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nonlinear transform of Gaussian random variable that preserves Gaussianity

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3

1

$begingroup$

I recently know that following results.

suppose that $x_1, x_2, x_3$ are independent real Gaussian random variables with $mathcalN(0, 1)$. Then

$$
fracx_1 + x_2 x_3sqrt1+x_3^2 sim mathcalN(0, 1)
$$

We can prove this result by direct computing. But I am wondering if there is a simpler way. Also, since this result is interesting. I am wondering if there is any generalization

Thanks

random-variables random

share|cite|improve this question

asked Sep 1 '13 at 18:04

Yan ZhuYan Zhu

314210

$endgroup$

add a comment | 

3

1

$begingroup$

I recently know that following results.

suppose that $x_1, x_2, x_3$ are independent real Gaussian random variables with $mathcalN(0, 1)$. Then

$$
fracx_1 + x_2 x_3sqrt1+x_3^2 sim mathcalN(0, 1)
$$

We can prove this result by direct computing. But I am wondering if there is a simpler way. Also, since this result is interesting. I am wondering if there is any generalization

Thanks

random-variables random

share|cite|improve this question

asked Sep 1 '13 at 18:04

Yan ZhuYan Zhu

314210

$endgroup$

add a comment | 

$begingroup$

I recently know that following results.

suppose that $x_1, x_2, x_3$ are independent real Gaussian random variables with $mathcalN(0, 1)$. Then

$$
fracx_1 + x_2 x_3sqrt1+x_3^2 sim mathcalN(0, 1)
$$

We can prove this result by direct computing. But I am wondering if there is a simpler way. Also, since this result is interesting. I am wondering if there is any generalization

Thanks

random-variables random

share|cite|improve this question

asked Sep 1 '13 at 18:04

Yan ZhuYan Zhu

314210

$endgroup$

share|cite|improve this question

asked Sep 1 '13 at 18:04

Yan ZhuYan Zhu

314210

share|cite|improve this question

asked Sep 1 '13 at 18:04

Yan ZhuYan Zhu

314210

asked Sep 1 '13 at 18:04

Yan ZhuYan Zhu

314210

asked Sep 1 '13 at 18:04

Yan ZhuYan Zhu

314210

2 Answers
2

active

oldest

votes

2

$begingroup$

The point is that the conditional distribution of your random variable given $x_3$ is always $cal N(0,1)$. One generalization is this. Suppose
$X_1, ldots, X_n$ are independent $cal N(0,1)$ random variables, and $bf Y = (Y_1, ldots, Y_n)$ is a vector-valued random variable independent of $X_1, ldots, X_n$ and supported on the sphere $Y_1^2 + ldots + Y_n^2 = 1$. Then
$bf X cdot bf Y = X_1 Y_1 + ldots + X_n Y_n sim cal N(0,1)$.

share|cite|improve this answer

answered Sep 1 '13 at 18:35

Robert IsraelRobert Israel

330k23219473

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow...It's simple...I am rusty after three years left school.
  $endgroup$
  – Yan Zhu
  Sep 1 '13 at 22:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hmm, but the OP has a non-linear combination.
  $endgroup$
  – Walter
  Dec 21 '13 at 13:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @walter: What's your point? Given $x_3$, it is a linear combination of $x_1$ and $x_2$. And if the conditional distribution given $x_3$ is the same for all values of $x_3$, then that conditional distribution is the (unconditional) distribution.
  $endgroup$
  – Robert Israel
  Dec 22 '13 at 4:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertIsrael I missed that your $boldsymbolY$ was a random variable.
  $endgroup$
  – Walter
  Dec 22 '13 at 15:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I extended this answer here math.stackexchange.com/q/1404329/17474 when there is no linear relationship between $x_1$ and $x_2$.
  $endgroup$
  – Léo Léopold Hertz 준영
  Aug 20 '15 at 21:14
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Following Robert's answer, I would like to clarify that if your RVs, let's say, $z$ and $y$ satisfy

$P(z|y)$ ~ Normal with fixed parameters

$P(y)$ ~ Normal with fixed parameters

then we can condition and marginalize to get $P(z)$

$$P(z) = int_Y^ P(z|y)P(y) dy$$
And it can be shown that $P(z)$ follows a Normal distribution with a certain mean and variance.

share|cite|improve this answer

edited Mar 29 at 17:31

Daniele Tampieri

2,66221022

answered Mar 29 at 15:57

Mariano BurichMariano Burich

213

$endgroup$

add a comment | 

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2

$begingroup$

The point is that the conditional distribution of your random variable given $x_3$ is always $cal N(0,1)$. One generalization is this. Suppose
$X_1, ldots, X_n$ are independent $cal N(0,1)$ random variables, and $bf Y = (Y_1, ldots, Y_n)$ is a vector-valued random variable independent of $X_1, ldots, X_n$ and supported on the sphere $Y_1^2 + ldots + Y_n^2 = 1$. Then
$bf X cdot bf Y = X_1 Y_1 + ldots + X_n Y_n sim cal N(0,1)$.

share|cite|improve this answer

answered Sep 1 '13 at 18:35

Robert IsraelRobert Israel

330k23219473

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow...It's simple...I am rusty after three years left school.
  $endgroup$
  – Yan Zhu
  Sep 1 '13 at 22:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hmm, but the OP has a non-linear combination.
  $endgroup$
  – Walter
  Dec 21 '13 at 13:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @walter: What's your point? Given $x_3$, it is a linear combination of $x_1$ and $x_2$. And if the conditional distribution given $x_3$ is the same for all values of $x_3$, then that conditional distribution is the (unconditional) distribution.
  $endgroup$
  – Robert Israel
  Dec 22 '13 at 4:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertIsrael I missed that your $boldsymbolY$ was a random variable.
  $endgroup$
  – Walter
  Dec 22 '13 at 15:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I extended this answer here math.stackexchange.com/q/1404329/17474 when there is no linear relationship between $x_1$ and $x_2$.
  $endgroup$
  – Léo Léopold Hertz 준영
  Aug 20 '15 at 21:14
  
  

  
  
  
  

add a comment | 

2

$begingroup$

The point is that the conditional distribution of your random variable given $x_3$ is always $cal N(0,1)$. One generalization is this. Suppose
$X_1, ldots, X_n$ are independent $cal N(0,1)$ random variables, and $bf Y = (Y_1, ldots, Y_n)$ is a vector-valued random variable independent of $X_1, ldots, X_n$ and supported on the sphere $Y_1^2 + ldots + Y_n^2 = 1$. Then
$bf X cdot bf Y = X_1 Y_1 + ldots + X_n Y_n sim cal N(0,1)$.

share|cite|improve this answer

answered Sep 1 '13 at 18:35

Robert IsraelRobert Israel

330k23219473

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow...It's simple...I am rusty after three years left school.
  $endgroup$
  – Yan Zhu
  Sep 1 '13 at 22:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hmm, but the OP has a non-linear combination.
  $endgroup$
  – Walter
  Dec 21 '13 at 13:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @walter: What's your point? Given $x_3$, it is a linear combination of $x_1$ and $x_2$. And if the conditional distribution given $x_3$ is the same for all values of $x_3$, then that conditional distribution is the (unconditional) distribution.
  $endgroup$
  – Robert Israel
  Dec 22 '13 at 4:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertIsrael I missed that your $boldsymbolY$ was a random variable.
  $endgroup$
  – Walter
  Dec 22 '13 at 15:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I extended this answer here math.stackexchange.com/q/1404329/17474 when there is no linear relationship between $x_1$ and $x_2$.
  $endgroup$
  – Léo Léopold Hertz 준영
  Aug 20 '15 at 21:14
  
  

  
  
  
  

add a comment | 

$begingroup$

The point is that the conditional distribution of your random variable given $x_3$ is always $cal N(0,1)$. One generalization is this. Suppose
$X_1, ldots, X_n$ are independent $cal N(0,1)$ random variables, and $bf Y = (Y_1, ldots, Y_n)$ is a vector-valued random variable independent of $X_1, ldots, X_n$ and supported on the sphere $Y_1^2 + ldots + Y_n^2 = 1$. Then
$bf X cdot bf Y = X_1 Y_1 + ldots + X_n Y_n sim cal N(0,1)$.

share|cite|improve this answer

answered Sep 1 '13 at 18:35

Robert IsraelRobert Israel

330k23219473

$endgroup$

share|cite|improve this answer

answered Sep 1 '13 at 18:35

Robert IsraelRobert Israel

330k23219473

answered Sep 1 '13 at 18:35

Robert IsraelRobert Israel

330k23219473

answered Sep 1 '13 at 18:35

Robert IsraelRobert Israel

330k23219473

0

$begingroup$

Following Robert's answer, I would like to clarify that if your RVs, let's say, $z$ and $y$ satisfy

$P(z|y)$ ~ Normal with fixed parameters

$P(y)$ ~ Normal with fixed parameters

then we can condition and marginalize to get $P(z)$

$$P(z) = int_Y^ P(z|y)P(y) dy$$
And it can be shown that $P(z)$ follows a Normal distribution with a certain mean and variance.

share|cite|improve this answer

edited Mar 29 at 17:31

Daniele Tampieri

2,66221022

answered Mar 29 at 15:57

Mariano BurichMariano Burich

213

$endgroup$

add a comment | 

0

$begingroup$

Following Robert's answer, I would like to clarify that if your RVs, let's say, $z$ and $y$ satisfy

$P(z|y)$ ~ Normal with fixed parameters

$P(y)$ ~ Normal with fixed parameters

then we can condition and marginalize to get $P(z)$

$$P(z) = int_Y^ P(z|y)P(y) dy$$
And it can be shown that $P(z)$ follows a Normal distribution with a certain mean and variance.

share|cite|improve this answer

edited Mar 29 at 17:31

Daniele Tampieri

2,66221022

answered Mar 29 at 15:57

Mariano BurichMariano Burich

213

$endgroup$

add a comment | 

$begingroup$

Following Robert's answer, I would like to clarify that if your RVs, let's say, $z$ and $y$ satisfy

$P(z|y)$ ~ Normal with fixed parameters

$P(y)$ ~ Normal with fixed parameters

then we can condition and marginalize to get $P(z)$

$$P(z) = int_Y^ P(z|y)P(y) dy$$
And it can be shown that $P(z)$ follows a Normal distribution with a certain mean and variance.

share|cite|improve this answer

edited Mar 29 at 17:31

Daniele Tampieri

2,66221022

answered Mar 29 at 15:57

Mariano BurichMariano Burich

213

$endgroup$

share|cite|improve this answer

edited Mar 29 at 17:31

Daniele Tampieri

2,66221022

answered Mar 29 at 15:57

Mariano BurichMariano Burich

213

edited Mar 29 at 17:31

Daniele Tampieri

2,66221022

edited Mar 29 at 17:31

Daniele Tampieri

2,66221022

answered Mar 29 at 15:57

Mariano BurichMariano Burich

213

answered Mar 29 at 15:57

Mariano BurichMariano Burich

213