Why is it necessary X and Y beeing bounded when having same moments to be equal
Let $X,Y$ random variables in $[0,1]$ with $E(X^n)=E(Y^n) ,forall nin mathbb N$. I want to show $Xoverset{d}{=} Y$.
$$E(e^{itX})=intsum frac{(itx)^k}{k!}dF_Xoverset{dom. conv.}{=}sumintfrac{(itx)^k}{k!}dF_x=sumfrac{(it)^k}{k!}int x^kdF_xoverset{EX^n=EY^n}{=}sumfrac{(it)^k}{k!}int x^kdF_Y=dots=E(e^{itY})$$
I am aware of the fact $EX^n=EY^n$ does not imply $Xoverset{d}{=}Y$ in general. I did not use that X and Y are bounded, so what am I missing?
probability-theory proof-verification probability-distributions characteristic-functions
asked Nov 27 '18 at 10:20
Joey Doey
1104
add a comment |
Let $X,Y$ random variables in $[0,1]$ with $E(X^n)=E(Y^n) ,forall nin mathbb N$. I want to show $Xoverset{d}{=} Y$.
$$E(e^{itX})=intsum frac{(itx)^k}{k!}dF_Xoverset{dom. conv.}{=}sumintfrac{(itx)^k}{k!}dF_x=sumfrac{(it)^k}{k!}int x^kdF_xoverset{EX^n=EY^n}{=}sumfrac{(it)^k}{k!}int x^kdF_Y=dots=E(e^{itY})$$
I am aware of the fact $EX^n=EY^n$ does not imply $Xoverset{d}{=}Y$ in general. I did not use that X and Y are bounded, so what am I missing?
probability-theory proof-verification probability-distributions characteristic-functions
asked Nov 27 '18 at 10:20
Joey Doey
1104
add a comment |
Let $X,Y$ random variables in $[0,1]$ with $E(X^n)=E(Y^n) ,forall nin mathbb N$. I want to show $Xoverset{d}{=} Y$.
$$E(e^{itX})=intsum frac{(itx)^k}{k!}dF_Xoverset{dom. conv.}{=}sumintfrac{(itx)^k}{k!}dF_x=sumfrac{(it)^k}{k!}int x^kdF_xoverset{EX^n=EY^n}{=}sumfrac{(it)^k}{k!}int x^kdF_Y=dots=E(e^{itY})$$
I am aware of the fact $EX^n=EY^n$ does not imply $Xoverset{d}{=}Y$ in general. I did not use that X and Y are bounded, so what am I missing?
probability-theory proof-verification probability-distributions characteristic-functions
asked Nov 27 '18 at 10:20
Joey Doey
1104
Let $X,Y$ random variables in $[0,1]$ with $E(X^n)=E(Y^n) ,forall nin mathbb N$. I want to show $Xoverset{d}{=} Y$.
$$E(e^{itX})=intsum frac{(itx)^k}{k!}dF_Xoverset{dom. conv.}{=}sumintfrac{(itx)^k}{k!}dF_x=sumfrac{(it)^k}{k!}int x^kdF_xoverset{EX^n=EY^n}{=}sumfrac{(it)^k}{k!}int x^kdF_Y=dots=E(e^{itY})$$
I am aware of the fact $EX^n=EY^n$ does not imply $Xoverset{d}{=}Y$ in general. I did not use that X and Y are bounded, so what am I missing?
probability-theory proof-verification probability-distributions characteristic-functions
probability-theory proof-verification probability-distributions characteristic-functions
asked Nov 27 '18 at 10:20
Joey Doey
1104
asked Nov 27 '18 at 10:20
Joey Doey
1104
edited Nov 27 '18 at 13:58
asked Nov 27 '18 at 10:20
Joey Doey
1104
asked Nov 27 '18 at 10:20
Joey Doey
1104
asked Nov 27 '18 at 10:20
Joey Doey
1104
1104
add a comment |
add a comment |
1 Answer
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You have to justify interchange of integral and sum. In this case $int sum |frac {(itx)^{k}} {k!}|d_F(x) leq e^{|t|}$ so we can use Fubini's Theorem.
answered Nov 27 '18 at 10:23
Kavi Rama Murthy
50.5k31854
you mean fubini? but dom. convergence would work aswell or since $int e^{itx} le 1$ we can change its order by dom. convergence
– Joey Doey
Nov 27 '18 at 10:29
You can use DCT or Funbini.
– Kavi Rama Murthy
Nov 27 '18 at 10:32
but I still do not understand why I need that X and Y are bounded? What I wrote in the comment above is always true.
– Joey Doey
Nov 27 '18 at 10:34
1
Application of DCT is not that simple. It will not work if $x$ is not restricted to a bounded set.
– Kavi Rama Murthy
Nov 27 '18 at 10:35
add a comment |
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1 Answer
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You have to justify interchange of integral and sum. In this case $int sum |frac {(itx)^{k}} {k!}|d_F(x) leq e^{|t|}$ so we can use Fubini's Theorem.
answered Nov 27 '18 at 10:23
Kavi Rama Murthy
50.5k31854
you mean fubini? but dom. convergence would work aswell or since $int e^{itx} le 1$ we can change its order by dom. convergence
– Joey Doey
Nov 27 '18 at 10:29
You can use DCT or Funbini.
– Kavi Rama Murthy
Nov 27 '18 at 10:32
but I still do not understand why I need that X and Y are bounded? What I wrote in the comment above is always true.
– Joey Doey
Nov 27 '18 at 10:34
1
Application of DCT is not that simple. It will not work if $x$ is not restricted to a bounded set.
– Kavi Rama Murthy
Nov 27 '18 at 10:35
add a comment |
You have to justify interchange of integral and sum. In this case $int sum |frac {(itx)^{k}} {k!}|d_F(x) leq e^{|t|}$ so we can use Fubini's Theorem.
answered Nov 27 '18 at 10:23
Kavi Rama Murthy
50.5k31854
you mean fubini? but dom. convergence would work aswell or since $int e^{itx} le 1$ we can change its order by dom. convergence
– Joey Doey
Nov 27 '18 at 10:29
You can use DCT or Funbini.
– Kavi Rama Murthy
Nov 27 '18 at 10:32
but I still do not understand why I need that X and Y are bounded? What I wrote in the comment above is always true.
– Joey Doey
Nov 27 '18 at 10:34
1
Application of DCT is not that simple. It will not work if $x$ is not restricted to a bounded set.
– Kavi Rama Murthy
Nov 27 '18 at 10:35
add a comment |
You have to justify interchange of integral and sum. In this case $int sum |frac {(itx)^{k}} {k!}|d_F(x) leq e^{|t|}$ so we can use Fubini's Theorem.
answered Nov 27 '18 at 10:23
Kavi Rama Murthy
50.5k31854
You have to justify interchange of integral and sum. In this case $int sum |frac {(itx)^{k}} {k!}|d_F(x) leq e^{|t|}$ so we can use Fubini's Theorem.
answered Nov 27 '18 at 10:23
Kavi Rama Murthy
50.5k31854
answered Nov 27 '18 at 10:23
Kavi Rama Murthy
50.5k31854
answered Nov 27 '18 at 10:23
Kavi Rama Murthy
50.5k31854
answered Nov 27 '18 at 10:23
Kavi Rama Murthy
50.5k31854
50.5k31854
you mean fubini? but dom. convergence would work aswell or since $int e^{itx} le 1$ we can change its order by dom. convergence
– Joey Doey
Nov 27 '18 at 10:29
You can use DCT or Funbini.
– Kavi Rama Murthy
Nov 27 '18 at 10:32
but I still do not understand why I need that X and Y are bounded? What I wrote in the comment above is always true.
– Joey Doey
Nov 27 '18 at 10:34
1
Application of DCT is not that simple. It will not work if $x$ is not restricted to a bounded set.
– Kavi Rama Murthy
Nov 27 '18 at 10:35
add a comment |
you mean fubini? but dom. convergence would work aswell or since $int e^{itx} le 1$ we can change its order by dom. convergence
– Joey Doey
Nov 27 '18 at 10:29
You can use DCT or Funbini.
– Kavi Rama Murthy
Nov 27 '18 at 10:32
but I still do not understand why I need that X and Y are bounded? What I wrote in the comment above is always true.
– Joey Doey
Nov 27 '18 at 10:34
1
Application of DCT is not that simple. It will not work if $x$ is not restricted to a bounded set.
– Kavi Rama Murthy
Nov 27 '18 at 10:35
you mean fubini? but dom. convergence would work aswell or since $int e^{itx} le 1$ we can change its order by dom. convergence
– Joey Doey
Nov 27 '18 at 10:29
you mean fubini? but dom. convergence would work aswell or since $int e^{itx} le 1$ we can change its order by dom. convergence
– Joey Doey
Nov 27 '18 at 10:29
You can use DCT or Funbini.
– Kavi Rama Murthy
Nov 27 '18 at 10:32
You can use DCT or Funbini.
– Kavi Rama Murthy
Nov 27 '18 at 10:32
but I still do not understand why I need that X and Y are bounded? What I wrote in the comment above is always true.
– Joey Doey
Nov 27 '18 at 10:34
but I still do not understand why I need that X and Y are bounded? What I wrote in the comment above is always true.
– Joey Doey
Nov 27 '18 at 10:34
1
1
Application of DCT is not that simple. It will not work if $x$ is not restricted to a bounded set.
– Kavi Rama Murthy
Nov 27 '18 at 10:35
Application of DCT is not that simple. It will not work if $x$ is not restricted to a bounded set.
– Kavi Rama Murthy
Nov 27 '18 at 10:35
add a comment |
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