Show that $frac{d}{dt} EBig[big |phi_X(tZ)big|^2Big ]=2 E left[ Z phi_X(-t Z) phi_X'(t Z) right] $.
$begingroup$
Let $Z$ be standard normal. Is the following sequence of steps correct?
begin{align}
frac{d}{dt} E Big[ big|phi_X(tZ)big|^2 Big] &= E left[ frac{d}{dt} |phi_X(tZ)|^2 right] \
&=E left[ frac{d}{dt} phi_X(tZ)phi_X^*(tZ) right] text{ using } |phi_X(tZ)|^2= phi_X(tZ)phi_X^*(tZ) \
&=E left[ frac{d}{dt} phi_X(tZ)phi_X(-tZ) right] text{ using } phi_X^*(tZ)=phi_X(-tZ) \
&=E left[ Z phi_X(-t Z) phi_X'(t Z) - Z phi_X(t Z) phi_X'(-t Z) right]\
&=E left[ Z phi_X(-t Z) phi_X'(t Z) right] - E left[ Z phi_X(t Z) phi_X'(-t Z) right]\
&=E left[ Z phi_X(-t Z) phi_X'(t Z) right] - E left[ (-Z) phi_X(-t Z) phi_X'(t Z) right] text{using symmetry of $Z$}\
&=2 E left[ Z phi_X(-t Z) phi_X'(t Z) right]
end{align}
Here, $phi_X(t)$ is the characteristic function of a random variable $X$.
Thank you for your comments.
probability-theory proof-verification normal-distribution characteristic-functions expected-value
edited Dec 14 '18 at 21:33
Batominovski
33.1k33293
asked Dec 14 '18 at 20:00
BobyBoby
1,0471929
$endgroup$
add a comment |
$begingroup$
Let $Z$ be standard normal. Is the following sequence of steps correct?
begin{align}
frac{d}{dt} E Big[ big|phi_X(tZ)big|^2 Big] &= E left[ frac{d}{dt} |phi_X(tZ)|^2 right] \
&=E left[ frac{d}{dt} phi_X(tZ)phi_X^*(tZ) right] text{ using } |phi_X(tZ)|^2= phi_X(tZ)phi_X^*(tZ) \
&=E left[ frac{d}{dt} phi_X(tZ)phi_X(-tZ) right] text{ using } phi_X^*(tZ)=phi_X(-tZ) \
&=E left[ Z phi_X(-t Z) phi_X'(t Z) - Z phi_X(t Z) phi_X'(-t Z) right]\
&=E left[ Z phi_X(-t Z) phi_X'(t Z) right] - E left[ Z phi_X(t Z) phi_X'(-t Z) right]\
&=E left[ Z phi_X(-t Z) phi_X'(t Z) right] - E left[ (-Z) phi_X(-t Z) phi_X'(t Z) right] text{using symmetry of $Z$}\
&=2 E left[ Z phi_X(-t Z) phi_X'(t Z) right]
end{align}
Here, $phi_X(t)$ is the characteristic function of a random variable $X$.
Thank you for your comments.
probability-theory proof-verification normal-distribution characteristic-functions expected-value
edited Dec 14 '18 at 21:33
Batominovski
33.1k33293
asked Dec 14 '18 at 20:00
BobyBoby
1,0471929
$endgroup$
2
$begingroup$
Yes. $ $ $ $ $ $
$endgroup$
– Did
Dec 14 '18 at 20:13
$begingroup$
Except for justification of first step.
$endgroup$
– Kavi Rama Murthy
Dec 15 '18 at 0:04
$begingroup$
@KaviRamaMurthy What would be justification for that?
$endgroup$
– Boby
Dec 15 '18 at 16:34
$begingroup$
$E|phi_X(tZ)|^{2}=Ee^{-t^{2}X^{2}}$. Try to apply DCT.
$endgroup$
– Kavi Rama Murthy
Dec 15 '18 at 23:48
add a comment |
$begingroup$
Let $Z$ be standard normal. Is the following sequence of steps correct?
begin{align}
frac{d}{dt} E Big[ big|phi_X(tZ)big|^2 Big] &= E left[ frac{d}{dt} |phi_X(tZ)|^2 right] \
&=E left[ frac{d}{dt} phi_X(tZ)phi_X^*(tZ) right] text{ using } |phi_X(tZ)|^2= phi_X(tZ)phi_X^*(tZ) \
&=E left[ frac{d}{dt} phi_X(tZ)phi_X(-tZ) right] text{ using } phi_X^*(tZ)=phi_X(-tZ) \
&=E left[ Z phi_X(-t Z) phi_X'(t Z) - Z phi_X(t Z) phi_X'(-t Z) right]\
&=E left[ Z phi_X(-t Z) phi_X'(t Z) right] - E left[ Z phi_X(t Z) phi_X'(-t Z) right]\
&=E left[ Z phi_X(-t Z) phi_X'(t Z) right] - E left[ (-Z) phi_X(-t Z) phi_X'(t Z) right] text{using symmetry of $Z$}\
&=2 E left[ Z phi_X(-t Z) phi_X'(t Z) right]
end{align}
Here, $phi_X(t)$ is the characteristic function of a random variable $X$.
Thank you for your comments.
probability-theory proof-verification normal-distribution characteristic-functions expected-value
edited Dec 14 '18 at 21:33
Batominovski
33.1k33293
asked Dec 14 '18 at 20:00
BobyBoby
1,0471929
$endgroup$
Let $Z$ be standard normal. Is the following sequence of steps correct?
begin{align}
frac{d}{dt} E Big[ big|phi_X(tZ)big|^2 Big] &= E left[ frac{d}{dt} |phi_X(tZ)|^2 right] \
&=E left[ frac{d}{dt} phi_X(tZ)phi_X^*(tZ) right] text{ using } |phi_X(tZ)|^2= phi_X(tZ)phi_X^*(tZ) \
&=E left[ frac{d}{dt} phi_X(tZ)phi_X(-tZ) right] text{ using } phi_X^*(tZ)=phi_X(-tZ) \
&=E left[ Z phi_X(-t Z) phi_X'(t Z) - Z phi_X(t Z) phi_X'(-t Z) right]\
&=E left[ Z phi_X(-t Z) phi_X'(t Z) right] - E left[ Z phi_X(t Z) phi_X'(-t Z) right]\
&=E left[ Z phi_X(-t Z) phi_X'(t Z) right] - E left[ (-Z) phi_X(-t Z) phi_X'(t Z) right] text{using symmetry of $Z$}\
&=2 E left[ Z phi_X(-t Z) phi_X'(t Z) right]
end{align}
Here, $phi_X(t)$ is the characteristic function of a random variable $X$.
Thank you for your comments.
probability-theory proof-verification normal-distribution characteristic-functions expected-value
probability-theory proof-verification normal-distribution characteristic-functions expected-value
edited Dec 14 '18 at 21:33
Batominovski
33.1k33293
asked Dec 14 '18 at 20:00
BobyBoby
1,0471929
edited Dec 14 '18 at 21:33
Batominovski
33.1k33293
asked Dec 14 '18 at 20:00
BobyBoby
1,0471929
edited Dec 14 '18 at 21:33
Batominovski
33.1k33293
edited Dec 14 '18 at 21:33
Batominovski
33.1k33293
edited Dec 14 '18 at 21:33
Batominovski
33.1k33293
33.1k33293
asked Dec 14 '18 at 20:00
BobyBoby
1,0471929
asked Dec 14 '18 at 20:00
BobyBoby
1,0471929
asked Dec 14 '18 at 20:00
BobyBoby
1,0471929
1,0471929
2
$begingroup$
Yes. $ $ $ $ $ $
$endgroup$
– Did
Dec 14 '18 at 20:13
$begingroup$
Except for justification of first step.
$endgroup$
– Kavi Rama Murthy
Dec 15 '18 at 0:04
$begingroup$
@KaviRamaMurthy What would be justification for that?
$endgroup$
– Boby
Dec 15 '18 at 16:34
$begingroup$
$E|phi_X(tZ)|^{2}=Ee^{-t^{2}X^{2}}$. Try to apply DCT.
$endgroup$
– Kavi Rama Murthy
Dec 15 '18 at 23:48
add a comment |
2
$begingroup$
Yes. $ $ $ $ $ $
$endgroup$
– Did
Dec 14 '18 at 20:13
$begingroup$
Except for justification of first step.
$endgroup$
– Kavi Rama Murthy
Dec 15 '18 at 0:04
$begingroup$
@KaviRamaMurthy What would be justification for that?
$endgroup$
– Boby
Dec 15 '18 at 16:34
$begingroup$
$E|phi_X(tZ)|^{2}=Ee^{-t^{2}X^{2}}$. Try to apply DCT.
$endgroup$
– Kavi Rama Murthy
Dec 15 '18 at 23:48
2
2
$begingroup$
Yes. $ $ $ $ $ $
$endgroup$
– Did
Dec 14 '18 at 20:13
$begingroup$
Yes. $ $ $ $ $ $
$endgroup$
– Did
Dec 14 '18 at 20:13
$begingroup$
Except for justification of first step.
$endgroup$
– Kavi Rama Murthy
Dec 15 '18 at 0:04
$begingroup$
Except for justification of first step.
$endgroup$
– Kavi Rama Murthy
Dec 15 '18 at 0:04
$begingroup$
@KaviRamaMurthy What would be justification for that?
$endgroup$
– Boby
Dec 15 '18 at 16:34
$begingroup$
@KaviRamaMurthy What would be justification for that?
$endgroup$
– Boby
Dec 15 '18 at 16:34
$begingroup$
$E|phi_X(tZ)|^{2}=Ee^{-t^{2}X^{2}}$. Try to apply DCT.
$endgroup$
– Kavi Rama Murthy
Dec 15 '18 at 23:48
$begingroup$
$E|phi_X(tZ)|^{2}=Ee^{-t^{2}X^{2}}$. Try to apply DCT.
$endgroup$
– Kavi Rama Murthy
Dec 15 '18 at 23:48
add a comment |
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2
$begingroup$
Yes. $ $ $ $ $ $
$endgroup$
– Did
Dec 14 '18 at 20:13
$begingroup$
Except for justification of first step.
$endgroup$
– Kavi Rama Murthy
Dec 15 '18 at 0:04
$begingroup$
@KaviRamaMurthy What would be justification for that?
$endgroup$
– Boby
Dec 15 '18 at 16:34
$begingroup$
$E|phi_X(tZ)|^{2}=Ee^{-t^{2}X^{2}}$. Try to apply DCT.
$endgroup$
– Kavi Rama Murthy
Dec 15 '18 at 23:48