Functional equation for distribution function
$begingroup$
I have next functional equation for some distribution:
$$overline F_xi^2(T) = overline F_xi(2T) forall T>0$$
If suggest, that it differentiable, we can do something like this:
$$2overline F_xi(T) cdot overline F'_xi(T) = 2overline F'_xi(2T)$$
$$overline F_xi(T) cdot overline F'_xi(T) = overline F'_xi(2T)$$
But how to solve this differential functional equation?
where $$overline F_xi = 1 - F_xi = P(xi > T)$$
Also, I have $mathbb Exi=1.$
ordinary-differential-equations probability-theory probability-distributions
asked Nov 26 '18 at 13:13
LisaLisa
255
$endgroup$
add a comment |
$begingroup$
I have next functional equation for some distribution:
$$overline F_xi^2(T) = overline F_xi(2T) forall T>0$$
If suggest, that it differentiable, we can do something like this:
$$2overline F_xi(T) cdot overline F'_xi(T) = 2overline F'_xi(2T)$$
$$overline F_xi(T) cdot overline F'_xi(T) = overline F'_xi(2T)$$
But how to solve this differential functional equation?
where $$overline F_xi = 1 - F_xi = P(xi > T)$$
Also, I have $mathbb Exi=1.$
ordinary-differential-equations probability-theory probability-distributions
asked Nov 26 '18 at 13:13
LisaLisa
255
$endgroup$
add a comment |
$begingroup$
I have next functional equation for some distribution:
$$overline F_xi^2(T) = overline F_xi(2T) forall T>0$$
If suggest, that it differentiable, we can do something like this:
$$2overline F_xi(T) cdot overline F'_xi(T) = 2overline F'_xi(2T)$$
$$overline F_xi(T) cdot overline F'_xi(T) = overline F'_xi(2T)$$
But how to solve this differential functional equation?
where $$overline F_xi = 1 - F_xi = P(xi > T)$$
Also, I have $mathbb Exi=1.$
ordinary-differential-equations probability-theory probability-distributions
asked Nov 26 '18 at 13:13
LisaLisa
255
$endgroup$
I have next functional equation for some distribution:
$$overline F_xi^2(T) = overline F_xi(2T) forall T>0$$
If suggest, that it differentiable, we can do something like this:
$$2overline F_xi(T) cdot overline F'_xi(T) = 2overline F'_xi(2T)$$
$$overline F_xi(T) cdot overline F'_xi(T) = overline F'_xi(2T)$$
But how to solve this differential functional equation?
where $$overline F_xi = 1 - F_xi = P(xi > T)$$
Also, I have $mathbb Exi=1.$
ordinary-differential-equations probability-theory probability-distributions
ordinary-differential-equations probability-theory probability-distributions
asked Nov 26 '18 at 13:13
LisaLisa
255
asked Nov 26 '18 at 13:13
LisaLisa
255
edited Nov 26 '18 at 18:24
Lisa
asked Nov 26 '18 at 13:13
LisaLisa
255
asked Nov 26 '18 at 13:13
LisaLisa
255
asked Nov 26 '18 at 13:13
LisaLisa
255
255
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
For brevity, put $f=bar F_{xi}$. Since $f$ is decreasing, if $f(x)=0$, a simple proof by induction shows $f(2^{-n}x)=0$ and therefore $fle 0$ everywhere, which contradicts $Bbb E[xi]=1$. Hence, $f>0$ and we can define $g(t)=ln f(t)$ so the hypothesis yields $2g(t)=g(2t)$. If we assume $f$ is analytic on $Bbb R$, $g'(t)=g'(2t)$ and induction shows $g^{(n)}(t)=2^{n-1}g^{(n)}(2t)Rightarrow g^{(n)}(0)=0$ for $n>1$ so the Taylor Series of $g$ reduces to $g(t)=a_0+a_1t$. Imposing the condition that $2a_0+2a_1t=2g(t)=g(2t)=a_0+2a_1t$, and equating coefficients forces $a_0=0$. The definition of $g$ shows $f(t)=e^{g(t)}=e^{at}$ so that $xisim text{Exp}(-a)$, but $Bbb E[xi]=-frac 1a=1$, so $a=-1$.
answered Nov 26 '18 at 22:30
Guacho PerezGuacho Perez
3,91411132
$endgroup$
add a comment |
$begingroup$
Hint: divide both sides by F(2T)
answered Nov 26 '18 at 15:48
user619894user619894
111
$endgroup$
$begingroup$
you must try to provide a bit more detailed Hint, or solution.
$endgroup$
– idea
Nov 26 '18 at 18:26
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
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active
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votes
$begingroup$
For brevity, put $f=bar F_{xi}$. Since $f$ is decreasing, if $f(x)=0$, a simple proof by induction shows $f(2^{-n}x)=0$ and therefore $fle 0$ everywhere, which contradicts $Bbb E[xi]=1$. Hence, $f>0$ and we can define $g(t)=ln f(t)$ so the hypothesis yields $2g(t)=g(2t)$. If we assume $f$ is analytic on $Bbb R$, $g'(t)=g'(2t)$ and induction shows $g^{(n)}(t)=2^{n-1}g^{(n)}(2t)Rightarrow g^{(n)}(0)=0$ for $n>1$ so the Taylor Series of $g$ reduces to $g(t)=a_0+a_1t$. Imposing the condition that $2a_0+2a_1t=2g(t)=g(2t)=a_0+2a_1t$, and equating coefficients forces $a_0=0$. The definition of $g$ shows $f(t)=e^{g(t)}=e^{at}$ so that $xisim text{Exp}(-a)$, but $Bbb E[xi]=-frac 1a=1$, so $a=-1$.
answered Nov 26 '18 at 22:30
Guacho PerezGuacho Perez
3,91411132
$endgroup$
add a comment |
$begingroup$
For brevity, put $f=bar F_{xi}$. Since $f$ is decreasing, if $f(x)=0$, a simple proof by induction shows $f(2^{-n}x)=0$ and therefore $fle 0$ everywhere, which contradicts $Bbb E[xi]=1$. Hence, $f>0$ and we can define $g(t)=ln f(t)$ so the hypothesis yields $2g(t)=g(2t)$. If we assume $f$ is analytic on $Bbb R$, $g'(t)=g'(2t)$ and induction shows $g^{(n)}(t)=2^{n-1}g^{(n)}(2t)Rightarrow g^{(n)}(0)=0$ for $n>1$ so the Taylor Series of $g$ reduces to $g(t)=a_0+a_1t$. Imposing the condition that $2a_0+2a_1t=2g(t)=g(2t)=a_0+2a_1t$, and equating coefficients forces $a_0=0$. The definition of $g$ shows $f(t)=e^{g(t)}=e^{at}$ so that $xisim text{Exp}(-a)$, but $Bbb E[xi]=-frac 1a=1$, so $a=-1$.
answered Nov 26 '18 at 22:30
Guacho PerezGuacho Perez
3,91411132
$endgroup$
add a comment |
$begingroup$
For brevity, put $f=bar F_{xi}$. Since $f$ is decreasing, if $f(x)=0$, a simple proof by induction shows $f(2^{-n}x)=0$ and therefore $fle 0$ everywhere, which contradicts $Bbb E[xi]=1$. Hence, $f>0$ and we can define $g(t)=ln f(t)$ so the hypothesis yields $2g(t)=g(2t)$. If we assume $f$ is analytic on $Bbb R$, $g'(t)=g'(2t)$ and induction shows $g^{(n)}(t)=2^{n-1}g^{(n)}(2t)Rightarrow g^{(n)}(0)=0$ for $n>1$ so the Taylor Series of $g$ reduces to $g(t)=a_0+a_1t$. Imposing the condition that $2a_0+2a_1t=2g(t)=g(2t)=a_0+2a_1t$, and equating coefficients forces $a_0=0$. The definition of $g$ shows $f(t)=e^{g(t)}=e^{at}$ so that $xisim text{Exp}(-a)$, but $Bbb E[xi]=-frac 1a=1$, so $a=-1$.
answered Nov 26 '18 at 22:30
Guacho PerezGuacho Perez
3,91411132
$endgroup$
For brevity, put $f=bar F_{xi}$. Since $f$ is decreasing, if $f(x)=0$, a simple proof by induction shows $f(2^{-n}x)=0$ and therefore $fle 0$ everywhere, which contradicts $Bbb E[xi]=1$. Hence, $f>0$ and we can define $g(t)=ln f(t)$ so the hypothesis yields $2g(t)=g(2t)$. If we assume $f$ is analytic on $Bbb R$, $g'(t)=g'(2t)$ and induction shows $g^{(n)}(t)=2^{n-1}g^{(n)}(2t)Rightarrow g^{(n)}(0)=0$ for $n>1$ so the Taylor Series of $g$ reduces to $g(t)=a_0+a_1t$. Imposing the condition that $2a_0+2a_1t=2g(t)=g(2t)=a_0+2a_1t$, and equating coefficients forces $a_0=0$. The definition of $g$ shows $f(t)=e^{g(t)}=e^{at}$ so that $xisim text{Exp}(-a)$, but $Bbb E[xi]=-frac 1a=1$, so $a=-1$.
answered Nov 26 '18 at 22:30
Guacho PerezGuacho Perez
3,91411132
answered Nov 26 '18 at 22:30
Guacho PerezGuacho Perez
3,91411132
answered Nov 26 '18 at 22:30
Guacho PerezGuacho Perez
3,91411132
answered Nov 26 '18 at 22:30
Guacho PerezGuacho Perez
3,91411132
3,91411132
add a comment |
add a comment |
$begingroup$
Hint: divide both sides by F(2T)
answered Nov 26 '18 at 15:48
user619894user619894
111
$endgroup$
$begingroup$
you must try to provide a bit more detailed Hint, or solution.
$endgroup$
– idea
Nov 26 '18 at 18:26
add a comment |
$begingroup$
Hint: divide both sides by F(2T)
answered Nov 26 '18 at 15:48
user619894user619894
111
$endgroup$
$begingroup$
you must try to provide a bit more detailed Hint, or solution.
$endgroup$
– idea
Nov 26 '18 at 18:26
add a comment |
$begingroup$
Hint: divide both sides by F(2T)
answered Nov 26 '18 at 15:48
user619894user619894
111
$endgroup$
Hint: divide both sides by F(2T)
answered Nov 26 '18 at 15:48
user619894user619894
111
answered Nov 26 '18 at 15:48
user619894user619894
111
answered Nov 26 '18 at 15:48
user619894user619894
111
answered Nov 26 '18 at 15:48
user619894user619894
111
111
$begingroup$
you must try to provide a bit more detailed Hint, or solution.
$endgroup$
– idea
Nov 26 '18 at 18:26
add a comment |
$begingroup$
you must try to provide a bit more detailed Hint, or solution.
$endgroup$
– idea
Nov 26 '18 at 18:26
$begingroup$
you must try to provide a bit more detailed Hint, or solution.
$endgroup$
– idea
Nov 26 '18 at 18:26
$begingroup$
you must try to provide a bit more detailed Hint, or solution.
$endgroup$
– idea
Nov 26 '18 at 18:26
add a comment |
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