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Let X and Y two independent random variables with exponential distribution of parameter a>0. U = X+ Y and V = X- Y are not independent

1

$begingroup$

Let X and Y two independent random variables with exponential distribution of parameter a>0.
Proof using characteristic functions, that U = X+ Y and V = X- Y are not independent.

1) I calculate the characteristic function of X and Y (is the same)

$varphi_X(t)$ = $varphi_Y(t)$=$left( 1- fracita right)^left( -1 right)$

2) I know that $varphi_left( X+Y right) (t)=varphi_X(t).varphi_Y(t)$, if X and Y are independent, that is:

$varphi_U(t)$=$varphi_left( X+Y right) (t)=left( 1- fracita right)^left( -2 right)$

3) I know that $varphi_left(-Y right) (t)=barvarphi_Y(t))$, where $barvarphi_Y(t))$, is the complex conjugate of $varphi_Y(t))$

4) I calculate $varphi_V(t)$=$varphi_left( X-Y right) (t)=varphi_X(t).barvarphi_Y(t)$

5) How can I proof that U and V are not independent?

probability characteristic-functions

share|cite|improve this question

asked Apr 2 at 17:59

MariaMaria

1544

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You need to work out a formula for the joint characteristic function of $(U,V)$, that is, $phi(s,t)=Eexp( sU +tV)$.
  $endgroup$
  – kimchi lover
  Apr 2 at 18:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I never work with a join characteristic function. The teacher has not taught us how to do it, has not done exercises about it and it is not in the given notes. Is there another way to do it?
  $endgroup$
  – Maria
  Apr 2 at 18:35
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I never work with a join characteristic function with characteristic functions.
  $endgroup$
  – Maria
  Apr 2 at 18:37
  

  
  
  
  

add a comment | 

1

$begingroup$

Let X and Y two independent random variables with exponential distribution of parameter a>0.
Proof using characteristic functions, that U = X+ Y and V = X- Y are not independent.

1) I calculate the characteristic function of X and Y (is the same)

$varphi_X(t)$ = $varphi_Y(t)$=$left( 1- fracita right)^left( -1 right)$

2) I know that $varphi_left( X+Y right) (t)=varphi_X(t).varphi_Y(t)$, if X and Y are independent, that is:

$varphi_U(t)$=$varphi_left( X+Y right) (t)=left( 1- fracita right)^left( -2 right)$

3) I know that $varphi_left(-Y right) (t)=barvarphi_Y(t))$, where $barvarphi_Y(t))$, is the complex conjugate of $varphi_Y(t))$

4) I calculate $varphi_V(t)$=$varphi_left( X-Y right) (t)=varphi_X(t).barvarphi_Y(t)$

5) How can I proof that U and V are not independent?

probability characteristic-functions

share|cite|improve this question

asked Apr 2 at 17:59

MariaMaria

1544

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You need to work out a formula for the joint characteristic function of $(U,V)$, that is, $phi(s,t)=Eexp( sU +tV)$.
  $endgroup$
  – kimchi lover
  Apr 2 at 18:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I never work with a join characteristic function. The teacher has not taught us how to do it, has not done exercises about it and it is not in the given notes. Is there another way to do it?
  $endgroup$
  – Maria
  Apr 2 at 18:35
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I never work with a join characteristic function with characteristic functions.
  $endgroup$
  – Maria
  Apr 2 at 18:37
  

  
  
  
  

add a comment | 

$begingroup$

Let X and Y two independent random variables with exponential distribution of parameter a>0.
Proof using characteristic functions, that U = X+ Y and V = X- Y are not independent.

1) I calculate the characteristic function of X and Y (is the same)

$varphi_X(t)$ = $varphi_Y(t)$=$left( 1- fracita right)^left( -1 right)$

2) I know that $varphi_left( X+Y right) (t)=varphi_X(t).varphi_Y(t)$, if X and Y are independent, that is:

$varphi_U(t)$=$varphi_left( X+Y right) (t)=left( 1- fracita right)^left( -2 right)$

3) I know that $varphi_left(-Y right) (t)=barvarphi_Y(t))$, where $barvarphi_Y(t))$, is the complex conjugate of $varphi_Y(t))$

4) I calculate $varphi_V(t)$=$varphi_left( X-Y right) (t)=varphi_X(t).barvarphi_Y(t)$

5) How can I proof that U and V are not independent?

probability characteristic-functions

share|cite|improve this question

asked Apr 2 at 17:59

MariaMaria

1544

$endgroup$

share|cite|improve this question

asked Apr 2 at 17:59

MariaMaria

1544

share|cite|improve this question

asked Apr 2 at 17:59

MariaMaria

1544

asked Apr 2 at 17:59

MariaMaria

1544

asked Apr 2 at 17:59

MariaMaria

1544

1 Answer
1

active

oldest

votes

1

$begingroup$

You show that they are not independent by showing that
$$
varphi_(X+Y) + (X-Y)(t) neq varphi_(X+Y)(t) varphi_(X-Y)(t)
$$
To see this, assume $a>0$ and note that
$$
varphi_(X+Y) + (X-Y)(t) = varphi_2X(t) = int_0^infty e^itx f_2a(x),dx = frac2a2a-it
$$
and the characteristic functions of $Xpm Y$ are
$$varphi_(X+Y)(t) = int_x=0^infty int_y=0^infty e^it(x+y)
f_a(y) f_a(x),dy,dx = fraca^2(a-it)^2 \
varphi_(X-Y)(t) = int_x=0^infty int_y=0^infty e^it(x-y)
f_a(y) f_a(x),dy,dx = fraca^2a^2+t^2
$$
whence
$$
varphi_(X+Y)(t) varphi_(X-Y)(t) = fraca^4(a-it)^3(a+it) neq frac2a2a-it = varphi_(X+Y) + (X-Y)(t)
$$
Contrast this with what you get doing the same steps for Gaussian-distributed variables sharing the same mean and $sigma$, where that last equality does turn out to be true.

share|cite|improve this answer

answered Apr 2 at 19:01

Mark FischlerMark Fischler

34.5k12552

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. One question. phi(2X) is a/(a-i2t) or 2a/ (2a + it)??
  $endgroup$
  – Maria
  Apr 2 at 20:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's $2a/(2a-it)$. Note that $$left( 1-fracita right)^-1 = left(fraca-itaright)^-1 = fracaa-it$$ and the $-it$ in the denominator of $varphi_2X$ comes about in the same way, replacing $a$ by $2X$.
  $endgroup$
  – Mark Fischler
  Apr 4 at 6:48
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I do not think that $phi_2X(t)=frac2a2a-it$. If $X$ is exponential with parameter $a$ and mean $1/a$, then $cX$ is exponential with parameter $a/c$ and mean $c/a$, provided $c>0$. The CF of $X$ is $fracaa-it$ and the CF of $cX$ is $fraca/ca/c-it=fracaa-ict$. This is also seen by $phi_cX(t)=phi_X(ct)$. Can you correct me or verify this?
  $endgroup$
  – LoveTooNap29
  Apr 7 at 18:04
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

You show that they are not independent by showing that
$$
varphi_(X+Y) + (X-Y)(t) neq varphi_(X+Y)(t) varphi_(X-Y)(t)
$$
To see this, assume $a>0$ and note that
$$
varphi_(X+Y) + (X-Y)(t) = varphi_2X(t) = int_0^infty e^itx f_2a(x),dx = frac2a2a-it
$$
and the characteristic functions of $Xpm Y$ are
$$varphi_(X+Y)(t) = int_x=0^infty int_y=0^infty e^it(x+y)
f_a(y) f_a(x),dy,dx = fraca^2(a-it)^2 \
varphi_(X-Y)(t) = int_x=0^infty int_y=0^infty e^it(x-y)
f_a(y) f_a(x),dy,dx = fraca^2a^2+t^2
$$
whence
$$
varphi_(X+Y)(t) varphi_(X-Y)(t) = fraca^4(a-it)^3(a+it) neq frac2a2a-it = varphi_(X+Y) + (X-Y)(t)
$$
Contrast this with what you get doing the same steps for Gaussian-distributed variables sharing the same mean and $sigma$, where that last equality does turn out to be true.

share|cite|improve this answer

answered Apr 2 at 19:01

Mark FischlerMark Fischler

34.5k12552

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. One question. phi(2X) is a/(a-i2t) or 2a/ (2a + it)??
  $endgroup$
  – Maria
  Apr 2 at 20:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's $2a/(2a-it)$. Note that $$left( 1-fracita right)^-1 = left(fraca-itaright)^-1 = fracaa-it$$ and the $-it$ in the denominator of $varphi_2X$ comes about in the same way, replacing $a$ by $2X$.
  $endgroup$
  – Mark Fischler
  Apr 4 at 6:48
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I do not think that $phi_2X(t)=frac2a2a-it$. If $X$ is exponential with parameter $a$ and mean $1/a$, then $cX$ is exponential with parameter $a/c$ and mean $c/a$, provided $c>0$. The CF of $X$ is $fracaa-it$ and the CF of $cX$ is $fraca/ca/c-it=fracaa-ict$. This is also seen by $phi_cX(t)=phi_X(ct)$. Can you correct me or verify this?
  $endgroup$
  – LoveTooNap29
  Apr 7 at 18:04
  
  

  
  
  
  

add a comment | 

1

$begingroup$

You show that they are not independent by showing that
$$
varphi_(X+Y) + (X-Y)(t) neq varphi_(X+Y)(t) varphi_(X-Y)(t)
$$
To see this, assume $a>0$ and note that
$$
varphi_(X+Y) + (X-Y)(t) = varphi_2X(t) = int_0^infty e^itx f_2a(x),dx = frac2a2a-it
$$
and the characteristic functions of $Xpm Y$ are
$$varphi_(X+Y)(t) = int_x=0^infty int_y=0^infty e^it(x+y)
f_a(y) f_a(x),dy,dx = fraca^2(a-it)^2 \
varphi_(X-Y)(t) = int_x=0^infty int_y=0^infty e^it(x-y)
f_a(y) f_a(x),dy,dx = fraca^2a^2+t^2
$$
whence
$$
varphi_(X+Y)(t) varphi_(X-Y)(t) = fraca^4(a-it)^3(a+it) neq frac2a2a-it = varphi_(X+Y) + (X-Y)(t)
$$
Contrast this with what you get doing the same steps for Gaussian-distributed variables sharing the same mean and $sigma$, where that last equality does turn out to be true.

share|cite|improve this answer

answered Apr 2 at 19:01

Mark FischlerMark Fischler

34.5k12552

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. One question. phi(2X) is a/(a-i2t) or 2a/ (2a + it)??
  $endgroup$
  – Maria
  Apr 2 at 20:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's $2a/(2a-it)$. Note that $$left( 1-fracita right)^-1 = left(fraca-itaright)^-1 = fracaa-it$$ and the $-it$ in the denominator of $varphi_2X$ comes about in the same way, replacing $a$ by $2X$.
  $endgroup$
  – Mark Fischler
  Apr 4 at 6:48
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I do not think that $phi_2X(t)=frac2a2a-it$. If $X$ is exponential with parameter $a$ and mean $1/a$, then $cX$ is exponential with parameter $a/c$ and mean $c/a$, provided $c>0$. The CF of $X$ is $fracaa-it$ and the CF of $cX$ is $fraca/ca/c-it=fracaa-ict$. This is also seen by $phi_cX(t)=phi_X(ct)$. Can you correct me or verify this?
  $endgroup$
  – LoveTooNap29
  Apr 7 at 18:04
  
  

  
  
  
  

add a comment | 

$begingroup$

You show that they are not independent by showing that
$$
varphi_(X+Y) + (X-Y)(t) neq varphi_(X+Y)(t) varphi_(X-Y)(t)
$$
To see this, assume $a>0$ and note that
$$
varphi_(X+Y) + (X-Y)(t) = varphi_2X(t) = int_0^infty e^itx f_2a(x),dx = frac2a2a-it
$$
and the characteristic functions of $Xpm Y$ are
$$varphi_(X+Y)(t) = int_x=0^infty int_y=0^infty e^it(x+y)
f_a(y) f_a(x),dy,dx = fraca^2(a-it)^2 \
varphi_(X-Y)(t) = int_x=0^infty int_y=0^infty e^it(x-y)
f_a(y) f_a(x),dy,dx = fraca^2a^2+t^2
$$
whence
$$
varphi_(X+Y)(t) varphi_(X-Y)(t) = fraca^4(a-it)^3(a+it) neq frac2a2a-it = varphi_(X+Y) + (X-Y)(t)
$$
Contrast this with what you get doing the same steps for Gaussian-distributed variables sharing the same mean and $sigma$, where that last equality does turn out to be true.

share|cite|improve this answer

answered Apr 2 at 19:01

Mark FischlerMark Fischler

34.5k12552

$endgroup$

share|cite|improve this answer

answered Apr 2 at 19:01

Mark FischlerMark Fischler

34.5k12552

answered Apr 2 at 19:01

Mark FischlerMark Fischler

34.5k12552

answered Apr 2 at 19:01

Mark FischlerMark Fischler

34.5k12552