Markov's Inequality and Probability distribution?
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I know that Markov's inequality provides an upper bound for probability of a random variable being greater than a certain value, but in what instances/distributions would the upper bound be the exact probability?
probability probability-distributions random-variables
asked Nov 20 at 5:57
Justin Dee
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up vote
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I know that Markov's inequality provides an upper bound for probability of a random variable being greater than a certain value, but in what instances/distributions would the upper bound be the exact probability?
probability probability-distributions random-variables
asked Nov 20 at 5:57
Justin Dee
615
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I know that Markov's inequality provides an upper bound for probability of a random variable being greater than a certain value, but in what instances/distributions would the upper bound be the exact probability?
probability probability-distributions random-variables
asked Nov 20 at 5:57
Justin Dee
615
I know that Markov's inequality provides an upper bound for probability of a random variable being greater than a certain value, but in what instances/distributions would the upper bound be the exact probability?
probability probability-distributions random-variables
probability probability-distributions random-variables
asked Nov 20 at 5:57
Justin Dee
615
asked Nov 20 at 5:57
Justin Dee
615
asked Nov 20 at 5:57
Justin Dee
615
asked Nov 20 at 5:57
Justin Dee
615
asked Nov 20 at 5:57
Justin Dee
615
615
add a comment |
add a comment |
1 Answer
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Markov's inequality says $EX geq aP{Xgeq a}$ for any non-negative random variablse $X$ and any $a geq 0$. Equality holds iff $X$ is equal to the constant $a$ with probability $1$. To see this assume that $EX = aP{Xgeq a}$. Then $aP{Xgeq a} =EX= EXI_{Xgeq a} +EXI_{X<a} geq aP{Xgeq a}+ EXI_{X<a}$ so $EXI_{X<a}=0$. This implies that $X geq a$ almost surely. Now $E(X-a)=aP{Xgeq a}-a=0$ and $X-a$ is non-negative, so $X=a$ almost surely.
answered Nov 20 at 7:32
Kavi Rama Murthy
45.1k31852
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Markov's inequality says $EX geq aP{Xgeq a}$ for any non-negative random variablse $X$ and any $a geq 0$. Equality holds iff $X$ is equal to the constant $a$ with probability $1$. To see this assume that $EX = aP{Xgeq a}$. Then $aP{Xgeq a} =EX= EXI_{Xgeq a} +EXI_{X<a} geq aP{Xgeq a}+ EXI_{X<a}$ so $EXI_{X<a}=0$. This implies that $X geq a$ almost surely. Now $E(X-a)=aP{Xgeq a}-a=0$ and $X-a$ is non-negative, so $X=a$ almost surely.
answered Nov 20 at 7:32
Kavi Rama Murthy
45.1k31852
add a comment |
up vote
1
down vote
Markov's inequality says $EX geq aP{Xgeq a}$ for any non-negative random variablse $X$ and any $a geq 0$. Equality holds iff $X$ is equal to the constant $a$ with probability $1$. To see this assume that $EX = aP{Xgeq a}$. Then $aP{Xgeq a} =EX= EXI_{Xgeq a} +EXI_{X<a} geq aP{Xgeq a}+ EXI_{X<a}$ so $EXI_{X<a}=0$. This implies that $X geq a$ almost surely. Now $E(X-a)=aP{Xgeq a}-a=0$ and $X-a$ is non-negative, so $X=a$ almost surely.
answered Nov 20 at 7:32
Kavi Rama Murthy
45.1k31852
add a comment |
up vote
1
down vote
up vote
1
down vote
Markov's inequality says $EX geq aP{Xgeq a}$ for any non-negative random variablse $X$ and any $a geq 0$. Equality holds iff $X$ is equal to the constant $a$ with probability $1$. To see this assume that $EX = aP{Xgeq a}$. Then $aP{Xgeq a} =EX= EXI_{Xgeq a} +EXI_{X<a} geq aP{Xgeq a}+ EXI_{X<a}$ so $EXI_{X<a}=0$. This implies that $X geq a$ almost surely. Now $E(X-a)=aP{Xgeq a}-a=0$ and $X-a$ is non-negative, so $X=a$ almost surely.
answered Nov 20 at 7:32
Kavi Rama Murthy
45.1k31852
Markov's inequality says $EX geq aP{Xgeq a}$ for any non-negative random variablse $X$ and any $a geq 0$. Equality holds iff $X$ is equal to the constant $a$ with probability $1$. To see this assume that $EX = aP{Xgeq a}$. Then $aP{Xgeq a} =EX= EXI_{Xgeq a} +EXI_{X<a} geq aP{Xgeq a}+ EXI_{X<a}$ so $EXI_{X<a}=0$. This implies that $X geq a$ almost surely. Now $E(X-a)=aP{Xgeq a}-a=0$ and $X-a$ is non-negative, so $X=a$ almost surely.
answered Nov 20 at 7:32
Kavi Rama Murthy
45.1k31852
answered Nov 20 at 7:32
Kavi Rama Murthy
45.1k31852
answered Nov 20 at 7:32
Kavi Rama Murthy
45.1k31852
answered Nov 20 at 7:32
Kavi Rama Murthy
45.1k31852
45.1k31852
add a comment |
add a comment |
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