Mixing
Proving that if $tX_1 + sX_2 stackrel{D}{=} sqrt{t^2 + s^2}X$ then $X stackrel{D}{=} N(0, delta^2)$
$begingroup$
Assume that for all $s, t in mathbb{R}$ the following property
$$tX_1 + sX_2 stackrel{D}{=} sqrt{t^2 + s^2}X tag{1}$$
is true.
Moreover $X_1, X_2, X$ are i. i. d.
My task is to prove that if $(1)$ stands then $X stackrel{D}{=}N(0, sigma^2)$. So $X$ is symmetrically normally distributed.
How can it be proved? I suppose that characteristic functions should be used but I have no idea how. I will appreciate any hints or tips.
probability probability-distributions characteristic-functions
asked Feb 2 at 23:01
HendrraHendrra
1,214516
$endgroup$
add a comment |
$begingroup$
Assume that for all $s, t in mathbb{R}$ the following property
$$tX_1 + sX_2 stackrel{D}{=} sqrt{t^2 + s^2}X tag{1}$$
is true.
Moreover $X_1, X_2, X$ are i. i. d.
My task is to prove that if $(1)$ stands then $X stackrel{D}{=}N(0, sigma^2)$. So $X$ is symmetrically normally distributed.
How can it be proved? I suppose that characteristic functions should be used but I have no idea how. I will appreciate any hints or tips.
probability probability-distributions characteristic-functions
asked Feb 2 at 23:01
HendrraHendrra
1,214516
$endgroup$
1
$begingroup$
Are you assuming something like $X_1$ and $X_2$ are iid, and have the same distribution as $X$?
$endgroup$
– kimchi lover
Feb 2 at 23:06
$begingroup$
@kimchilover Yes. I will edit my post. Thanks!
$endgroup$
– Hendrra
Feb 2 at 23:15
$begingroup$
Check out en.wikipedia.org/wiki/Maxwell%27s_theorem
$endgroup$
– kimchi lover
Feb 4 at 14:30
add a comment |
$begingroup$
Assume that for all $s, t in mathbb{R}$ the following property
$$tX_1 + sX_2 stackrel{D}{=} sqrt{t^2 + s^2}X tag{1}$$
is true.
Moreover $X_1, X_2, X$ are i. i. d.
My task is to prove that if $(1)$ stands then $X stackrel{D}{=}N(0, sigma^2)$. So $X$ is symmetrically normally distributed.
How can it be proved? I suppose that characteristic functions should be used but I have no idea how. I will appreciate any hints or tips.
probability probability-distributions characteristic-functions
asked Feb 2 at 23:01
HendrraHendrra
1,214516
$endgroup$
Assume that for all $s, t in mathbb{R}$ the following property
$$tX_1 + sX_2 stackrel{D}{=} sqrt{t^2 + s^2}X tag{1}$$
is true.
Moreover $X_1, X_2, X$ are i. i. d.
My task is to prove that if $(1)$ stands then $X stackrel{D}{=}N(0, sigma^2)$. So $X$ is symmetrically normally distributed.
How can it be proved? I suppose that characteristic functions should be used but I have no idea how. I will appreciate any hints or tips.
probability probability-distributions characteristic-functions
probability probability-distributions characteristic-functions
asked Feb 2 at 23:01
HendrraHendrra
1,214516
asked Feb 2 at 23:01
HendrraHendrra
1,214516
edited Feb 2 at 23:15
Hendrra
asked Feb 2 at 23:01
HendrraHendrra
1,214516
asked Feb 2 at 23:01
HendrraHendrra
1,214516
asked Feb 2 at 23:01
HendrraHendrra
1,214516
1,214516
1
$begingroup$
Are you assuming something like $X_1$ and $X_2$ are iid, and have the same distribution as $X$?
$endgroup$
– kimchi lover
Feb 2 at 23:06
$begingroup$
@kimchilover Yes. I will edit my post. Thanks!
$endgroup$
– Hendrra
Feb 2 at 23:15
$begingroup$
Check out en.wikipedia.org/wiki/Maxwell%27s_theorem
$endgroup$
– kimchi lover
Feb 4 at 14:30
add a comment |
1
$begingroup$
Are you assuming something like $X_1$ and $X_2$ are iid, and have the same distribution as $X$?
$endgroup$
– kimchi lover
Feb 2 at 23:06
$begingroup$
@kimchilover Yes. I will edit my post. Thanks!
$endgroup$
– Hendrra
Feb 2 at 23:15
$begingroup$
Check out en.wikipedia.org/wiki/Maxwell%27s_theorem
$endgroup$
– kimchi lover
Feb 4 at 14:30
1
1
$begingroup$
Are you assuming something like $X_1$ and $X_2$ are iid, and have the same distribution as $X$?
$endgroup$
– kimchi lover
Feb 2 at 23:06
$begingroup$
Are you assuming something like $X_1$ and $X_2$ are iid, and have the same distribution as $X$?
$endgroup$
– kimchi lover
Feb 2 at 23:06
$begingroup$
@kimchilover Yes. I will edit my post. Thanks!
$endgroup$
– Hendrra
Feb 2 at 23:15
$begingroup$
@kimchilover Yes. I will edit my post. Thanks!
$endgroup$
– Hendrra
Feb 2 at 23:15
$begingroup$
Check out en.wikipedia.org/wiki/Maxwell%27s_theorem
$endgroup$
– kimchi lover
Feb 4 at 14:30
$begingroup$
Check out en.wikipedia.org/wiki/Maxwell%27s_theorem
$endgroup$
– kimchi lover
Feb 4 at 14:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There are 2 ingredients. As you guessed, characteristic functions are involved. Here is a hint outline.
Step 1: derive the identity $phi(s)phi(t)=phi(sqrt{s^2+t^2})$ obeyed by the characteristic function $phi$. Then, somehow manipulate that to obtain a condition like $f(s)f(t)=f(s+t)$ or $g(s)+g(t)=g(s+t)$, valid for all real $s,t$.
Step 2: realize that this puts a severe limitation of what $f$ or $g$ (and ultimately $phi$) can be.
answered Feb 2 at 23:29
kimchi loverkimchi lover
11.8k31229
$endgroup$
$begingroup$
Thanks for your answer. How can $phi(s)phi(t)=phi(sqrt{s^2+t^2})$ be manipulated? It's hard to do anything with those functions.
$endgroup$
– Hendrra
Feb 4 at 12:45
$begingroup$
You have $phi(t)phi(0)=phi(sqrt{t^2})$ so $phi(t)$ depends on $t^2$ alone. Write it as $phi(t)=f(t^2)$. Now $f(a+b)=f(a+b)$ for all $ge0$ arguments. And so on. As spelled out in Murthy's answer below.
$endgroup$
– kimchi lover
Feb 4 at 12:51
add a comment |
$begingroup$
Hints: Assume that $X_1$ is not $0$ with positive probability. let $phi$ be the characteristic function of $X_1$. Then $phi (t) phi (s)=phi (sqrt {t^{2}+s^{2}})$. If $phi (t)=0$ for some $t$ then we get $phi (sqrt {t^{2}+s^{2}})=0$ for all $s$ and hence $phi (u)=0$ for all $u >|t|$. S how that this leads to a contradiction to the given property. Otherwise, by a well known fact there is a continuous function $psi$ such that $e^{psi (t)}=phi (t)$ and $psi (0)=0$. Let $g(t)=psi (sqrt t)$ for $t >0$. Show that $g(t+s)=g(t)+g(s)$. I will leave the rest of the argument to you.
answered Feb 2 at 23:26
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
There are 2 ingredients. As you guessed, characteristic functions are involved. Here is a hint outline.
Step 1: derive the identity $phi(s)phi(t)=phi(sqrt{s^2+t^2})$ obeyed by the characteristic function $phi$. Then, somehow manipulate that to obtain a condition like $f(s)f(t)=f(s+t)$ or $g(s)+g(t)=g(s+t)$, valid for all real $s,t$.
Step 2: realize that this puts a severe limitation of what $f$ or $g$ (and ultimately $phi$) can be.
answered Feb 2 at 23:29
kimchi loverkimchi lover
11.8k31229
$endgroup$
$begingroup$
Thanks for your answer. How can $phi(s)phi(t)=phi(sqrt{s^2+t^2})$ be manipulated? It's hard to do anything with those functions.
$endgroup$
– Hendrra
Feb 4 at 12:45
$begingroup$
You have $phi(t)phi(0)=phi(sqrt{t^2})$ so $phi(t)$ depends on $t^2$ alone. Write it as $phi(t)=f(t^2)$. Now $f(a+b)=f(a+b)$ for all $ge0$ arguments. And so on. As spelled out in Murthy's answer below.
$endgroup$
– kimchi lover
Feb 4 at 12:51
add a comment |
$begingroup$
There are 2 ingredients. As you guessed, characteristic functions are involved. Here is a hint outline.
Step 1: derive the identity $phi(s)phi(t)=phi(sqrt{s^2+t^2})$ obeyed by the characteristic function $phi$. Then, somehow manipulate that to obtain a condition like $f(s)f(t)=f(s+t)$ or $g(s)+g(t)=g(s+t)$, valid for all real $s,t$.
Step 2: realize that this puts a severe limitation of what $f$ or $g$ (and ultimately $phi$) can be.
answered Feb 2 at 23:29
kimchi loverkimchi lover
11.8k31229
$endgroup$
$begingroup$
Thanks for your answer. How can $phi(s)phi(t)=phi(sqrt{s^2+t^2})$ be manipulated? It's hard to do anything with those functions.
$endgroup$
– Hendrra
Feb 4 at 12:45
$begingroup$
You have $phi(t)phi(0)=phi(sqrt{t^2})$ so $phi(t)$ depends on $t^2$ alone. Write it as $phi(t)=f(t^2)$. Now $f(a+b)=f(a+b)$ for all $ge0$ arguments. And so on. As spelled out in Murthy's answer below.
$endgroup$
– kimchi lover
Feb 4 at 12:51
add a comment |
$begingroup$
There are 2 ingredients. As you guessed, characteristic functions are involved. Here is a hint outline.
Step 1: derive the identity $phi(s)phi(t)=phi(sqrt{s^2+t^2})$ obeyed by the characteristic function $phi$. Then, somehow manipulate that to obtain a condition like $f(s)f(t)=f(s+t)$ or $g(s)+g(t)=g(s+t)$, valid for all real $s,t$.
Step 2: realize that this puts a severe limitation of what $f$ or $g$ (and ultimately $phi$) can be.
answered Feb 2 at 23:29
kimchi loverkimchi lover
11.8k31229
$endgroup$
There are 2 ingredients. As you guessed, characteristic functions are involved. Here is a hint outline.
Step 1: derive the identity $phi(s)phi(t)=phi(sqrt{s^2+t^2})$ obeyed by the characteristic function $phi$. Then, somehow manipulate that to obtain a condition like $f(s)f(t)=f(s+t)$ or $g(s)+g(t)=g(s+t)$, valid for all real $s,t$.
Step 2: realize that this puts a severe limitation of what $f$ or $g$ (and ultimately $phi$) can be.
answered Feb 2 at 23:29
kimchi loverkimchi lover
11.8k31229
answered Feb 2 at 23:29
kimchi loverkimchi lover
11.8k31229
answered Feb 2 at 23:29
kimchi loverkimchi lover
11.8k31229
answered Feb 2 at 23:29
kimchi loverkimchi lover
11.8k31229
11.8k31229
$begingroup$
Thanks for your answer. How can $phi(s)phi(t)=phi(sqrt{s^2+t^2})$ be manipulated? It's hard to do anything with those functions.
$endgroup$
– Hendrra
Feb 4 at 12:45
$begingroup$
You have $phi(t)phi(0)=phi(sqrt{t^2})$ so $phi(t)$ depends on $t^2$ alone. Write it as $phi(t)=f(t^2)$. Now $f(a+b)=f(a+b)$ for all $ge0$ arguments. And so on. As spelled out in Murthy's answer below.
$endgroup$
– kimchi lover
Feb 4 at 12:51
add a comment |
$begingroup$
Thanks for your answer. How can $phi(s)phi(t)=phi(sqrt{s^2+t^2})$ be manipulated? It's hard to do anything with those functions.
$endgroup$
– Hendrra
Feb 4 at 12:45
$begingroup$
You have $phi(t)phi(0)=phi(sqrt{t^2})$ so $phi(t)$ depends on $t^2$ alone. Write it as $phi(t)=f(t^2)$. Now $f(a+b)=f(a+b)$ for all $ge0$ arguments. And so on. As spelled out in Murthy's answer below.
$endgroup$
– kimchi lover
Feb 4 at 12:51
$begingroup$
Thanks for your answer. How can $phi(s)phi(t)=phi(sqrt{s^2+t^2})$ be manipulated? It's hard to do anything with those functions.
$endgroup$
– Hendrra
Feb 4 at 12:45
$begingroup$
Thanks for your answer. How can $phi(s)phi(t)=phi(sqrt{s^2+t^2})$ be manipulated? It's hard to do anything with those functions.
$endgroup$
– Hendrra
Feb 4 at 12:45
$begingroup$
You have $phi(t)phi(0)=phi(sqrt{t^2})$ so $phi(t)$ depends on $t^2$ alone. Write it as $phi(t)=f(t^2)$. Now $f(a+b)=f(a+b)$ for all $ge0$ arguments. And so on. As spelled out in Murthy's answer below.
$endgroup$
– kimchi lover
Feb 4 at 12:51
$begingroup$
You have $phi(t)phi(0)=phi(sqrt{t^2})$ so $phi(t)$ depends on $t^2$ alone. Write it as $phi(t)=f(t^2)$. Now $f(a+b)=f(a+b)$ for all $ge0$ arguments. And so on. As spelled out in Murthy's answer below.
$endgroup$
– kimchi lover
Feb 4 at 12:51
add a comment |
$begingroup$
Hints: Assume that $X_1$ is not $0$ with positive probability. let $phi$ be the characteristic function of $X_1$. Then $phi (t) phi (s)=phi (sqrt {t^{2}+s^{2}})$. If $phi (t)=0$ for some $t$ then we get $phi (sqrt {t^{2}+s^{2}})=0$ for all $s$ and hence $phi (u)=0$ for all $u >|t|$. S how that this leads to a contradiction to the given property. Otherwise, by a well known fact there is a continuous function $psi$ such that $e^{psi (t)}=phi (t)$ and $psi (0)=0$. Let $g(t)=psi (sqrt t)$ for $t >0$. Show that $g(t+s)=g(t)+g(s)$. I will leave the rest of the argument to you.
answered Feb 2 at 23:26
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
$endgroup$
add a comment |
$begingroup$
Hints: Assume that $X_1$ is not $0$ with positive probability. let $phi$ be the characteristic function of $X_1$. Then $phi (t) phi (s)=phi (sqrt {t^{2}+s^{2}})$. If $phi (t)=0$ for some $t$ then we get $phi (sqrt {t^{2}+s^{2}})=0$ for all $s$ and hence $phi (u)=0$ for all $u >|t|$. S how that this leads to a contradiction to the given property. Otherwise, by a well known fact there is a continuous function $psi$ such that $e^{psi (t)}=phi (t)$ and $psi (0)=0$. Let $g(t)=psi (sqrt t)$ for $t >0$. Show that $g(t+s)=g(t)+g(s)$. I will leave the rest of the argument to you.
answered Feb 2 at 23:26
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
$endgroup$
add a comment |
$begingroup$
Hints: Assume that $X_1$ is not $0$ with positive probability. let $phi$ be the characteristic function of $X_1$. Then $phi (t) phi (s)=phi (sqrt {t^{2}+s^{2}})$. If $phi (t)=0$ for some $t$ then we get $phi (sqrt {t^{2}+s^{2}})=0$ for all $s$ and hence $phi (u)=0$ for all $u >|t|$. S how that this leads to a contradiction to the given property. Otherwise, by a well known fact there is a continuous function $psi$ such that $e^{psi (t)}=phi (t)$ and $psi (0)=0$. Let $g(t)=psi (sqrt t)$ for $t >0$. Show that $g(t+s)=g(t)+g(s)$. I will leave the rest of the argument to you.
answered Feb 2 at 23:26
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
$endgroup$
Hints: Assume that $X_1$ is not $0$ with positive probability. let $phi$ be the characteristic function of $X_1$. Then $phi (t) phi (s)=phi (sqrt {t^{2}+s^{2}})$. If $phi (t)=0$ for some $t$ then we get $phi (sqrt {t^{2}+s^{2}})=0$ for all $s$ and hence $phi (u)=0$ for all $u >|t|$. S how that this leads to a contradiction to the given property. Otherwise, by a well known fact there is a continuous function $psi$ such that $e^{psi (t)}=phi (t)$ and $psi (0)=0$. Let $g(t)=psi (sqrt t)$ for $t >0$. Show that $g(t+s)=g(t)+g(s)$. I will leave the rest of the argument to you.
answered Feb 2 at 23:26
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
answered Feb 2 at 23:26
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
answered Feb 2 at 23:26
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
answered Feb 2 at 23:26
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
74.8k53270
add a comment |
add a comment |
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1
$begingroup$
Are you assuming something like $X_1$ and $X_2$ are iid, and have the same distribution as $X$?
$endgroup$
– kimchi lover
Feb 2 at 23:06
$begingroup$
@kimchilover Yes. I will edit my post. Thanks!
$endgroup$
– Hendrra
Feb 2 at 23:15
$begingroup$
Check out en.wikipedia.org/wiki/Maxwell%27s_theorem
$endgroup$
– kimchi lover
Feb 4 at 14:30