Mixing
Prove that if $P[Xleq Y] =1$ then $E[X]leq E[Y]$
$begingroup$
I would like to prove/disprove that following claim:
Prove that if $P[Xleq Y] =1$ then $E[X]leq E[Y]$
How can it be done?
probability expected-value
asked Feb 1 at 17:25
superuser123superuser123
48628
$endgroup$
add a comment |
$begingroup$
I would like to prove/disprove that following claim:
Prove that if $P[Xleq Y] =1$ then $E[X]leq E[Y]$
How can it be done?
probability expected-value
asked Feb 1 at 17:25
superuser123superuser123
48628
$endgroup$
1
$begingroup$
Equivalent to "if $P(Y-X ge 0)=1$ then $E[Y]-E[X] ge 0$" which, since $E[Y]-E[X] =E[Y-X]$, is equivalent to "if $P(Z ge 0)=1$ then $E[Z] ge 0$"
$endgroup$
– Henry
Feb 1 at 18:20
$begingroup$
On a set of full measure, $X leq Y$ then the integrals will have the same inequality.
$endgroup$
– Will M.
Feb 2 at 8:11
add a comment |
$begingroup$
I would like to prove/disprove that following claim:
Prove that if $P[Xleq Y] =1$ then $E[X]leq E[Y]$
How can it be done?
probability expected-value
asked Feb 1 at 17:25
superuser123superuser123
48628
$endgroup$
I would like to prove/disprove that following claim:
Prove that if $P[Xleq Y] =1$ then $E[X]leq E[Y]$
How can it be done?
probability expected-value
probability expected-value
asked Feb 1 at 17:25
superuser123superuser123
48628
asked Feb 1 at 17:25
superuser123superuser123
48628
asked Feb 1 at 17:25
superuser123superuser123
48628
asked Feb 1 at 17:25
superuser123superuser123
48628
asked Feb 1 at 17:25
superuser123superuser123
48628
48628
1
$begingroup$
Equivalent to "if $P(Y-X ge 0)=1$ then $E[Y]-E[X] ge 0$" which, since $E[Y]-E[X] =E[Y-X]$, is equivalent to "if $P(Z ge 0)=1$ then $E[Z] ge 0$"
$endgroup$
– Henry
Feb 1 at 18:20
$begingroup$
On a set of full measure, $X leq Y$ then the integrals will have the same inequality.
$endgroup$
– Will M.
Feb 2 at 8:11
add a comment |
1
$begingroup$
Equivalent to "if $P(Y-X ge 0)=1$ then $E[Y]-E[X] ge 0$" which, since $E[Y]-E[X] =E[Y-X]$, is equivalent to "if $P(Z ge 0)=1$ then $E[Z] ge 0$"
$endgroup$
– Henry
Feb 1 at 18:20
$begingroup$
On a set of full measure, $X leq Y$ then the integrals will have the same inequality.
$endgroup$
– Will M.
Feb 2 at 8:11
1
1
$begingroup$
Equivalent to "if $P(Y-X ge 0)=1$ then $E[Y]-E[X] ge 0$" which, since $E[Y]-E[X] =E[Y-X]$, is equivalent to "if $P(Z ge 0)=1$ then $E[Z] ge 0$"
$endgroup$
– Henry
Feb 1 at 18:20
$begingroup$
Equivalent to "if $P(Y-X ge 0)=1$ then $E[Y]-E[X] ge 0$" which, since $E[Y]-E[X] =E[Y-X]$, is equivalent to "if $P(Z ge 0)=1$ then $E[Z] ge 0$"
$endgroup$
– Henry
Feb 1 at 18:20
$begingroup$
On a set of full measure, $X leq Y$ then the integrals will have the same inequality.
$endgroup$
– Will M.
Feb 2 at 8:11
$begingroup$
On a set of full measure, $X leq Y$ then the integrals will have the same inequality.
$endgroup$
– Will M.
Feb 2 at 8:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $X$ and $Y$ are random variables defined on a probability space $(Omega ,Sigma ,P )$, and $$S={omegainOmega text{ such that }Y(omega)geq X(omega)}$$
Then $P(Omegabackslash S)=P(X> Y)=0$,
$$begin{split}
E(Y)&=int_{Omega}Y(omega)dP(omega) ,,,text {(by definition)}\
&= int_{S}Y(omega)dP(omega) + int_{Omegabackslash S}Y(omega)dP(omega) \
&= int_{S}Y(omega)dP(omega) + 0 \
&geq int_{S}X(omega)dP(omega) ,,,text {(by definition of $S$)} \
&geq E(X)
end{split}$$
answered Feb 1 at 17:50
Stefan LafonStefan Lafon
3,005212
$endgroup$
$begingroup$
Your last inequality should be equality.
$endgroup$
– yurnero
Feb 1 at 17:56
$begingroup$
Well, depends on how you tabulate the equations. I agree that the last two expressions are equal, but when I tabulate inequalities like this, I carry over the last inequality sign. In other words, the inequality sign is between the RHS, and the original LHS, that is $E(Y)$.
$endgroup$
– Stefan Lafon
Feb 1 at 18:00
1
$begingroup$
As it is written, it is not wrong, as $=$ implies $geq$ but it looks strange stylistically IMO considering you already have $int_Omega Y=int_SY$ earlier. Put differently, if you are to write the same argument "horizontally": $$E(Y)=int_Omega Y=int_SYgeqint_SX=int_Omega X=E(X),$$ then having $geq$ in the next-to-last step just looks off.
$endgroup$
– yurnero
Feb 1 at 18:12
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
If $X$ and $Y$ are random variables defined on a probability space $(Omega ,Sigma ,P )$, and $$S={omegainOmega text{ such that }Y(omega)geq X(omega)}$$
Then $P(Omegabackslash S)=P(X> Y)=0$,
$$begin{split}
E(Y)&=int_{Omega}Y(omega)dP(omega) ,,,text {(by definition)}\
&= int_{S}Y(omega)dP(omega) + int_{Omegabackslash S}Y(omega)dP(omega) \
&= int_{S}Y(omega)dP(omega) + 0 \
&geq int_{S}X(omega)dP(omega) ,,,text {(by definition of $S$)} \
&geq E(X)
end{split}$$
answered Feb 1 at 17:50
Stefan LafonStefan Lafon
3,005212
$endgroup$
$begingroup$
Your last inequality should be equality.
$endgroup$
– yurnero
Feb 1 at 17:56
$begingroup$
Well, depends on how you tabulate the equations. I agree that the last two expressions are equal, but when I tabulate inequalities like this, I carry over the last inequality sign. In other words, the inequality sign is between the RHS, and the original LHS, that is $E(Y)$.
$endgroup$
– Stefan Lafon
Feb 1 at 18:00
1
$begingroup$
As it is written, it is not wrong, as $=$ implies $geq$ but it looks strange stylistically IMO considering you already have $int_Omega Y=int_SY$ earlier. Put differently, if you are to write the same argument "horizontally": $$E(Y)=int_Omega Y=int_SYgeqint_SX=int_Omega X=E(X),$$ then having $geq$ in the next-to-last step just looks off.
$endgroup$
– yurnero
Feb 1 at 18:12
add a comment |
$begingroup$
If $X$ and $Y$ are random variables defined on a probability space $(Omega ,Sigma ,P )$, and $$S={omegainOmega text{ such that }Y(omega)geq X(omega)}$$
Then $P(Omegabackslash S)=P(X> Y)=0$,
$$begin{split}
E(Y)&=int_{Omega}Y(omega)dP(omega) ,,,text {(by definition)}\
&= int_{S}Y(omega)dP(omega) + int_{Omegabackslash S}Y(omega)dP(omega) \
&= int_{S}Y(omega)dP(omega) + 0 \
&geq int_{S}X(omega)dP(omega) ,,,text {(by definition of $S$)} \
&geq E(X)
end{split}$$
answered Feb 1 at 17:50
Stefan LafonStefan Lafon
3,005212
$endgroup$
$begingroup$
Your last inequality should be equality.
$endgroup$
– yurnero
Feb 1 at 17:56
$begingroup$
Well, depends on how you tabulate the equations. I agree that the last two expressions are equal, but when I tabulate inequalities like this, I carry over the last inequality sign. In other words, the inequality sign is between the RHS, and the original LHS, that is $E(Y)$.
$endgroup$
– Stefan Lafon
Feb 1 at 18:00
1
$begingroup$
As it is written, it is not wrong, as $=$ implies $geq$ but it looks strange stylistically IMO considering you already have $int_Omega Y=int_SY$ earlier. Put differently, if you are to write the same argument "horizontally": $$E(Y)=int_Omega Y=int_SYgeqint_SX=int_Omega X=E(X),$$ then having $geq$ in the next-to-last step just looks off.
$endgroup$
– yurnero
Feb 1 at 18:12
add a comment |
$begingroup$
If $X$ and $Y$ are random variables defined on a probability space $(Omega ,Sigma ,P )$, and $$S={omegainOmega text{ such that }Y(omega)geq X(omega)}$$
Then $P(Omegabackslash S)=P(X> Y)=0$,
$$begin{split}
E(Y)&=int_{Omega}Y(omega)dP(omega) ,,,text {(by definition)}\
&= int_{S}Y(omega)dP(omega) + int_{Omegabackslash S}Y(omega)dP(omega) \
&= int_{S}Y(omega)dP(omega) + 0 \
&geq int_{S}X(omega)dP(omega) ,,,text {(by definition of $S$)} \
&geq E(X)
end{split}$$
answered Feb 1 at 17:50
Stefan LafonStefan Lafon
3,005212
$endgroup$
If $X$ and $Y$ are random variables defined on a probability space $(Omega ,Sigma ,P )$, and $$S={omegainOmega text{ such that }Y(omega)geq X(omega)}$$
Then $P(Omegabackslash S)=P(X> Y)=0$,
$$begin{split}
E(Y)&=int_{Omega}Y(omega)dP(omega) ,,,text {(by definition)}\
&= int_{S}Y(omega)dP(omega) + int_{Omegabackslash S}Y(omega)dP(omega) \
&= int_{S}Y(omega)dP(omega) + 0 \
&geq int_{S}X(omega)dP(omega) ,,,text {(by definition of $S$)} \
&geq E(X)
end{split}$$
answered Feb 1 at 17:50
Stefan LafonStefan Lafon
3,005212
answered Feb 1 at 17:50
Stefan LafonStefan Lafon
3,005212
answered Feb 1 at 17:50
Stefan LafonStefan Lafon
3,005212
answered Feb 1 at 17:50
Stefan LafonStefan Lafon
3,005212
3,005212
$begingroup$
Your last inequality should be equality.
$endgroup$
– yurnero
Feb 1 at 17:56
$begingroup$
Well, depends on how you tabulate the equations. I agree that the last two expressions are equal, but when I tabulate inequalities like this, I carry over the last inequality sign. In other words, the inequality sign is between the RHS, and the original LHS, that is $E(Y)$.
$endgroup$
– Stefan Lafon
Feb 1 at 18:00
1
$begingroup$
As it is written, it is not wrong, as $=$ implies $geq$ but it looks strange stylistically IMO considering you already have $int_Omega Y=int_SY$ earlier. Put differently, if you are to write the same argument "horizontally": $$E(Y)=int_Omega Y=int_SYgeqint_SX=int_Omega X=E(X),$$ then having $geq$ in the next-to-last step just looks off.
$endgroup$
– yurnero
Feb 1 at 18:12
add a comment |
$begingroup$
Your last inequality should be equality.
$endgroup$
– yurnero
Feb 1 at 17:56
$begingroup$
Well, depends on how you tabulate the equations. I agree that the last two expressions are equal, but when I tabulate inequalities like this, I carry over the last inequality sign. In other words, the inequality sign is between the RHS, and the original LHS, that is $E(Y)$.
$endgroup$
– Stefan Lafon
Feb 1 at 18:00
1
$begingroup$
As it is written, it is not wrong, as $=$ implies $geq$ but it looks strange stylistically IMO considering you already have $int_Omega Y=int_SY$ earlier. Put differently, if you are to write the same argument "horizontally": $$E(Y)=int_Omega Y=int_SYgeqint_SX=int_Omega X=E(X),$$ then having $geq$ in the next-to-last step just looks off.
$endgroup$
– yurnero
Feb 1 at 18:12
$begingroup$
Your last inequality should be equality.
$endgroup$
– yurnero
Feb 1 at 17:56
$begingroup$
Your last inequality should be equality.
$endgroup$
– yurnero
Feb 1 at 17:56
$begingroup$
Well, depends on how you tabulate the equations. I agree that the last two expressions are equal, but when I tabulate inequalities like this, I carry over the last inequality sign. In other words, the inequality sign is between the RHS, and the original LHS, that is $E(Y)$.
$endgroup$
– Stefan Lafon
Feb 1 at 18:00
$begingroup$
Well, depends on how you tabulate the equations. I agree that the last two expressions are equal, but when I tabulate inequalities like this, I carry over the last inequality sign. In other words, the inequality sign is between the RHS, and the original LHS, that is $E(Y)$.
$endgroup$
– Stefan Lafon
Feb 1 at 18:00
1
1
$begingroup$
As it is written, it is not wrong, as $=$ implies $geq$ but it looks strange stylistically IMO considering you already have $int_Omega Y=int_SY$ earlier. Put differently, if you are to write the same argument "horizontally": $$E(Y)=int_Omega Y=int_SYgeqint_SX=int_Omega X=E(X),$$ then having $geq$ in the next-to-last step just looks off.
$endgroup$
– yurnero
Feb 1 at 18:12
$begingroup$
As it is written, it is not wrong, as $=$ implies $geq$ but it looks strange stylistically IMO considering you already have $int_Omega Y=int_SY$ earlier. Put differently, if you are to write the same argument "horizontally": $$E(Y)=int_Omega Y=int_SYgeqint_SX=int_Omega X=E(X),$$ then having $geq$ in the next-to-last step just looks off.
$endgroup$
– yurnero
Feb 1 at 18:12
add a comment |
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1
$begingroup$
Equivalent to "if $P(Y-X ge 0)=1$ then $E[Y]-E[X] ge 0$" which, since $E[Y]-E[X] =E[Y-X]$, is equivalent to "if $P(Z ge 0)=1$ then $E[Z] ge 0$"
$endgroup$
– Henry
Feb 1 at 18:20
$begingroup$
On a set of full measure, $X leq Y$ then the integrals will have the same inequality.
$endgroup$
– Will M.
Feb 2 at 8:11