Find $ E(X mid X^2 + Y^2) $ for $X,Y$ independent standard normal [closed]
$begingroup$
Let $X,Y$ independent random variables with $ X,Ysim mathcal{N}(0,1) $.
I want to find: $$ E(X mid X^2 + Y^2) $$
How can it be found? I would appreciate any tips or hints.
probability normal-distribution expected-value
asked Dec 3 '18 at 0:14
WywanaWywana
585
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closed as off-topic by Saad, Leucippus, Cesareo, John B, choco_addicted Dec 3 '18 at 11:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Leucippus, Cesareo, John B, choco_addicted
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $X,Y$ independent random variables with $ X,Ysim mathcal{N}(0,1) $.
I want to find: $$ E(X mid X^2 + Y^2) $$
How can it be found? I would appreciate any tips or hints.
probability normal-distribution expected-value
asked Dec 3 '18 at 0:14
WywanaWywana
585
$endgroup$
closed as off-topic by Saad, Leucippus, Cesareo, John B, choco_addicted Dec 3 '18 at 11:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Leucippus, Cesareo, John B, choco_addicted
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Wouldn't that be $0$ by symmetry? Since $mathbb{E}[Xmid X^2+Y^2] = mathbb{E}[(-X)mid (-X)^2+Y^2]=-mathbb{E}[Xmid X^2+Y^2]$ (the first equality as $X$ and $-X$ have same distribution, and are both independent of $Y$)
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– Clement C.
Dec 3 '18 at 0:24
1
$begingroup$
My hint would be to think in terms of symmetry. $X^2+Y^2$ tells you the distance from the origin, which doesn't tell you anything about which side of $X=0$ it is on.
$endgroup$
– spaceisdarkgreen
Dec 3 '18 at 0:25
add a comment |
$begingroup$
Let $X,Y$ independent random variables with $ X,Ysim mathcal{N}(0,1) $.
I want to find: $$ E(X mid X^2 + Y^2) $$
How can it be found? I would appreciate any tips or hints.
probability normal-distribution expected-value
asked Dec 3 '18 at 0:14
WywanaWywana
585
$endgroup$
Let $X,Y$ independent random variables with $ X,Ysim mathcal{N}(0,1) $.
I want to find: $$ E(X mid X^2 + Y^2) $$
How can it be found? I would appreciate any tips or hints.
probability normal-distribution expected-value
probability normal-distribution expected-value
asked Dec 3 '18 at 0:14
WywanaWywana
585
asked Dec 3 '18 at 0:14
WywanaWywana
585
asked Dec 3 '18 at 0:14
WywanaWywana
585
asked Dec 3 '18 at 0:14
WywanaWywana
585
asked Dec 3 '18 at 0:14
WywanaWywana
585
585
closed as off-topic by Saad, Leucippus, Cesareo, John B, choco_addicted Dec 3 '18 at 11:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Leucippus, Cesareo, John B, choco_addicted
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Leucippus, Cesareo, John B, choco_addicted Dec 3 '18 at 11:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Leucippus, Cesareo, John B, choco_addicted
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Wouldn't that be $0$ by symmetry? Since $mathbb{E}[Xmid X^2+Y^2] = mathbb{E}[(-X)mid (-X)^2+Y^2]=-mathbb{E}[Xmid X^2+Y^2]$ (the first equality as $X$ and $-X$ have same distribution, and are both independent of $Y$)
$endgroup$
– Clement C.
Dec 3 '18 at 0:24
1
$begingroup$
My hint would be to think in terms of symmetry. $X^2+Y^2$ tells you the distance from the origin, which doesn't tell you anything about which side of $X=0$ it is on.
$endgroup$
– spaceisdarkgreen
Dec 3 '18 at 0:25
add a comment |
1
$begingroup$
Wouldn't that be $0$ by symmetry? Since $mathbb{E}[Xmid X^2+Y^2] = mathbb{E}[(-X)mid (-X)^2+Y^2]=-mathbb{E}[Xmid X^2+Y^2]$ (the first equality as $X$ and $-X$ have same distribution, and are both independent of $Y$)
$endgroup$
– Clement C.
Dec 3 '18 at 0:24
1
$begingroup$
My hint would be to think in terms of symmetry. $X^2+Y^2$ tells you the distance from the origin, which doesn't tell you anything about which side of $X=0$ it is on.
$endgroup$
– spaceisdarkgreen
Dec 3 '18 at 0:25
1
1
$begingroup$
Wouldn't that be $0$ by symmetry? Since $mathbb{E}[Xmid X^2+Y^2] = mathbb{E}[(-X)mid (-X)^2+Y^2]=-mathbb{E}[Xmid X^2+Y^2]$ (the first equality as $X$ and $-X$ have same distribution, and are both independent of $Y$)
$endgroup$
– Clement C.
Dec 3 '18 at 0:24
$begingroup$
Wouldn't that be $0$ by symmetry? Since $mathbb{E}[Xmid X^2+Y^2] = mathbb{E}[(-X)mid (-X)^2+Y^2]=-mathbb{E}[Xmid X^2+Y^2]$ (the first equality as $X$ and $-X$ have same distribution, and are both independent of $Y$)
$endgroup$
– Clement C.
Dec 3 '18 at 0:24
1
1
$begingroup$
My hint would be to think in terms of symmetry. $X^2+Y^2$ tells you the distance from the origin, which doesn't tell you anything about which side of $X=0$ it is on.
$endgroup$
– spaceisdarkgreen
Dec 3 '18 at 0:25
$begingroup$
My hint would be to think in terms of symmetry. $X^2+Y^2$ tells you the distance from the origin, which doesn't tell you anything about which side of $X=0$ it is on.
$endgroup$
– spaceisdarkgreen
Dec 3 '18 at 0:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
By symmetry, you'll have
$$
mathbb{E}[Xmid X^2+Y^2] = 0,.$$
This is because $mathbb{E}[Xmid X^2+Y^2] = mathbb{E}[(-X)mid (-X)^2+Y^2]=-mathbb{E}[Xmid X^2+Y^2]$ (the first equality as $X$ and $-X$ have same distribution, and are both independent of $Y$; the second is linearity).
answered Dec 3 '18 at 0:27
Clement C.Clement C.
49.8k33886
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add a comment |
$begingroup$
The answer is $0$. To show this what you have to show is $int_{{X^{2}+Y^{2} leq t}}XdP =0$ for all $t in mathbb R$. But this is immediate from the fact that $(X,Y)$ has same distribution as $(-X,-Y)$.
answered Dec 3 '18 at 0:25
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By symmetry, you'll have
$$
mathbb{E}[Xmid X^2+Y^2] = 0,.$$
This is because $mathbb{E}[Xmid X^2+Y^2] = mathbb{E}[(-X)mid (-X)^2+Y^2]=-mathbb{E}[Xmid X^2+Y^2]$ (the first equality as $X$ and $-X$ have same distribution, and are both independent of $Y$; the second is linearity).
answered Dec 3 '18 at 0:27
Clement C.Clement C.
49.8k33886
$endgroup$
add a comment |
$begingroup$
By symmetry, you'll have
$$
mathbb{E}[Xmid X^2+Y^2] = 0,.$$
This is because $mathbb{E}[Xmid X^2+Y^2] = mathbb{E}[(-X)mid (-X)^2+Y^2]=-mathbb{E}[Xmid X^2+Y^2]$ (the first equality as $X$ and $-X$ have same distribution, and are both independent of $Y$; the second is linearity).
answered Dec 3 '18 at 0:27
Clement C.Clement C.
49.8k33886
$endgroup$
add a comment |
$begingroup$
By symmetry, you'll have
$$
mathbb{E}[Xmid X^2+Y^2] = 0,.$$
This is because $mathbb{E}[Xmid X^2+Y^2] = mathbb{E}[(-X)mid (-X)^2+Y^2]=-mathbb{E}[Xmid X^2+Y^2]$ (the first equality as $X$ and $-X$ have same distribution, and are both independent of $Y$; the second is linearity).
answered Dec 3 '18 at 0:27
Clement C.Clement C.
49.8k33886
$endgroup$
By symmetry, you'll have
$$
mathbb{E}[Xmid X^2+Y^2] = 0,.$$
This is because $mathbb{E}[Xmid X^2+Y^2] = mathbb{E}[(-X)mid (-X)^2+Y^2]=-mathbb{E}[Xmid X^2+Y^2]$ (the first equality as $X$ and $-X$ have same distribution, and are both independent of $Y$; the second is linearity).
answered Dec 3 '18 at 0:27
Clement C.Clement C.
49.8k33886
answered Dec 3 '18 at 0:27
Clement C.Clement C.
49.8k33886
answered Dec 3 '18 at 0:27
Clement C.Clement C.
49.8k33886
answered Dec 3 '18 at 0:27
Clement C.Clement C.
49.8k33886
49.8k33886
add a comment |
add a comment |
$begingroup$
The answer is $0$. To show this what you have to show is $int_{{X^{2}+Y^{2} leq t}}XdP =0$ for all $t in mathbb R$. But this is immediate from the fact that $(X,Y)$ has same distribution as $(-X,-Y)$.
answered Dec 3 '18 at 0:25
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
$endgroup$
add a comment |
$begingroup$
The answer is $0$. To show this what you have to show is $int_{{X^{2}+Y^{2} leq t}}XdP =0$ for all $t in mathbb R$. But this is immediate from the fact that $(X,Y)$ has same distribution as $(-X,-Y)$.
answered Dec 3 '18 at 0:25
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
$endgroup$
add a comment |
$begingroup$
The answer is $0$. To show this what you have to show is $int_{{X^{2}+Y^{2} leq t}}XdP =0$ for all $t in mathbb R$. But this is immediate from the fact that $(X,Y)$ has same distribution as $(-X,-Y)$.
answered Dec 3 '18 at 0:25
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
$endgroup$
The answer is $0$. To show this what you have to show is $int_{{X^{2}+Y^{2} leq t}}XdP =0$ for all $t in mathbb R$. But this is immediate from the fact that $(X,Y)$ has same distribution as $(-X,-Y)$.
answered Dec 3 '18 at 0:25
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
answered Dec 3 '18 at 0:25
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
answered Dec 3 '18 at 0:25
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
answered Dec 3 '18 at 0:25
Kavi Rama MurthyKavi Rama Murthy
53.9k32055
53.9k32055
add a comment |
add a comment |
1
$begingroup$
Wouldn't that be $0$ by symmetry? Since $mathbb{E}[Xmid X^2+Y^2] = mathbb{E}[(-X)mid (-X)^2+Y^2]=-mathbb{E}[Xmid X^2+Y^2]$ (the first equality as $X$ and $-X$ have same distribution, and are both independent of $Y$)
$endgroup$
– Clement C.
Dec 3 '18 at 0:24
1
$begingroup$
My hint would be to think in terms of symmetry. $X^2+Y^2$ tells you the distance from the origin, which doesn't tell you anything about which side of $X=0$ it is on.
$endgroup$
– spaceisdarkgreen
Dec 3 '18 at 0:25