Question involving Characteristic Functions and the Existence of a Distribution
$begingroup$
Question
Is it possible for $X$, $Y$ and $Z$ to have the same distribution and satisfy $X=U(Y+Z)$ where $U$ is uniform on $[0,1]$ and $Y$, $Z$ are independent of $U$ and of one another?
The above question is from Grimmett and Stirzaker.
My attempt
We translate the condition into characteristic functions. Let $phi(t)=Ee^{it X}$ be the characteristic function of $X$. Then
$$
phi(t)=Ee^{itUY}Ee^{itUZ}=(Ee^{itUX})^2=left[int_0^1 int e^{itux},dF(x), duright]^2=left[int_0^1 phi(tu), duright]^2
$$
using the independence and equality of distribution assumptions. We can write the above equation as
$$
phi (t)=frac{1}{t^2}left[int_0^t phi(y), dyright]^2
$$
but I am not sure where to proceed from here. I guess we have to solve a differential equation. Put $Phi(t)=int_0^t phi(y), dy$. Then we have that
$$
Phi'(t)=frac{1}{t^2}Phi(t)^2
$$
but I am unable to solve this differential equation.
Any help is appreciated.
real-analysis probability probability-theory characteristic-functions
asked Dec 9 '18 at 17:17
Foobaz JohnFoobaz John
21.8k41352
$endgroup$
add a comment |
$begingroup$
Question
Is it possible for $X$, $Y$ and $Z$ to have the same distribution and satisfy $X=U(Y+Z)$ where $U$ is uniform on $[0,1]$ and $Y$, $Z$ are independent of $U$ and of one another?
The above question is from Grimmett and Stirzaker.
My attempt
We translate the condition into characteristic functions. Let $phi(t)=Ee^{it X}$ be the characteristic function of $X$. Then
$$
phi(t)=Ee^{itUY}Ee^{itUZ}=(Ee^{itUX})^2=left[int_0^1 int e^{itux},dF(x), duright]^2=left[int_0^1 phi(tu), duright]^2
$$
using the independence and equality of distribution assumptions. We can write the above equation as
$$
phi (t)=frac{1}{t^2}left[int_0^t phi(y), dyright]^2
$$
but I am not sure where to proceed from here. I guess we have to solve a differential equation. Put $Phi(t)=int_0^t phi(y), dy$. Then we have that
$$
Phi'(t)=frac{1}{t^2}Phi(t)^2
$$
but I am unable to solve this differential equation.
Any help is appreciated.
real-analysis probability probability-theory characteristic-functions
asked Dec 9 '18 at 17:17
Foobaz JohnFoobaz John
21.8k41352
$endgroup$
$begingroup$
Incredibly enough, that ODE can be solved setting "$u = Phi(t)$" so that "$dfrac{du}{u^2} = dfrac{dt}{t^2}.$"
$endgroup$
– Will M.
Dec 9 '18 at 17:36
add a comment |
$begingroup$
Question
Is it possible for $X$, $Y$ and $Z$ to have the same distribution and satisfy $X=U(Y+Z)$ where $U$ is uniform on $[0,1]$ and $Y$, $Z$ are independent of $U$ and of one another?
The above question is from Grimmett and Stirzaker.
My attempt
We translate the condition into characteristic functions. Let $phi(t)=Ee^{it X}$ be the characteristic function of $X$. Then
$$
phi(t)=Ee^{itUY}Ee^{itUZ}=(Ee^{itUX})^2=left[int_0^1 int e^{itux},dF(x), duright]^2=left[int_0^1 phi(tu), duright]^2
$$
using the independence and equality of distribution assumptions. We can write the above equation as
$$
phi (t)=frac{1}{t^2}left[int_0^t phi(y), dyright]^2
$$
but I am not sure where to proceed from here. I guess we have to solve a differential equation. Put $Phi(t)=int_0^t phi(y), dy$. Then we have that
$$
Phi'(t)=frac{1}{t^2}Phi(t)^2
$$
but I am unable to solve this differential equation.
Any help is appreciated.
real-analysis probability probability-theory characteristic-functions
asked Dec 9 '18 at 17:17
Foobaz JohnFoobaz John
21.8k41352
$endgroup$
Question
Is it possible for $X$, $Y$ and $Z$ to have the same distribution and satisfy $X=U(Y+Z)$ where $U$ is uniform on $[0,1]$ and $Y$, $Z$ are independent of $U$ and of one another?
The above question is from Grimmett and Stirzaker.
My attempt
We translate the condition into characteristic functions. Let $phi(t)=Ee^{it X}$ be the characteristic function of $X$. Then
$$
phi(t)=Ee^{itUY}Ee^{itUZ}=(Ee^{itUX})^2=left[int_0^1 int e^{itux},dF(x), duright]^2=left[int_0^1 phi(tu), duright]^2
$$
using the independence and equality of distribution assumptions. We can write the above equation as
$$
phi (t)=frac{1}{t^2}left[int_0^t phi(y), dyright]^2
$$
but I am not sure where to proceed from here. I guess we have to solve a differential equation. Put $Phi(t)=int_0^t phi(y), dy$. Then we have that
$$
Phi'(t)=frac{1}{t^2}Phi(t)^2
$$
but I am unable to solve this differential equation.
Any help is appreciated.
real-analysis probability probability-theory characteristic-functions
real-analysis probability probability-theory characteristic-functions
asked Dec 9 '18 at 17:17
Foobaz JohnFoobaz John
21.8k41352
asked Dec 9 '18 at 17:17
Foobaz JohnFoobaz John
21.8k41352
edited Dec 9 '18 at 17:23
Foobaz John
asked Dec 9 '18 at 17:17
Foobaz JohnFoobaz John
21.8k41352
asked Dec 9 '18 at 17:17
Foobaz JohnFoobaz John
21.8k41352
asked Dec 9 '18 at 17:17
Foobaz JohnFoobaz John
21.8k41352
21.8k41352
$begingroup$
Incredibly enough, that ODE can be solved setting "$u = Phi(t)$" so that "$dfrac{du}{u^2} = dfrac{dt}{t^2}.$"
$endgroup$
– Will M.
Dec 9 '18 at 17:36
add a comment |
$begingroup$
Incredibly enough, that ODE can be solved setting "$u = Phi(t)$" so that "$dfrac{du}{u^2} = dfrac{dt}{t^2}.$"
$endgroup$
– Will M.
Dec 9 '18 at 17:36
$begingroup$
Incredibly enough, that ODE can be solved setting "$u = Phi(t)$" so that "$dfrac{du}{u^2} = dfrac{dt}{t^2}.$"
$endgroup$
– Will M.
Dec 9 '18 at 17:36
$begingroup$
Incredibly enough, that ODE can be solved setting "$u = Phi(t)$" so that "$dfrac{du}{u^2} = dfrac{dt}{t^2}.$"
$endgroup$
– Will M.
Dec 9 '18 at 17:36
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Rearranging your differential equation gives $dPhi/Phi^2=dt/t^2$ so $1/Phi=1/t+C$, i.e. $Phi=frac{t}{1+Ct}$. Hence $phi=frac{1}{(1+Ct)^2}$. Thus $C=0$ (otherwise $|phi|le 1$ would fail for some $tinBbb R$). This implies $X,,Y,,Z$ are identically $0$, which works.
answered Dec 9 '18 at 17:41
J.G.J.G.
24.6k22539
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Rearranging your differential equation gives $dPhi/Phi^2=dt/t^2$ so $1/Phi=1/t+C$, i.e. $Phi=frac{t}{1+Ct}$. Hence $phi=frac{1}{(1+Ct)^2}$. Thus $C=0$ (otherwise $|phi|le 1$ would fail for some $tinBbb R$). This implies $X,,Y,,Z$ are identically $0$, which works.
answered Dec 9 '18 at 17:41
J.G.J.G.
24.6k22539
$endgroup$
add a comment |
$begingroup$
Rearranging your differential equation gives $dPhi/Phi^2=dt/t^2$ so $1/Phi=1/t+C$, i.e. $Phi=frac{t}{1+Ct}$. Hence $phi=frac{1}{(1+Ct)^2}$. Thus $C=0$ (otherwise $|phi|le 1$ would fail for some $tinBbb R$). This implies $X,,Y,,Z$ are identically $0$, which works.
answered Dec 9 '18 at 17:41
J.G.J.G.
24.6k22539
$endgroup$
add a comment |
$begingroup$
Rearranging your differential equation gives $dPhi/Phi^2=dt/t^2$ so $1/Phi=1/t+C$, i.e. $Phi=frac{t}{1+Ct}$. Hence $phi=frac{1}{(1+Ct)^2}$. Thus $C=0$ (otherwise $|phi|le 1$ would fail for some $tinBbb R$). This implies $X,,Y,,Z$ are identically $0$, which works.
answered Dec 9 '18 at 17:41
J.G.J.G.
24.6k22539
$endgroup$
Rearranging your differential equation gives $dPhi/Phi^2=dt/t^2$ so $1/Phi=1/t+C$, i.e. $Phi=frac{t}{1+Ct}$. Hence $phi=frac{1}{(1+Ct)^2}$. Thus $C=0$ (otherwise $|phi|le 1$ would fail for some $tinBbb R$). This implies $X,,Y,,Z$ are identically $0$, which works.
answered Dec 9 '18 at 17:41
J.G.J.G.
24.6k22539
answered Dec 9 '18 at 17:41
J.G.J.G.
24.6k22539
answered Dec 9 '18 at 17:41
J.G.J.G.
24.6k22539
answered Dec 9 '18 at 17:41
J.G.J.G.
24.6k22539
24.6k22539
add a comment |
add a comment |
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$begingroup$
Incredibly enough, that ODE can be solved setting "$u = Phi(t)$" so that "$dfrac{du}{u^2} = dfrac{dt}{t^2}.$"
$endgroup$
– Will M.
Dec 9 '18 at 17:36