Characteristic function equal to 1
0
$begingroup$
Is the fact that the characteristic function of $X$ is equal to 1 enough to say that $X=0$ almost surely?
Is this the correct proof?
The characteristic function is determined uniquely. The random variable $X=0$ has the characteristic function equal to 1. Hence X=0. But why only almost surely? Can anyone relate?
probability probability-distributions characteristic-functions
asked Dec 4 '18 at 17:35
ryszard egginkryszard eggink
308110
$endgroup$
add a comment |
0
$begingroup$
Is the fact that the characteristic function of $X$ is equal to 1 enough to say that $X=0$ almost surely?
Is this the correct proof?
The characteristic function is determined uniquely. The random variable $X=0$ has the characteristic function equal to 1. Hence X=0. But why only almost surely? Can anyone relate?
probability probability-distributions characteristic-functions
asked Dec 4 '18 at 17:35
ryszard egginkryszard eggink
308110
$endgroup$
$begingroup$
The distribution is unique. If $X = Y$ ae. then their distributions are the same.
$endgroup$
– copper.hat
Dec 4 '18 at 17:53
1
$begingroup$
I see no proof in your post. For a proof that, if $E(e^{itX})=1$ for every $t$, then $X=0$ almost surely, note that $E(e^{itX})=1$ if and only if $tXin2pimathbb Z$ almost surely. Thus, if $E(e^{itX})=E(e^{isX})=1$ for some nonzero $s$ and $t$, then $Xin(2pimathbb Z/t)cap(2pimathbb Z/s)$ almost surely. Finally, if $s/t$ is irrational, then $(2pimathbb Z/t)cap(2pimathbb Z/s)={0}$ hence you are done.
$endgroup$
– Did
Dec 4 '18 at 18:26
$begingroup$
For the characteristic function to be $1$ then $P(X=0)=1$, but $X$ could have other values as long as the probability is $0$.
$endgroup$
– herb steinberg
Dec 4 '18 at 18:58
$begingroup$
@herbsteinberg you mean their probability tends to zero or is zero, when it is zero, they do not exist right?
$endgroup$
– ryszard eggink
Dec 4 '18 at 19:07
$begingroup$
It is $0$. Simple example $0le Xle 1$, where $X=x$ for rational $x$ and $X=0$ otherwise, where the probability space is the unit interval, with ordinary measure.
$endgroup$
– herb steinberg
Dec 4 '18 at 22:40
add a comment |
0
0
0
$begingroup$
Is the fact that the characteristic function of $X$ is equal to 1 enough to say that $X=0$ almost surely?
Is this the correct proof?
The characteristic function is determined uniquely. The random variable $X=0$ has the characteristic function equal to 1. Hence X=0. But why only almost surely? Can anyone relate?
probability probability-distributions characteristic-functions
asked Dec 4 '18 at 17:35
ryszard egginkryszard eggink
308110
$endgroup$
Is the fact that the characteristic function of $X$ is equal to 1 enough to say that $X=0$ almost surely?
Is this the correct proof?
The characteristic function is determined uniquely. The random variable $X=0$ has the characteristic function equal to 1. Hence X=0. But why only almost surely? Can anyone relate?
probability probability-distributions characteristic-functions
probability probability-distributions characteristic-functions
asked Dec 4 '18 at 17:35
ryszard egginkryszard eggink
308110
asked Dec 4 '18 at 17:35
ryszard egginkryszard eggink
308110
asked Dec 4 '18 at 17:35
ryszard egginkryszard eggink
308110
asked Dec 4 '18 at 17:35
ryszard egginkryszard eggink
308110
asked Dec 4 '18 at 17:35
ryszard egginkryszard eggink
308110
308110
$begingroup$
The distribution is unique. If $X = Y$ ae. then their distributions are the same.
$endgroup$
– copper.hat
Dec 4 '18 at 17:53
1
$begingroup$
I see no proof in your post. For a proof that, if $E(e^{itX})=1$ for every $t$, then $X=0$ almost surely, note that $E(e^{itX})=1$ if and only if $tXin2pimathbb Z$ almost surely. Thus, if $E(e^{itX})=E(e^{isX})=1$ for some nonzero $s$ and $t$, then $Xin(2pimathbb Z/t)cap(2pimathbb Z/s)$ almost surely. Finally, if $s/t$ is irrational, then $(2pimathbb Z/t)cap(2pimathbb Z/s)={0}$ hence you are done.
$endgroup$
– Did
Dec 4 '18 at 18:26
$begingroup$
For the characteristic function to be $1$ then $P(X=0)=1$, but $X$ could have other values as long as the probability is $0$.
$endgroup$
– herb steinberg
Dec 4 '18 at 18:58
$begingroup$
@herbsteinberg you mean their probability tends to zero or is zero, when it is zero, they do not exist right?
$endgroup$
– ryszard eggink
Dec 4 '18 at 19:07
$begingroup$
It is $0$. Simple example $0le Xle 1$, where $X=x$ for rational $x$ and $X=0$ otherwise, where the probability space is the unit interval, with ordinary measure.
$endgroup$
– herb steinberg
Dec 4 '18 at 22:40
add a comment |
$begingroup$
The distribution is unique. If $X = Y$ ae. then their distributions are the same.
$endgroup$
– copper.hat
Dec 4 '18 at 17:53
1
$begingroup$
I see no proof in your post. For a proof that, if $E(e^{itX})=1$ for every $t$, then $X=0$ almost surely, note that $E(e^{itX})=1$ if and only if $tXin2pimathbb Z$ almost surely. Thus, if $E(e^{itX})=E(e^{isX})=1$ for some nonzero $s$ and $t$, then $Xin(2pimathbb Z/t)cap(2pimathbb Z/s)$ almost surely. Finally, if $s/t$ is irrational, then $(2pimathbb Z/t)cap(2pimathbb Z/s)={0}$ hence you are done.
$endgroup$
– Did
Dec 4 '18 at 18:26
$begingroup$
For the characteristic function to be $1$ then $P(X=0)=1$, but $X$ could have other values as long as the probability is $0$.
$endgroup$
– herb steinberg
Dec 4 '18 at 18:58
$begingroup$
@herbsteinberg you mean their probability tends to zero or is zero, when it is zero, they do not exist right?
$endgroup$
– ryszard eggink
Dec 4 '18 at 19:07
$begingroup$
It is $0$. Simple example $0le Xle 1$, where $X=x$ for rational $x$ and $X=0$ otherwise, where the probability space is the unit interval, with ordinary measure.
$endgroup$
– herb steinberg
Dec 4 '18 at 22:40
$begingroup$
The distribution is unique. If $X = Y$ ae. then their distributions are the same.
$endgroup$
– copper.hat
Dec 4 '18 at 17:53
$begingroup$
The distribution is unique. If $X = Y$ ae. then their distributions are the same.
$endgroup$
– copper.hat
Dec 4 '18 at 17:53
1
1
$begingroup$
I see no proof in your post. For a proof that, if $E(e^{itX})=1$ for every $t$, then $X=0$ almost surely, note that $E(e^{itX})=1$ if and only if $tXin2pimathbb Z$ almost surely. Thus, if $E(e^{itX})=E(e^{isX})=1$ for some nonzero $s$ and $t$, then $Xin(2pimathbb Z/t)cap(2pimathbb Z/s)$ almost surely. Finally, if $s/t$ is irrational, then $(2pimathbb Z/t)cap(2pimathbb Z/s)={0}$ hence you are done.
$endgroup$
– Did
Dec 4 '18 at 18:26
$begingroup$
I see no proof in your post. For a proof that, if $E(e^{itX})=1$ for every $t$, then $X=0$ almost surely, note that $E(e^{itX})=1$ if and only if $tXin2pimathbb Z$ almost surely. Thus, if $E(e^{itX})=E(e^{isX})=1$ for some nonzero $s$ and $t$, then $Xin(2pimathbb Z/t)cap(2pimathbb Z/s)$ almost surely. Finally, if $s/t$ is irrational, then $(2pimathbb Z/t)cap(2pimathbb Z/s)={0}$ hence you are done.
$endgroup$
– Did
Dec 4 '18 at 18:26
$begingroup$
For the characteristic function to be $1$ then $P(X=0)=1$, but $X$ could have other values as long as the probability is $0$.
$endgroup$
– herb steinberg
Dec 4 '18 at 18:58
$begingroup$
For the characteristic function to be $1$ then $P(X=0)=1$, but $X$ could have other values as long as the probability is $0$.
$endgroup$
– herb steinberg
Dec 4 '18 at 18:58
$begingroup$
@herbsteinberg you mean their probability tends to zero or is zero, when it is zero, they do not exist right?
$endgroup$
– ryszard eggink
Dec 4 '18 at 19:07
$begingroup$
@herbsteinberg you mean their probability tends to zero or is zero, when it is zero, they do not exist right?
$endgroup$
– ryszard eggink
Dec 4 '18 at 19:07
$begingroup$
It is $0$. Simple example $0le Xle 1$, where $X=x$ for rational $x$ and $X=0$ otherwise, where the probability space is the unit interval, with ordinary measure.
$endgroup$
– herb steinberg
Dec 4 '18 at 22:40
$begingroup$
It is $0$. Simple example $0le Xle 1$, where $X=x$ for rational $x$ and $X=0$ otherwise, where the probability space is the unit interval, with ordinary measure.
$endgroup$
– herb steinberg
Dec 4 '18 at 22:40
add a comment |
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$begingroup$
The distribution is unique. If $X = Y$ ae. then their distributions are the same.
$endgroup$
– copper.hat
Dec 4 '18 at 17:53
1
$begingroup$
I see no proof in your post. For a proof that, if $E(e^{itX})=1$ for every $t$, then $X=0$ almost surely, note that $E(e^{itX})=1$ if and only if $tXin2pimathbb Z$ almost surely. Thus, if $E(e^{itX})=E(e^{isX})=1$ for some nonzero $s$ and $t$, then $Xin(2pimathbb Z/t)cap(2pimathbb Z/s)$ almost surely. Finally, if $s/t$ is irrational, then $(2pimathbb Z/t)cap(2pimathbb Z/s)={0}$ hence you are done.
$endgroup$
– Did
Dec 4 '18 at 18:26
$begingroup$
For the characteristic function to be $1$ then $P(X=0)=1$, but $X$ could have other values as long as the probability is $0$.
$endgroup$
– herb steinberg
Dec 4 '18 at 18:58
$begingroup$
@herbsteinberg you mean their probability tends to zero or is zero, when it is zero, they do not exist right?
$endgroup$
– ryszard eggink
Dec 4 '18 at 19:07
$begingroup$
It is $0$. Simple example $0le Xle 1$, where $X=x$ for rational $x$ and $X=0$ otherwise, where the probability space is the unit interval, with ordinary measure.
$endgroup$
– herb steinberg
Dec 4 '18 at 22:40