Distribution of difference of two normally distributed random variables divided by square root of 2

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In Statistical Inference by Casella & Berger, in the proof of a theorem (theorem 5.3.1, page 220 in particular) I came across with the following proposition:

"the distribution of $fracX_2 - X_1sqrt 2$ is $N(0,1)$" where $X_2, X_1$ are iid normally random variables with the same mean $mu$ and variance $sigma^2$.

But despite of hours of trying, I couldn't prove this proposition even if it seems quite straightforward. I tried to use MGFs:

beginalign*
M_fracX_2-X_1sqrt 2(t)&=Eleft[e^tfracX_2-X_1sqrt 2right]\
&=Eleft[e^tfracX_2sqrt 2right]Eleft[e^tfrac-X_1sqrt 2right]\
&=left(e^frac1sqrt 2mu t+frac14t^2sigma ^2right)left(e^frac-1sqrt 2mu t+frac14t^2sigma ^2right)\
&=e^0t+frac12t^2sigma ^2
endalign*

thus $fracX_2-X_1sqrt 2 sim N(0,sigma ^2)$, so I found the variance not equal to $1$. What's my mistake here?

probability normal-distribution

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edited Jul 19 at 11:06

psmears

70549

asked Jul 19 at 7:58

Osman Bulut

536

add a comment | 

up vote
2
down vote

favorite

In Statistical Inference by Casella & Berger, in the proof of a theorem (theorem 5.3.1, page 220 in particular) I came across with the following proposition:

"the distribution of $fracX_2 - X_1sqrt 2$ is $N(0,1)$" where $X_2, X_1$ are iid normally random variables with the same mean $mu$ and variance $sigma^2$.

But despite of hours of trying, I couldn't prove this proposition even if it seems quite straightforward. I tried to use MGFs:

beginalign*
M_fracX_2-X_1sqrt 2(t)&=Eleft[e^tfracX_2-X_1sqrt 2right]\
&=Eleft[e^tfracX_2sqrt 2right]Eleft[e^tfrac-X_1sqrt 2right]\
&=left(e^frac1sqrt 2mu t+frac14t^2sigma ^2right)left(e^frac-1sqrt 2mu t+frac14t^2sigma ^2right)\
&=e^0t+frac12t^2sigma ^2
endalign*

thus $fracX_2-X_1sqrt 2 sim N(0,sigma ^2)$, so I found the variance not equal to $1$. What's my mistake here?

probability normal-distribution

share|cite|improve this question

edited Jul 19 at 11:06

psmears

70549

asked Jul 19 at 7:58

Osman Bulut

536

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

In Statistical Inference by Casella & Berger, in the proof of a theorem (theorem 5.3.1, page 220 in particular) I came across with the following proposition:

"the distribution of $fracX_2 - X_1sqrt 2$ is $N(0,1)$" where $X_2, X_1$ are iid normally random variables with the same mean $mu$ and variance $sigma^2$.

But despite of hours of trying, I couldn't prove this proposition even if it seems quite straightforward. I tried to use MGFs:

beginalign*
M_fracX_2-X_1sqrt 2(t)&=Eleft[e^tfracX_2-X_1sqrt 2right]\
&=Eleft[e^tfracX_2sqrt 2right]Eleft[e^tfrac-X_1sqrt 2right]\
&=left(e^frac1sqrt 2mu t+frac14t^2sigma ^2right)left(e^frac-1sqrt 2mu t+frac14t^2sigma ^2right)\
&=e^0t+frac12t^2sigma ^2
endalign*

thus $fracX_2-X_1sqrt 2 sim N(0,sigma ^2)$, so I found the variance not equal to $1$. What's my mistake here?

probability normal-distribution

share|cite|improve this question

edited Jul 19 at 11:06

psmears

70549

asked Jul 19 at 7:58

Osman Bulut

536

In Statistical Inference by Casella & Berger, in the proof of a theorem (theorem 5.3.1, page 220 in particular) I came across with the following proposition:

"the distribution of $fracX_2 - X_1sqrt 2$ is $N(0,1)$" where $X_2, X_1$ are iid normally random variables with the same mean $mu$ and variance $sigma^2$.

But despite of hours of trying, I couldn't prove this proposition even if it seems quite straightforward. I tried to use MGFs:

beginalign*
M_fracX_2-X_1sqrt 2(t)&=Eleft[e^tfracX_2-X_1sqrt 2right]\
&=Eleft[e^tfracX_2sqrt 2right]Eleft[e^tfrac-X_1sqrt 2right]\
&=left(e^frac1sqrt 2mu t+frac14t^2sigma ^2right)left(e^frac-1sqrt 2mu t+frac14t^2sigma ^2right)\
&=e^0t+frac12t^2sigma ^2
endalign*

thus $fracX_2-X_1sqrt 2 sim N(0,sigma ^2)$, so I found the variance not equal to $1$. What's my mistake here?

probability normal-distribution

share|cite|improve this question

edited Jul 19 at 11:06

psmears

70549

asked Jul 19 at 7:58

Osman Bulut

536

share|cite|improve this question

share|cite|improve this question

edited Jul 19 at 11:06

psmears

70549

edited Jul 19 at 11:06

psmears

70549

edited Jul 19 at 11:06

psmears

70549

70549

asked Jul 19 at 7:58

Osman Bulut

536

asked Jul 19 at 7:58

Osman Bulut

536

asked Jul 19 at 7:58

Osman Bulut

536

536

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
3
down vote



accepted

Your calculation is correct. What is true is $frac X_1-X_2 sqrt 2 sigma$ is normal with mean $0$ and variance $1$.

share|cite|improve this answer

answered Jul 19 at 8:03

Kavi Rama Murthy

20.9k2830

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks! So I must assume that this was a spelling error in the book.
  – Osman Bulut
  Jul 19 at 8:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, the statement in the book is surely wrong.
  – Kavi Rama Murthy
  Jul 19 at 8:15
  

  
  
  
  

add a comment | 

up vote
3
down vote

In general, one has that for independent $X_i, i=1,2dots n$ which are independent $N(mu_i, sigma^2_i)$
$$alpha_1X_1+alpha_2X_2dots a_nX_n text is N(sum_i=1^nalpha_imu_i, sum_i=1^nalpha^2_isigma^2_i)$$

and this is easily proven using MGFs.

In particular, taking $n=2, alpha_1=frac1sqrt2, alpha_2=-frac1sqrt2$ gives your answer

share|cite|improve this answer

answered Jul 19 at 8:09

asdf

3,378519

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks! So I must assume that this was a spelling error in the book.
  – Osman Bulut
  Jul 19 at 8:14
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

Your calculation is correct. What is true is $frac X_1-X_2 sqrt 2 sigma$ is normal with mean $0$ and variance $1$.

share|cite|improve this answer

answered Jul 19 at 8:03

Kavi Rama Murthy

20.9k2830

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks! So I must assume that this was a spelling error in the book.
  – Osman Bulut
  Jul 19 at 8:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, the statement in the book is surely wrong.
  – Kavi Rama Murthy
  Jul 19 at 8:15
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

Your calculation is correct. What is true is $frac X_1-X_2 sqrt 2 sigma$ is normal with mean $0$ and variance $1$.

share|cite|improve this answer

answered Jul 19 at 8:03

Kavi Rama Murthy

20.9k2830

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks! So I must assume that this was a spelling error in the book.
  – Osman Bulut
  Jul 19 at 8:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, the statement in the book is surely wrong.
  – Kavi Rama Murthy
  Jul 19 at 8:15
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

Your calculation is correct. What is true is $frac X_1-X_2 sqrt 2 sigma$ is normal with mean $0$ and variance $1$.

share|cite|improve this answer

answered Jul 19 at 8:03

Kavi Rama Murthy

20.9k2830

Your calculation is correct. What is true is $frac X_1-X_2 sqrt 2 sigma$ is normal with mean $0$ and variance $1$.

share|cite|improve this answer

answered Jul 19 at 8:03

Kavi Rama Murthy

20.9k2830

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 8:03

Kavi Rama Murthy

20.9k2830

answered Jul 19 at 8:03

Kavi Rama Murthy

20.9k2830

answered Jul 19 at 8:03

Kavi Rama Murthy

20.9k2830

20.9k2830

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks! So I must assume that this was a spelling error in the book.
  – Osman Bulut
  Jul 19 at 8:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, the statement in the book is surely wrong.
  – Kavi Rama Murthy
  Jul 19 at 8:15
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks! So I must assume that this was a spelling error in the book.
  – Osman Bulut
  Jul 19 at 8:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, the statement in the book is surely wrong.
  – Kavi Rama Murthy
  Jul 19 at 8:15
  

  
  
  
  

thanks! So I must assume that this was a spelling error in the book.
– Osman Bulut
Jul 19 at 8:14

thanks! So I must assume that this was a spelling error in the book.
– Osman Bulut
Jul 19 at 8:14

Yes, the statement in the book is surely wrong.
– Kavi Rama Murthy
Jul 19 at 8:15

Yes, the statement in the book is surely wrong.
– Kavi Rama Murthy
Jul 19 at 8:15

add a comment | 

up vote
3
down vote

In general, one has that for independent $X_i, i=1,2dots n$ which are independent $N(mu_i, sigma^2_i)$
$$alpha_1X_1+alpha_2X_2dots a_nX_n text is N(sum_i=1^nalpha_imu_i, sum_i=1^nalpha^2_isigma^2_i)$$

and this is easily proven using MGFs.

In particular, taking $n=2, alpha_1=frac1sqrt2, alpha_2=-frac1sqrt2$ gives your answer

share|cite|improve this answer

answered Jul 19 at 8:09

asdf

3,378519

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks! So I must assume that this was a spelling error in the book.
  – Osman Bulut
  Jul 19 at 8:14
  

  
  
  
  

add a comment | 

up vote
3
down vote

In general, one has that for independent $X_i, i=1,2dots n$ which are independent $N(mu_i, sigma^2_i)$
$$alpha_1X_1+alpha_2X_2dots a_nX_n text is N(sum_i=1^nalpha_imu_i, sum_i=1^nalpha^2_isigma^2_i)$$

and this is easily proven using MGFs.

In particular, taking $n=2, alpha_1=frac1sqrt2, alpha_2=-frac1sqrt2$ gives your answer

share|cite|improve this answer

answered Jul 19 at 8:09

asdf

3,378519

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks! So I must assume that this was a spelling error in the book.
  – Osman Bulut
  Jul 19 at 8:14
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

In general, one has that for independent $X_i, i=1,2dots n$ which are independent $N(mu_i, sigma^2_i)$
$$alpha_1X_1+alpha_2X_2dots a_nX_n text is N(sum_i=1^nalpha_imu_i, sum_i=1^nalpha^2_isigma^2_i)$$

and this is easily proven using MGFs.

In particular, taking $n=2, alpha_1=frac1sqrt2, alpha_2=-frac1sqrt2$ gives your answer

share|cite|improve this answer

answered Jul 19 at 8:09

asdf

3,378519

In general, one has that for independent $X_i, i=1,2dots n$ which are independent $N(mu_i, sigma^2_i)$
$$alpha_1X_1+alpha_2X_2dots a_nX_n text is N(sum_i=1^nalpha_imu_i, sum_i=1^nalpha^2_isigma^2_i)$$

and this is easily proven using MGFs.

In particular, taking $n=2, alpha_1=frac1sqrt2, alpha_2=-frac1sqrt2$ gives your answer

share|cite|improve this answer

answered Jul 19 at 8:09

asdf

3,378519

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 8:09

asdf

3,378519

answered Jul 19 at 8:09

asdf

3,378519

answered Jul 19 at 8:09

asdf

3,378519

3,378519

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks! So I must assume that this was a spelling error in the book.
  – Osman Bulut
  Jul 19 at 8:14
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks! So I must assume that this was a spelling error in the book.
  – Osman Bulut
  Jul 19 at 8:14
  

  
  
  
  

thanks! So I must assume that this was a spelling error in the book.
– Osman Bulut
Jul 19 at 8:14

thanks! So I must assume that this was a spelling error in the book.
– Osman Bulut
Jul 19 at 8:14

add a comment | 

 

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