Mixing
Proof involving the difference of iid normal random variables
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I am trying to prove that, for two i.i.d. random variables $X_1,X_2 sim N(mu, sigma^2)$, their difference divided by $sqrt 2$ is standard normal, i.e. $frac {X_1-X_2}{sqrt 2} sim N(0,1)$.
I am trying to show it using convolution, but am getting caught up in the algebra of trying to get it into a recognizable form. I think there is a more simple way using the definition of expectation, but when I try it, it doesn't quite come out right as I get $N(0, sigma^2)$ instead of $N(0,1)$.
probability statistics
edited Jan 29 at 5:09
Mohammad Zuhair Khan
1,6792625
asked Jan 29 at 4:55
eyehearyoueyehearyou
306
$endgroup$
add a comment |
$begingroup$
I am trying to prove that, for two i.i.d. random variables $X_1,X_2 sim N(mu, sigma^2)$, their difference divided by $sqrt 2$ is standard normal, i.e. $frac {X_1-X_2}{sqrt 2} sim N(0,1)$.
I am trying to show it using convolution, but am getting caught up in the algebra of trying to get it into a recognizable form. I think there is a more simple way using the definition of expectation, but when I try it, it doesn't quite come out right as I get $N(0, sigma^2)$ instead of $N(0,1)$.
probability statistics
edited Jan 29 at 5:09
Mohammad Zuhair Khan
1,6792625
asked Jan 29 at 4:55
eyehearyoueyehearyou
306
$endgroup$
$begingroup$
@angryavian So the problem I am having with the linear combination approach comes down to the variance. I am trying to calculate as follows: Var[(X_1-X_2)/sqrt(2)]=(1/2)Var[X_1-X_2]=2(1/2)Var[X_1]=sigma^2 So I'm getting sigma^2 instead of 1. Can you help me find my mistake?
$endgroup$
– eyehearyou
Jan 29 at 5:13
1
$begingroup$
It's not a mistake. You need to divide by $sqrt{2}sigma$ to get $N(0,1)$.
$endgroup$
– d.k.o.
Jan 29 at 5:20
$begingroup$
The simplest way to prove this is to use characteristic functions. Are you allowed to do that?. (As already pointed out you will not get $N(0,1)$ if you divide by $sqrt 2$).
$endgroup$
– Kavi Rama Murthy
Jan 29 at 5:31
$begingroup$
@KaviRamaMurthy I'm allowed to do anything. This is not for a homework or school assignment, I'm just reading my way through a textbook and came upon an something that I couldn't immediately see how to prove. Thanks!
$endgroup$
– eyehearyou
Jan 29 at 5:34
add a comment |
$begingroup$
I am trying to prove that, for two i.i.d. random variables $X_1,X_2 sim N(mu, sigma^2)$, their difference divided by $sqrt 2$ is standard normal, i.e. $frac {X_1-X_2}{sqrt 2} sim N(0,1)$.
I am trying to show it using convolution, but am getting caught up in the algebra of trying to get it into a recognizable form. I think there is a more simple way using the definition of expectation, but when I try it, it doesn't quite come out right as I get $N(0, sigma^2)$ instead of $N(0,1)$.
probability statistics
edited Jan 29 at 5:09
Mohammad Zuhair Khan
1,6792625
asked Jan 29 at 4:55
eyehearyoueyehearyou
306
$endgroup$
I am trying to prove that, for two i.i.d. random variables $X_1,X_2 sim N(mu, sigma^2)$, their difference divided by $sqrt 2$ is standard normal, i.e. $frac {X_1-X_2}{sqrt 2} sim N(0,1)$.
I am trying to show it using convolution, but am getting caught up in the algebra of trying to get it into a recognizable form. I think there is a more simple way using the definition of expectation, but when I try it, it doesn't quite come out right as I get $N(0, sigma^2)$ instead of $N(0,1)$.
probability statistics
probability statistics
edited Jan 29 at 5:09
Mohammad Zuhair Khan
1,6792625
asked Jan 29 at 4:55
eyehearyoueyehearyou
306
edited Jan 29 at 5:09
Mohammad Zuhair Khan
1,6792625
asked Jan 29 at 4:55
eyehearyoueyehearyou
306
edited Jan 29 at 5:09
Mohammad Zuhair Khan
1,6792625
edited Jan 29 at 5:09
Mohammad Zuhair Khan
1,6792625
edited Jan 29 at 5:09
Mohammad Zuhair Khan
1,6792625
1,6792625
asked Jan 29 at 4:55
eyehearyoueyehearyou
306
asked Jan 29 at 4:55
eyehearyoueyehearyou
306
asked Jan 29 at 4:55
eyehearyoueyehearyou
306
306
$begingroup$
@angryavian So the problem I am having with the linear combination approach comes down to the variance. I am trying to calculate as follows: Var[(X_1-X_2)/sqrt(2)]=(1/2)Var[X_1-X_2]=2(1/2)Var[X_1]=sigma^2 So I'm getting sigma^2 instead of 1. Can you help me find my mistake?
$endgroup$
– eyehearyou
Jan 29 at 5:13
1
$begingroup$
It's not a mistake. You need to divide by $sqrt{2}sigma$ to get $N(0,1)$.
$endgroup$
– d.k.o.
Jan 29 at 5:20
$begingroup$
The simplest way to prove this is to use characteristic functions. Are you allowed to do that?. (As already pointed out you will not get $N(0,1)$ if you divide by $sqrt 2$).
$endgroup$
– Kavi Rama Murthy
Jan 29 at 5:31
$begingroup$
@KaviRamaMurthy I'm allowed to do anything. This is not for a homework or school assignment, I'm just reading my way through a textbook and came upon an something that I couldn't immediately see how to prove. Thanks!
$endgroup$
– eyehearyou
Jan 29 at 5:34
add a comment |
$begingroup$
@angryavian So the problem I am having with the linear combination approach comes down to the variance. I am trying to calculate as follows: Var[(X_1-X_2)/sqrt(2)]=(1/2)Var[X_1-X_2]=2(1/2)Var[X_1]=sigma^2 So I'm getting sigma^2 instead of 1. Can you help me find my mistake?
$endgroup$
– eyehearyou
Jan 29 at 5:13
1
$begingroup$
It's not a mistake. You need to divide by $sqrt{2}sigma$ to get $N(0,1)$.
$endgroup$
– d.k.o.
Jan 29 at 5:20
$begingroup$
The simplest way to prove this is to use characteristic functions. Are you allowed to do that?. (As already pointed out you will not get $N(0,1)$ if you divide by $sqrt 2$).
$endgroup$
– Kavi Rama Murthy
Jan 29 at 5:31
$begingroup$
@KaviRamaMurthy I'm allowed to do anything. This is not for a homework or school assignment, I'm just reading my way through a textbook and came upon an something that I couldn't immediately see how to prove. Thanks!
$endgroup$
– eyehearyou
Jan 29 at 5:34
$begingroup$
@angryavian So the problem I am having with the linear combination approach comes down to the variance. I am trying to calculate as follows: Var[(X_1-X_2)/sqrt(2)]=(1/2)Var[X_1-X_2]=2(1/2)Var[X_1]=sigma^2 So I'm getting sigma^2 instead of 1. Can you help me find my mistake?
$endgroup$
– eyehearyou
Jan 29 at 5:13
$begingroup$
@angryavian So the problem I am having with the linear combination approach comes down to the variance. I am trying to calculate as follows: Var[(X_1-X_2)/sqrt(2)]=(1/2)Var[X_1-X_2]=2(1/2)Var[X_1]=sigma^2 So I'm getting sigma^2 instead of 1. Can you help me find my mistake?
$endgroup$
– eyehearyou
Jan 29 at 5:13
1
1
$begingroup$
It's not a mistake. You need to divide by $sqrt{2}sigma$ to get $N(0,1)$.
$endgroup$
– d.k.o.
Jan 29 at 5:20
$begingroup$
It's not a mistake. You need to divide by $sqrt{2}sigma$ to get $N(0,1)$.
$endgroup$
– d.k.o.
Jan 29 at 5:20
$begingroup$
The simplest way to prove this is to use characteristic functions. Are you allowed to do that?. (As already pointed out you will not get $N(0,1)$ if you divide by $sqrt 2$).
$endgroup$
– Kavi Rama Murthy
Jan 29 at 5:31
$begingroup$
The simplest way to prove this is to use characteristic functions. Are you allowed to do that?. (As already pointed out you will not get $N(0,1)$ if you divide by $sqrt 2$).
$endgroup$
– Kavi Rama Murthy
Jan 29 at 5:31
$begingroup$
@KaviRamaMurthy I'm allowed to do anything. This is not for a homework or school assignment, I'm just reading my way through a textbook and came upon an something that I couldn't immediately see how to prove. Thanks!
$endgroup$
– eyehearyou
Jan 29 at 5:34
$begingroup$
@KaviRamaMurthy I'm allowed to do anything. This is not for a homework or school assignment, I'm just reading my way through a textbook and came upon an something that I couldn't immediately see how to prove. Thanks!
$endgroup$
– eyehearyou
Jan 29 at 5:34
add a comment |
1 Answer
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$begingroup$
$Ee^{itX_1}=e^{itmu}e^{-t^{2}sigma^{2}/2}$. Hence $Ee^{it(X_1-X_2)/sqrt2 sigma)}=Ee^{itX_1/sqrt2 sigma)}Ee^{-itX_2/sqrt2 sigma)}=(e^{-t^{2}/4})^{2}=e^{-t^{2}/2}$ which is the characteristic function of $N(0,1)$.
answered Jan 29 at 5:42
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
$begingroup$
Thanks so much Kavi! I see how this works when we have that factor of $sigma$ in the denominator. Unfortunately, the book I have just has $frac{X_1-X_2}{sqrt{2}}$ :(. Maybe it's an error, but there is an important proof that seems to rest on it, so that is rather discouraging.
$endgroup$
– eyehearyou
Jan 29 at 5:45
add a comment |
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$begingroup$
$Ee^{itX_1}=e^{itmu}e^{-t^{2}sigma^{2}/2}$. Hence $Ee^{it(X_1-X_2)/sqrt2 sigma)}=Ee^{itX_1/sqrt2 sigma)}Ee^{-itX_2/sqrt2 sigma)}=(e^{-t^{2}/4})^{2}=e^{-t^{2}/2}$ which is the characteristic function of $N(0,1)$.
answered Jan 29 at 5:42
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
$begingroup$
Thanks so much Kavi! I see how this works when we have that factor of $sigma$ in the denominator. Unfortunately, the book I have just has $frac{X_1-X_2}{sqrt{2}}$ :(. Maybe it's an error, but there is an important proof that seems to rest on it, so that is rather discouraging.
$endgroup$
– eyehearyou
Jan 29 at 5:45
add a comment |
$begingroup$
$Ee^{itX_1}=e^{itmu}e^{-t^{2}sigma^{2}/2}$. Hence $Ee^{it(X_1-X_2)/sqrt2 sigma)}=Ee^{itX_1/sqrt2 sigma)}Ee^{-itX_2/sqrt2 sigma)}=(e^{-t^{2}/4})^{2}=e^{-t^{2}/2}$ which is the characteristic function of $N(0,1)$.
answered Jan 29 at 5:42
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
$begingroup$
Thanks so much Kavi! I see how this works when we have that factor of $sigma$ in the denominator. Unfortunately, the book I have just has $frac{X_1-X_2}{sqrt{2}}$ :(. Maybe it's an error, but there is an important proof that seems to rest on it, so that is rather discouraging.
$endgroup$
– eyehearyou
Jan 29 at 5:45
add a comment |
$begingroup$
$Ee^{itX_1}=e^{itmu}e^{-t^{2}sigma^{2}/2}$. Hence $Ee^{it(X_1-X_2)/sqrt2 sigma)}=Ee^{itX_1/sqrt2 sigma)}Ee^{-itX_2/sqrt2 sigma)}=(e^{-t^{2}/4})^{2}=e^{-t^{2}/2}$ which is the characteristic function of $N(0,1)$.
answered Jan 29 at 5:42
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
$Ee^{itX_1}=e^{itmu}e^{-t^{2}sigma^{2}/2}$. Hence $Ee^{it(X_1-X_2)/sqrt2 sigma)}=Ee^{itX_1/sqrt2 sigma)}Ee^{-itX_2/sqrt2 sigma)}=(e^{-t^{2}/4})^{2}=e^{-t^{2}/2}$ which is the characteristic function of $N(0,1)$.
answered Jan 29 at 5:42
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
answered Jan 29 at 5:42
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
answered Jan 29 at 5:42
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
answered Jan 29 at 5:42
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
71.2k53170
$begingroup$
Thanks so much Kavi! I see how this works when we have that factor of $sigma$ in the denominator. Unfortunately, the book I have just has $frac{X_1-X_2}{sqrt{2}}$ :(. Maybe it's an error, but there is an important proof that seems to rest on it, so that is rather discouraging.
$endgroup$
– eyehearyou
Jan 29 at 5:45
add a comment |
$begingroup$
Thanks so much Kavi! I see how this works when we have that factor of $sigma$ in the denominator. Unfortunately, the book I have just has $frac{X_1-X_2}{sqrt{2}}$ :(. Maybe it's an error, but there is an important proof that seems to rest on it, so that is rather discouraging.
$endgroup$
– eyehearyou
Jan 29 at 5:45
$begingroup$
Thanks so much Kavi! I see how this works when we have that factor of $sigma$ in the denominator. Unfortunately, the book I have just has $frac{X_1-X_2}{sqrt{2}}$ :(. Maybe it's an error, but there is an important proof that seems to rest on it, so that is rather discouraging.
$endgroup$
– eyehearyou
Jan 29 at 5:45
$begingroup$
Thanks so much Kavi! I see how this works when we have that factor of $sigma$ in the denominator. Unfortunately, the book I have just has $frac{X_1-X_2}{sqrt{2}}$ :(. Maybe it's an error, but there is an important proof that seems to rest on it, so that is rather discouraging.
$endgroup$
– eyehearyou
Jan 29 at 5:45
add a comment |
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$begingroup$
@angryavian So the problem I am having with the linear combination approach comes down to the variance. I am trying to calculate as follows: Var[(X_1-X_2)/sqrt(2)]=(1/2)Var[X_1-X_2]=2(1/2)Var[X_1]=sigma^2 So I'm getting sigma^2 instead of 1. Can you help me find my mistake?
$endgroup$
– eyehearyou
Jan 29 at 5:13
1
$begingroup$
It's not a mistake. You need to divide by $sqrt{2}sigma$ to get $N(0,1)$.
$endgroup$
– d.k.o.
Jan 29 at 5:20
$begingroup$
The simplest way to prove this is to use characteristic functions. Are you allowed to do that?. (As already pointed out you will not get $N(0,1)$ if you divide by $sqrt 2$).
$endgroup$
– Kavi Rama Murthy
Jan 29 at 5:31
$begingroup$
@KaviRamaMurthy I'm allowed to do anything. This is not for a homework or school assignment, I'm just reading my way through a textbook and came upon an something that I couldn't immediately see how to prove. Thanks!
$endgroup$
– eyehearyou
Jan 29 at 5:34