Mixing
Expected weight split in a sum of independent variables
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Let $x,y geq 0$ be independent random variables with the property that $E[x] = E[y]$. Can I infer that
$$P(x geq E[x] mid x + y geq E[x+y]) = 1/2~?$$
My heuristic reasoning is that, since $x$ and $y$ have equal expectations, the weight should be split "fairly" between them. But that, of course, falls far short of a solution.
How can I analyze this rigorously?
probability expectation conditional-expectation
asked Jul 18 at 20:41
Elliot Gorokhovsky
1,051623
add a comment |Â
up vote
0
down vote
favorite
Let $x,y geq 0$ be independent random variables with the property that $E[x] = E[y]$. Can I infer that
$$P(x geq E[x] mid x + y geq E[x+y]) = 1/2~?$$
My heuristic reasoning is that, since $x$ and $y$ have equal expectations, the weight should be split "fairly" between them. But that, of course, falls far short of a solution.
How can I analyze this rigorously?
probability expectation conditional-expectation
asked Jul 18 at 20:41
Elliot Gorokhovsky
1,051623
Are you sure you want $t$ as a probability? That doesn't seem to make sense. If we rescale $x$, $y$ and $t$ by some factor, everything is invariant and the probability should be the same, but as you've written it, it scales with $t$. It seems that the right-hand side of the equation should be, if anything, $frac12$?
– joriki
Jul 18 at 20:48
@joriki You're right, I made an error. See corrected question.
– Elliot Gorokhovsky
Jul 18 at 20:50
@joriki It was still wrong. I just got rid of $t$ altogether; now I have 1/2 like you said.
– Elliot Gorokhovsky
Jul 18 at 20:53
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $x,y geq 0$ be independent random variables with the property that $E[x] = E[y]$. Can I infer that
$$P(x geq E[x] mid x + y geq E[x+y]) = 1/2~?$$
My heuristic reasoning is that, since $x$ and $y$ have equal expectations, the weight should be split "fairly" between them. But that, of course, falls far short of a solution.
How can I analyze this rigorously?
probability expectation conditional-expectation
asked Jul 18 at 20:41
Elliot Gorokhovsky
1,051623
Let $x,y geq 0$ be independent random variables with the property that $E[x] = E[y]$. Can I infer that
$$P(x geq E[x] mid x + y geq E[x+y]) = 1/2~?$$
My heuristic reasoning is that, since $x$ and $y$ have equal expectations, the weight should be split "fairly" between them. But that, of course, falls far short of a solution.
How can I analyze this rigorously?
probability expectation conditional-expectation
asked Jul 18 at 20:41
Elliot Gorokhovsky
1,051623
edited Jul 18 at 21:13
asked Jul 18 at 20:41
Elliot Gorokhovsky
1,051623
asked Jul 18 at 20:41
Elliot Gorokhovsky
1,051623
asked Jul 18 at 20:41
Elliot Gorokhovsky
1,051623
1,051623
Are you sure you want $t$ as a probability? That doesn't seem to make sense. If we rescale $x$, $y$ and $t$ by some factor, everything is invariant and the probability should be the same, but as you've written it, it scales with $t$. It seems that the right-hand side of the equation should be, if anything, $frac12$?
– joriki
Jul 18 at 20:48
@joriki You're right, I made an error. See corrected question.
– Elliot Gorokhovsky
Jul 18 at 20:50
@joriki It was still wrong. I just got rid of $t$ altogether; now I have 1/2 like you said.
– Elliot Gorokhovsky
Jul 18 at 20:53
add a comment |Â
Are you sure you want $t$ as a probability? That doesn't seem to make sense. If we rescale $x$, $y$ and $t$ by some factor, everything is invariant and the probability should be the same, but as you've written it, it scales with $t$. It seems that the right-hand side of the equation should be, if anything, $frac12$?
– joriki
Jul 18 at 20:48
@joriki You're right, I made an error. See corrected question.
– Elliot Gorokhovsky
Jul 18 at 20:50
@joriki It was still wrong. I just got rid of $t$ altogether; now I have 1/2 like you said.
– Elliot Gorokhovsky
Jul 18 at 20:53
Are you sure you want $t$ as a probability? That doesn't seem to make sense. If we rescale $x$, $y$ and $t$ by some factor, everything is invariant and the probability should be the same, but as you've written it, it scales with $t$. It seems that the right-hand side of the equation should be, if anything, $frac12$?
– joriki
Jul 18 at 20:48
Are you sure you want $t$ as a probability? That doesn't seem to make sense. If we rescale $x$, $y$ and $t$ by some factor, everything is invariant and the probability should be the same, but as you've written it, it scales with $t$. It seems that the right-hand side of the equation should be, if anything, $frac12$?
– joriki
Jul 18 at 20:48
@joriki You're right, I made an error. See corrected question.
– Elliot Gorokhovsky
Jul 18 at 20:50
@joriki You're right, I made an error. See corrected question.
– Elliot Gorokhovsky
Jul 18 at 20:50
@joriki It was still wrong. I just got rid of $t$ altogether; now I have 1/2 like you said.
– Elliot Gorokhovsky
Jul 18 at 20:53
@joriki It was still wrong. I just got rid of $t$ altogether; now I have 1/2 like you said.
– Elliot Gorokhovsky
Jul 18 at 20:53
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
An even simpler example would be to take a deterministic $X$, i.e., take for example $X=c$ (a.s), you could take $c$ any number you want. Then, $mathbbE[X] = c$ and $mathbbP(X geq mathbbE[X]) = 1$ which implies that $mathbbP(Xgeq mathbbE[X] | X+Y geq mathbbE[X+Y]) =1$ because $Xgeq mathbbE[X]$ always holds (in the deterministic case).
answered Jul 18 at 21:59
Stan Tendijck
1,277110
d'oh! thanks. [fill.]
– Elliot Gorokhovsky
Jul 18 at 22:34
add a comment |Â
up vote
3
down vote
No, you can't infer that. For simplicity, assume $E[x]=E[y]=0$. Then your equation reads
$$
P(xge0mid x+yge0)=frac12;.
$$
Consider $x=pm2$ with probability $frac12$ each, and $y=pm1$ with probability $frac12$ each. Then
$$
P(xge0mid x+yge0)=1;.
$$
answered Jul 18 at 21:01
joriki
164k10180328
Oh, I forgot about cancellation. I'm actually thinking of $x$ and $y$ positive; is there a counterexample satisfying that? I edited the question to reflect this.
– Elliot Gorokhovsky
Jul 18 at 21:13
add a comment |Â
up vote
2
down vote
No. Take the case of $X$ being a Bernoulli with parameter $1/3$, and $Y$ being constant equal to $1/3$.
We have $$mathbbP X geq mathbbE[X] = mathbbE[X]= mathbbE[Y] = 1/3,,$$
but
$$
mathbbP X geq mathbbE[X] mid X+Y geq mathbbE[X+Y]
= mathbbP X geq 1/3 mid X geq 1/3
=1,.
$$
answered Jul 18 at 21:40
Clement C.
47.2k33682
In fact, if $Y$ is constant, then for any random variable $X$ with finite expectation, $$Pr[X ge operatornameE[X] mid X+Y ge operatornameE[X+Y]] = Pr[X ge operatornameE[X] mid X ge operatornameE[X]] = 1.$$
– heropup
Jul 18 at 21:59
@heropup True. But I believe that a clear down-to-earth counterexample is better, as in more "impactful."
– Clement C.
Jul 18 at 22:01
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
An even simpler example would be to take a deterministic $X$, i.e., take for example $X=c$ (a.s), you could take $c$ any number you want. Then, $mathbbE[X] = c$ and $mathbbP(X geq mathbbE[X]) = 1$ which implies that $mathbbP(Xgeq mathbbE[X] | X+Y geq mathbbE[X+Y]) =1$ because $Xgeq mathbbE[X]$ always holds (in the deterministic case).
answered Jul 18 at 21:59
Stan Tendijck
1,277110
d'oh! thanks. [fill.]
– Elliot Gorokhovsky
Jul 18 at 22:34
add a comment |Â
up vote
1
down vote
accepted
An even simpler example would be to take a deterministic $X$, i.e., take for example $X=c$ (a.s), you could take $c$ any number you want. Then, $mathbbE[X] = c$ and $mathbbP(X geq mathbbE[X]) = 1$ which implies that $mathbbP(Xgeq mathbbE[X] | X+Y geq mathbbE[X+Y]) =1$ because $Xgeq mathbbE[X]$ always holds (in the deterministic case).
answered Jul 18 at 21:59
Stan Tendijck
1,277110
d'oh! thanks. [fill.]
– Elliot Gorokhovsky
Jul 18 at 22:34
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
An even simpler example would be to take a deterministic $X$, i.e., take for example $X=c$ (a.s), you could take $c$ any number you want. Then, $mathbbE[X] = c$ and $mathbbP(X geq mathbbE[X]) = 1$ which implies that $mathbbP(Xgeq mathbbE[X] | X+Y geq mathbbE[X+Y]) =1$ because $Xgeq mathbbE[X]$ always holds (in the deterministic case).
answered Jul 18 at 21:59
Stan Tendijck
1,277110
An even simpler example would be to take a deterministic $X$, i.e., take for example $X=c$ (a.s), you could take $c$ any number you want. Then, $mathbbE[X] = c$ and $mathbbP(X geq mathbbE[X]) = 1$ which implies that $mathbbP(Xgeq mathbbE[X] | X+Y geq mathbbE[X+Y]) =1$ because $Xgeq mathbbE[X]$ always holds (in the deterministic case).
answered Jul 18 at 21:59
Stan Tendijck
1,277110
answered Jul 18 at 21:59
Stan Tendijck
1,277110
answered Jul 18 at 21:59
Stan Tendijck
1,277110
answered Jul 18 at 21:59
Stan Tendijck
1,277110
1,277110
d'oh! thanks. [fill.]
– Elliot Gorokhovsky
Jul 18 at 22:34
add a comment |Â
d'oh! thanks. [fill.]
– Elliot Gorokhovsky
Jul 18 at 22:34
d'oh! thanks. [fill.]
– Elliot Gorokhovsky
Jul 18 at 22:34
d'oh! thanks. [fill.]
– Elliot Gorokhovsky
Jul 18 at 22:34
add a comment |Â
up vote
3
down vote
No, you can't infer that. For simplicity, assume $E[x]=E[y]=0$. Then your equation reads
$$
P(xge0mid x+yge0)=frac12;.
$$
Consider $x=pm2$ with probability $frac12$ each, and $y=pm1$ with probability $frac12$ each. Then
$$
P(xge0mid x+yge0)=1;.
$$
answered Jul 18 at 21:01
joriki
164k10180328
Oh, I forgot about cancellation. I'm actually thinking of $x$ and $y$ positive; is there a counterexample satisfying that? I edited the question to reflect this.
– Elliot Gorokhovsky
Jul 18 at 21:13
add a comment |Â
up vote
3
down vote
No, you can't infer that. For simplicity, assume $E[x]=E[y]=0$. Then your equation reads
$$
P(xge0mid x+yge0)=frac12;.
$$
Consider $x=pm2$ with probability $frac12$ each, and $y=pm1$ with probability $frac12$ each. Then
$$
P(xge0mid x+yge0)=1;.
$$
answered Jul 18 at 21:01
joriki
164k10180328
Oh, I forgot about cancellation. I'm actually thinking of $x$ and $y$ positive; is there a counterexample satisfying that? I edited the question to reflect this.
– Elliot Gorokhovsky
Jul 18 at 21:13
add a comment |Â
up vote
3
down vote
up vote
3
down vote
No, you can't infer that. For simplicity, assume $E[x]=E[y]=0$. Then your equation reads
$$
P(xge0mid x+yge0)=frac12;.
$$
Consider $x=pm2$ with probability $frac12$ each, and $y=pm1$ with probability $frac12$ each. Then
$$
P(xge0mid x+yge0)=1;.
$$
answered Jul 18 at 21:01
joriki
164k10180328
No, you can't infer that. For simplicity, assume $E[x]=E[y]=0$. Then your equation reads
$$
P(xge0mid x+yge0)=frac12;.
$$
Consider $x=pm2$ with probability $frac12$ each, and $y=pm1$ with probability $frac12$ each. Then
$$
P(xge0mid x+yge0)=1;.
$$
answered Jul 18 at 21:01
joriki
164k10180328
answered Jul 18 at 21:01
joriki
164k10180328
answered Jul 18 at 21:01
joriki
164k10180328
answered Jul 18 at 21:01
joriki
164k10180328
164k10180328
Oh, I forgot about cancellation. I'm actually thinking of $x$ and $y$ positive; is there a counterexample satisfying that? I edited the question to reflect this.
– Elliot Gorokhovsky
Jul 18 at 21:13
add a comment |Â
Oh, I forgot about cancellation. I'm actually thinking of $x$ and $y$ positive; is there a counterexample satisfying that? I edited the question to reflect this.
– Elliot Gorokhovsky
Jul 18 at 21:13
Oh, I forgot about cancellation. I'm actually thinking of $x$ and $y$ positive; is there a counterexample satisfying that? I edited the question to reflect this.
– Elliot Gorokhovsky
Jul 18 at 21:13
Oh, I forgot about cancellation. I'm actually thinking of $x$ and $y$ positive; is there a counterexample satisfying that? I edited the question to reflect this.
– Elliot Gorokhovsky
Jul 18 at 21:13
add a comment |Â
up vote
2
down vote
No. Take the case of $X$ being a Bernoulli with parameter $1/3$, and $Y$ being constant equal to $1/3$.
We have $$mathbbP X geq mathbbE[X] = mathbbE[X]= mathbbE[Y] = 1/3,,$$
but
$$
mathbbP X geq mathbbE[X] mid X+Y geq mathbbE[X+Y]
= mathbbP X geq 1/3 mid X geq 1/3
=1,.
$$
answered Jul 18 at 21:40
Clement C.
47.2k33682
In fact, if $Y$ is constant, then for any random variable $X$ with finite expectation, $$Pr[X ge operatornameE[X] mid X+Y ge operatornameE[X+Y]] = Pr[X ge operatornameE[X] mid X ge operatornameE[X]] = 1.$$
– heropup
Jul 18 at 21:59
@heropup True. But I believe that a clear down-to-earth counterexample is better, as in more "impactful."
– Clement C.
Jul 18 at 22:01
add a comment |Â
up vote
2
down vote
No. Take the case of $X$ being a Bernoulli with parameter $1/3$, and $Y$ being constant equal to $1/3$.
We have $$mathbbP X geq mathbbE[X] = mathbbE[X]= mathbbE[Y] = 1/3,,$$
but
$$
mathbbP X geq mathbbE[X] mid X+Y geq mathbbE[X+Y]
= mathbbP X geq 1/3 mid X geq 1/3
=1,.
$$
answered Jul 18 at 21:40
Clement C.
47.2k33682
In fact, if $Y$ is constant, then for any random variable $X$ with finite expectation, $$Pr[X ge operatornameE[X] mid X+Y ge operatornameE[X+Y]] = Pr[X ge operatornameE[X] mid X ge operatornameE[X]] = 1.$$
– heropup
Jul 18 at 21:59
@heropup True. But I believe that a clear down-to-earth counterexample is better, as in more "impactful."
– Clement C.
Jul 18 at 22:01
add a comment |Â
up vote
2
down vote
up vote
2
down vote
No. Take the case of $X$ being a Bernoulli with parameter $1/3$, and $Y$ being constant equal to $1/3$.
We have $$mathbbP X geq mathbbE[X] = mathbbE[X]= mathbbE[Y] = 1/3,,$$
but
$$
mathbbP X geq mathbbE[X] mid X+Y geq mathbbE[X+Y]
= mathbbP X geq 1/3 mid X geq 1/3
=1,.
$$
answered Jul 18 at 21:40
Clement C.
47.2k33682
No. Take the case of $X$ being a Bernoulli with parameter $1/3$, and $Y$ being constant equal to $1/3$.
We have $$mathbbP X geq mathbbE[X] = mathbbE[X]= mathbbE[Y] = 1/3,,$$
but
$$
mathbbP X geq mathbbE[X] mid X+Y geq mathbbE[X+Y]
= mathbbP X geq 1/3 mid X geq 1/3
=1,.
$$
answered Jul 18 at 21:40
Clement C.
47.2k33682
answered Jul 18 at 21:40
Clement C.
47.2k33682
answered Jul 18 at 21:40
Clement C.
47.2k33682
answered Jul 18 at 21:40
Clement C.
47.2k33682
47.2k33682
In fact, if $Y$ is constant, then for any random variable $X$ with finite expectation, $$Pr[X ge operatornameE[X] mid X+Y ge operatornameE[X+Y]] = Pr[X ge operatornameE[X] mid X ge operatornameE[X]] = 1.$$
– heropup
Jul 18 at 21:59
@heropup True. But I believe that a clear down-to-earth counterexample is better, as in more "impactful."
– Clement C.
Jul 18 at 22:01
add a comment |Â
In fact, if $Y$ is constant, then for any random variable $X$ with finite expectation, $$Pr[X ge operatornameE[X] mid X+Y ge operatornameE[X+Y]] = Pr[X ge operatornameE[X] mid X ge operatornameE[X]] = 1.$$
– heropup
Jul 18 at 21:59
@heropup True. But I believe that a clear down-to-earth counterexample is better, as in more "impactful."
– Clement C.
Jul 18 at 22:01
In fact, if $Y$ is constant, then for any random variable $X$ with finite expectation, $$Pr[X ge operatornameE[X] mid X+Y ge operatornameE[X+Y]] = Pr[X ge operatornameE[X] mid X ge operatornameE[X]] = 1.$$
– heropup
Jul 18 at 21:59
In fact, if $Y$ is constant, then for any random variable $X$ with finite expectation, $$Pr[X ge operatornameE[X] mid X+Y ge operatornameE[X+Y]] = Pr[X ge operatornameE[X] mid X ge operatornameE[X]] = 1.$$
– heropup
Jul 18 at 21:59
@heropup True. But I believe that a clear down-to-earth counterexample is better, as in more "impactful."
– Clement C.
Jul 18 at 22:01
@heropup True. But I believe that a clear down-to-earth counterexample is better, as in more "impactful."
– Clement C.
Jul 18 at 22:01
add a comment |Â
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Are you sure you want $t$ as a probability? That doesn't seem to make sense. If we rescale $x$, $y$ and $t$ by some factor, everything is invariant and the probability should be the same, but as you've written it, it scales with $t$. It seems that the right-hand side of the equation should be, if anything, $frac12$?
– joriki
Jul 18 at 20:48
@joriki You're right, I made an error. See corrected question.
– Elliot Gorokhovsky
Jul 18 at 20:50
@joriki It was still wrong. I just got rid of $t$ altogether; now I have 1/2 like you said.
– Elliot Gorokhovsky
Jul 18 at 20:53