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 Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Properties of characteristic functionsBernoulli distribution: expectation problem with independent random variablesCharacteristic function of seriesIf $X_1,X_2,X$ are iid random variables with $X_1+X_2$ has the distribution of $aX$, find all characteristic functions of $X$.Prove that $X,Y$ are independent iff the characteristic function of $(X,Y)$ equals the product of the characteristic functions of $X$ and $Y$Prove that random variables are independentFind the distribution of the average of exponential random variablesDifference between two iid random variables is not uniformly distributedFinding a characteristic function of a product of two normal random variablesLet $X,Y $ be two independent random variables with exponential distribution and parameter $lambda > 0$.Characteristic function of independent Poisson random variables
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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
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Properties of characteristic functionsBernoulli distribution: expectation problem with independent random variablesCharacteristic function of seriesIf $X_1,X_2,X$ are iid random variables with $X_1+X_2$ has the distribution of $aX$, find all characteristic functions of $X$.Prove that $X,Y$ are independent iff the characteristic function of $(X,Y)$ equals the product of the characteristic functions of $X$ and $Y$Prove that random variables are independentFind the distribution of the average of exponential random variablesDifference between two iid random variables is not uniformly distributedFinding a characteristic function of a product of two normal random variablesLet $X,Y $ be two independent random variables with exponential distribution and parameter $lambda > 0$.Characteristic function of independent Poisson random variables
$begingroup$
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
asked Apr 2 at 17:59
MariaMaria
1544
$endgroup$
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
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
asked Apr 2 at 17:59
MariaMaria
1544
$endgroup$
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
probability characteristic-functions
asked Apr 2 at 17:59
MariaMaria
1544
asked Apr 2 at 17:59
MariaMaria
1544
asked Apr 2 at 17:59
MariaMaria
1544
asked Apr 2 at 17:59
MariaMaria
1544
asked Apr 2 at 17:59
MariaMaria
1544
1544
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
$begingroup$
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
1
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
$begingroup$
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. 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
$begingroup$
I never work with a join characteristic function with characteristic functions.
$endgroup$
– Maria
Apr 2 at 18:37
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
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 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
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.
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.
answered Apr 2 at 19:01
Mark FischlerMark Fischler
34.5k12552
$endgroup$
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.
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
answered Apr 2 at 19:01
Mark FischlerMark Fischler
34.5k12552
34.5k12552
$begingroup$
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 |
$begingroup$
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
$begingroup$
Thanks. One question. phi(2X) is a/(a-i2t) or 2a/ (2a + it)??
$endgroup$
– Maria
Apr 2 at 20:23
$begingroup$
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
$begingroup$
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
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
$begingroup$
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
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$begingroup$
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