Total variation distance of probaiblity measures












3












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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










  • $begingroup$
    @KaviRamaMurthy I'll check your answer soon. I'm not the downvoter.
    $endgroup$
    – 0xbadf00d
    Dec 30 '18 at 0:14
















3












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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










  • $begingroup$
    @KaviRamaMurthy I'll check your answer soon. I'm not the downvoter.
    $endgroup$
    – 0xbadf00d
    Dec 30 '18 at 0:14














3












3








3


1



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 29 '18 at 11:57









0xbadf00d0xbadf00d

1,97541531




1,97541531












  • $begingroup$
    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 I'll check your answer soon. I'm not the downvoter.
    $endgroup$
    – 0xbadf00d
    Dec 30 '18 at 0:14


















  • $begingroup$
    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 I'll check your answer soon. I'm not the downvoter.
    $endgroup$
    – 0xbadf00d
    Dec 30 '18 at 0:14
















$begingroup$
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 I'll check your answer soon. I'm not the downvoter.
$endgroup$
– 0xbadf00d
Dec 30 '18 at 0:14




$begingroup$
@KaviRamaMurthy I'll check your answer soon. I'm not the downvoter.
$endgroup$
– 0xbadf00d
Dec 30 '18 at 0:14










2 Answers
2






active

oldest

votes


















-1












$begingroup$

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 $mu(E)-nu(E)=0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















-1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can someone tell me what is wrong with this answer?
    $endgroup$
    – Kavi Rama Murthy
    Dec 29 '18 at 23:13











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









-1












$begingroup$

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 $mu(E)-nu(E)=0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















-1












$begingroup$

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 $mu(E)-nu(E)=0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














-1












-1








-1





$begingroup$

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 $mu(E)-nu(E)=0$.






share|cite|improve this answer









$endgroup$



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 $mu(E)-nu(E)=0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 29 '18 at 18:30









John DawkinsJohn Dawkins

13.1k11017




13.1k11017












  • $begingroup$
    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


















  • $begingroup$
    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
















$begingroup$
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











-1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can someone tell me what is wrong with this answer?
    $endgroup$
    – Kavi Rama Murthy
    Dec 29 '18 at 23:13
















-1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can someone tell me what is wrong with this answer?
    $endgroup$
    – Kavi Rama Murthy
    Dec 29 '18 at 23:13














-1












-1








-1





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 30 '18 at 4:40

























answered Dec 29 '18 at 12:09









Kavi Rama MurthyKavi Rama Murthy

62.2k42262




62.2k42262












  • $begingroup$
    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 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


















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