=|mu-nu|.
$$
This shows that $|mu-nu|ge 2sup_{Binmathcal E}|mu(B)-nu(B)|$.
For the reverse inequality,
$$
eqalign{
|mu-nu|&=int_E|f-g|,dlambdacr
&=int_{{f>g}}(f-g),dlambda+int_{{f<g}}(g-f),dlambdacr
&=(mu-nu)({f>g})+(nu-mu)({f<g})cr
&=|(mu-nu)({f>g})|+|(mu-nu)({f<g})|cr
&le 2sup_{Binmathcal E}|mu(B)-nu(B)|.
}
$$
answered Dec 29 '18 at 18:30
John DawkinsJohn Dawkins
13.4k11017
edited Feb 25 at 20:06
answered Dec 29 '18 at 18:30
John DawkinsJohn Dawkins
13.4k11017
answered Dec 29 '18 at 18:30
John DawkinsJohn Dawkins
13.4k11017
answered Dec 29 '18 at 18:30
John DawkinsJohn Dawkins
13.4k11017
13.4k11017
$begingroup$
Maybe I forgot the factor $2$, but how is $mu(E)-mu(E)=1-1=0$ answering my question? Could you provide a proof for the identity? (I didn't downvote your answer.)
$endgroup$
– 0xbadf00d
Dec 30 '18 at 0:16
$begingroup$
It answers your question in the sense that it points out that you have the wrong formula except in the trivial case when both sides of the = sign are 0. So the answer to your question, put plainly, is NO.
$endgroup$
– John Dawkins
Jan 4 at 23:04
$begingroup$
Yes, okay. I didn't ask explicitly how we can prove the correct formula if mine would turn out to be wrong. However, can you tell me how we can prove it?
$endgroup$
– 0xbadf00d
Feb 24 at 18:19
$begingroup$
I've added a proof to my original answer.
$endgroup$
– John Dawkins
Feb 25 at 20:06
add a comment |
$begingroup$
Maybe I forgot the factor $2$, but how is $mu(E)-mu(E)=1-1=0$ answering my question? Could you provide a proof for the identity? (I didn't downvote your answer.)
$endgroup$
– 0xbadf00d
Dec 30 '18 at 0:16
$begingroup$
It answers your question in the sense that it points out that you have the wrong formula except in the trivial case when both sides of the = sign are 0. So the answer to your question, put plainly, is NO.
$endgroup$
– John Dawkins
Jan 4 at 23:04
$begingroup$
Yes, okay. I didn't ask explicitly how we can prove the correct formula if mine would turn out to be wrong. However, can you tell me how we can prove it?
$endgroup$
– 0xbadf00d
Feb 24 at 18:19
$begingroup$
I've added a proof to my original answer.
$endgroup$
– John Dawkins
Feb 25 at 20:06
$begingroup$
Maybe I forgot the factor $2$, but how is $mu(E)-mu(E)=1-1=0$ answering my question? Could you provide a proof for the identity? (I didn't downvote your answer.)
$endgroup$
– 0xbadf00d
Dec 30 '18 at 0:16
$begingroup$
Maybe I forgot the factor $2$, but how is $mu(E)-mu(E)=1-1=0$ answering my question? Could you provide a proof for the identity? (I didn't downvote your answer.)
$endgroup$
– 0xbadf00d
Dec 30 '18 at 0:16
$begingroup$
It answers your question in the sense that it points out that you have the wrong formula except in the trivial case when both sides of the = sign are 0. So the answer to your question, put plainly, is NO.
$endgroup$
– John Dawkins
Jan 4 at 23:04
$begingroup$
It answers your question in the sense that it points out that you have the wrong formula except in the trivial case when both sides of the = sign are 0. So the answer to your question, put plainly, is NO.
$endgroup$
– John Dawkins
Jan 4 at 23:04
$begingroup$
Yes, okay. I didn't ask explicitly how we can prove the correct formula if mine would turn out to be wrong. However, can you tell me how we can prove it?
$endgroup$
– 0xbadf00d
Feb 24 at 18:19
$begingroup$
Yes, okay. I didn't ask explicitly how we can prove the correct formula if mine would turn out to be wrong. However, can you tell me how we can prove it?
$endgroup$
– 0xbadf00d
Feb 24 at 18:19
$begingroup$
I've added a proof to my original answer.
$endgroup$
– John Dawkins
Feb 25 at 20:06
$begingroup$
I've added a proof to my original answer.
$endgroup$
– John Dawkins
Feb 25 at 20:06
add a comment |
$begingroup$
Your claim that the equality holds when the measures are absolutely continuous w.r.t. some measure is wrong. You can check that it fails even in simple cases like $mu (A)=m(Acap (0,1))$, $nu (A)= m(Acap (-1,0))$ where $m$ is Lebesgue measure. [Any two probability measures are absolutely continuous w.r.t some probability measure, e.g. $frac {mu +nu} 2$].
If $mu =delta_a$ and $nu =delta_b$ with $a neq b$ then $|mu-nu|=2$ and $|mu(B)-nu(B)| leq 1$ for all $B$.
[$|mu-nu| geq |(mu-nu)({a})|+|(mu-nu)({b})|=2$].
More generally if $P$ and $Q$ are any two probability measures with $Pperp Q$ the $|P-Q|=2$ and $|P(B)-Q(B)| leq 1$ for all $B$.
What is true in general is RHS $leq$ LHS $leq 2$RHS.
answered Dec 29 '18 at 12:09
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
Can someone tell me what is wrong with this answer?
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:13
$begingroup$
Can we even show that "$text{LHS}=2text{RHS}$"?
$endgroup$
– 0xbadf00d
Feb 24 at 19:27
add a comment |
$begingroup$
Your claim that the equality holds when the measures are absolutely continuous w.r.t. some measure is wrong. You can check that it fails even in simple cases like $mu (A)=m(Acap (0,1))$, $nu (A)= m(Acap (-1,0))$ where $m$ is Lebesgue measure. [Any two probability measures are absolutely continuous w.r.t some probability measure, e.g. $frac {mu +nu} 2$].
If $mu =delta_a$ and $nu =delta_b$ with $a neq b$ then $|mu-nu|=2$ and $|mu(B)-nu(B)| leq 1$ for all $B$.
[$|mu-nu| geq |(mu-nu)({a})|+|(mu-nu)({b})|=2$].
More generally if $P$ and $Q$ are any two probability measures with $Pperp Q$ the $|P-Q|=2$ and $|P(B)-Q(B)| leq 1$ for all $B$.
What is true in general is RHS $leq$ LHS $leq 2$RHS.
answered Dec 29 '18 at 12:09
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
Can someone tell me what is wrong with this answer?
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:13
$begingroup$
Can we even show that "$text{LHS}=2text{RHS}$"?
$endgroup$
– 0xbadf00d
Feb 24 at 19:27
add a comment |
$begingroup$
Your claim that the equality holds when the measures are absolutely continuous w.r.t. some measure is wrong. You can check that it fails even in simple cases like $mu (A)=m(Acap (0,1))$, $nu (A)= m(Acap (-1,0))$ where $m$ is Lebesgue measure. [Any two probability measures are absolutely continuous w.r.t some probability measure, e.g. $frac {mu +nu} 2$].
If $mu =delta_a$ and $nu =delta_b$ with $a neq b$ then $|mu-nu|=2$ and $|mu(B)-nu(B)| leq 1$ for all $B$.
[$|mu-nu| geq |(mu-nu)({a})|+|(mu-nu)({b})|=2$].
More generally if $P$ and $Q$ are any two probability measures with $Pperp Q$ the $|P-Q|=2$ and $|P(B)-Q(B)| leq 1$ for all $B$.
What is true in general is RHS $leq$ LHS $leq 2$RHS.
answered Dec 29 '18 at 12:09
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
Your claim that the equality holds when the measures are absolutely continuous w.r.t. some measure is wrong. You can check that it fails even in simple cases like $mu (A)=m(Acap (0,1))$, $nu (A)= m(Acap (-1,0))$ where $m$ is Lebesgue measure. [Any two probability measures are absolutely continuous w.r.t some probability measure, e.g. $frac {mu +nu} 2$].
If $mu =delta_a$ and $nu =delta_b$ with $a neq b$ then $|mu-nu|=2$ and $|mu(B)-nu(B)| leq 1$ for all $B$.
[$|mu-nu| geq |(mu-nu)({a})|+|(mu-nu)({b})|=2$].
More generally if $P$ and $Q$ are any two probability measures with $Pperp Q$ the $|P-Q|=2$ and $|P(B)-Q(B)| leq 1$ for all $B$.
What is true in general is RHS $leq$ LHS $leq 2$RHS.
answered Dec 29 '18 at 12:09
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
edited Dec 30 '18 at 4:40
answered Dec 29 '18 at 12:09
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Dec 29 '18 at 12:09
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Dec 29 '18 at 12:09
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
74.9k53270
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Can someone tell me what is wrong with this answer?
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– Kavi Rama Murthy
Dec 29 '18 at 23:13
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Can we even show that "$text{LHS}=2text{RHS}$"?
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– 0xbadf00d
Feb 24 at 19:27
add a comment |
$begingroup$
Can someone tell me what is wrong with this answer?
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:13
$begingroup$
Can we even show that "$text{LHS}=2text{RHS}$"?
$endgroup$
– 0xbadf00d
Feb 24 at 19:27
$begingroup$
Can someone tell me what is wrong with this answer?
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:13
$begingroup$
Can someone tell me what is wrong with this answer?
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:13
$begingroup$
Can we even show that "$text{LHS}=2text{RHS}$"?
$endgroup$
– 0xbadf00d
Feb 24 at 19:27
$begingroup$
Can we even show that "$text{LHS}=2text{RHS}$"?
$endgroup$
– 0xbadf00d
Feb 24 at 19:27
add a comment |
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$begingroup$
Your guess is wrong. The equation fails when $mu$ and $nu$ are singular.
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:15
$begingroup$
If you downvoted me please read my revised answer.
$endgroup$
– Kavi Rama Murthy
Dec 29 '18 at 23:25
$begingroup$
@KaviRamaMurthy Why did you delete your answer?
$endgroup$
– 0xbadf00d
Feb 25 at 17:48