Mixing
![QGIS: list of coordinates to polygon [duplicate]](https://lh6.googleusercontent.com/-lBG_D1yi7Z8/AAAAAAAAAAI/AAAAAAAAAHA/kwX564gJ5fM/s72-c/photo.jpg?sz=32)
Logarithm of martingale to the power of a Poisson random variable is a martingale
$begingroup$
Let $(Z_n)_n$ be independently Poisson(1) distributed random variables.
Let $X_1=Z_1+1$ and
$$X_n=(log X_{n-1}+1)^{Z_n}$$
for $n>1$ and define $F_n=sigma(Z_1,...,Z_n)$.
How do I show that
a) $X_n$ is a $F_n$-martingale
b) its limit is $1$ almost surely
c) it is not bounded in $L^p$ for any $p>1$?
For a) we need to show that $mathbb{E}|X_n|<infty$, but how do I write $mathbb{E}|(log X_{n-1}+1)^{Z_n}|$?
Further we want $mathbb{E}[X_n|f_{n-1}]=X_{n-1}$, but again I don't know what to do with the power of $Z_n$?
EDIT:
Based on Davide Giraudo's reply, I write for the martingale condition
$$begin{align*}
mathbb{E}[(log X_{n-1}+1)^{Z_n}|F_n]&=mathbb{E}[sumlimits_{kinmathbb{N}}frac{1}{ecdot k!}(log X_{n-1}+1)^k|F_n]\
&=e^{-1}mathbb{E}[exp(log X_{n-1}+1)|F_n]\
&=mathbb{E}[X_{n-1}|F_{n-1}]\
&=X_{n-1}
end{align*}$$
convergence martingales
asked Jan 28 at 11:18
Joachim DoyleJoachim Doyle
818
$endgroup$
add a comment |
$begingroup$
Let $(Z_n)_n$ be independently Poisson(1) distributed random variables.
Let $X_1=Z_1+1$ and
$$X_n=(log X_{n-1}+1)^{Z_n}$$
for $n>1$ and define $F_n=sigma(Z_1,...,Z_n)$.
How do I show that
a) $X_n$ is a $F_n$-martingale
b) its limit is $1$ almost surely
c) it is not bounded in $L^p$ for any $p>1$?
For a) we need to show that $mathbb{E}|X_n|<infty$, but how do I write $mathbb{E}|(log X_{n-1}+1)^{Z_n}|$?
Further we want $mathbb{E}[X_n|f_{n-1}]=X_{n-1}$, but again I don't know what to do with the power of $Z_n$?
EDIT:
Based on Davide Giraudo's reply, I write for the martingale condition
$$begin{align*}
mathbb{E}[(log X_{n-1}+1)^{Z_n}|F_n]&=mathbb{E}[sumlimits_{kinmathbb{N}}frac{1}{ecdot k!}(log X_{n-1}+1)^k|F_n]\
&=e^{-1}mathbb{E}[exp(log X_{n-1}+1)|F_n]\
&=mathbb{E}[X_{n-1}|F_{n-1}]\
&=X_{n-1}
end{align*}$$
convergence martingales
asked Jan 28 at 11:18
Joachim DoyleJoachim Doyle
818
$endgroup$
add a comment |
$begingroup$
Let $(Z_n)_n$ be independently Poisson(1) distributed random variables.
Let $X_1=Z_1+1$ and
$$X_n=(log X_{n-1}+1)^{Z_n}$$
for $n>1$ and define $F_n=sigma(Z_1,...,Z_n)$.
How do I show that
a) $X_n$ is a $F_n$-martingale
b) its limit is $1$ almost surely
c) it is not bounded in $L^p$ for any $p>1$?
For a) we need to show that $mathbb{E}|X_n|<infty$, but how do I write $mathbb{E}|(log X_{n-1}+1)^{Z_n}|$?
Further we want $mathbb{E}[X_n|f_{n-1}]=X_{n-1}$, but again I don't know what to do with the power of $Z_n$?
EDIT:
Based on Davide Giraudo's reply, I write for the martingale condition
$$begin{align*}
mathbb{E}[(log X_{n-1}+1)^{Z_n}|F_n]&=mathbb{E}[sumlimits_{kinmathbb{N}}frac{1}{ecdot k!}(log X_{n-1}+1)^k|F_n]\
&=e^{-1}mathbb{E}[exp(log X_{n-1}+1)|F_n]\
&=mathbb{E}[X_{n-1}|F_{n-1}]\
&=X_{n-1}
end{align*}$$
convergence martingales
asked Jan 28 at 11:18
Joachim DoyleJoachim Doyle
818
$endgroup$
Let $(Z_n)_n$ be independently Poisson(1) distributed random variables.
Let $X_1=Z_1+1$ and
$$X_n=(log X_{n-1}+1)^{Z_n}$$
for $n>1$ and define $F_n=sigma(Z_1,...,Z_n)$.
How do I show that
a) $X_n$ is a $F_n$-martingale
b) its limit is $1$ almost surely
c) it is not bounded in $L^p$ for any $p>1$?
For a) we need to show that $mathbb{E}|X_n|<infty$, but how do I write $mathbb{E}|(log X_{n-1}+1)^{Z_n}|$?
Further we want $mathbb{E}[X_n|f_{n-1}]=X_{n-1}$, but again I don't know what to do with the power of $Z_n$?
EDIT:
Based on Davide Giraudo's reply, I write for the martingale condition
$$begin{align*}
mathbb{E}[(log X_{n-1}+1)^{Z_n}|F_n]&=mathbb{E}[sumlimits_{kinmathbb{N}}frac{1}{ecdot k!}(log X_{n-1}+1)^k|F_n]\
&=e^{-1}mathbb{E}[exp(log X_{n-1}+1)|F_n]\
&=mathbb{E}[X_{n-1}|F_{n-1}]\
&=X_{n-1}
end{align*}$$
convergence martingales
convergence martingales
asked Jan 28 at 11:18
Joachim DoyleJoachim Doyle
818
asked Jan 28 at 11:18
Joachim DoyleJoachim Doyle
818
edited Jan 28 at 12:12
Joachim Doyle
asked Jan 28 at 11:18
Joachim DoyleJoachim Doyle
818
asked Jan 28 at 11:18
Joachim DoyleJoachim Doyle
818
asked Jan 28 at 11:18
Joachim DoyleJoachim Doyle
818
818
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
- Part (a):
As @Davide pointed out, we can see $X_{n-1}in F_{n-1} implies X_nin F_n$ easily. By induction, $(X_n,F_n)$ is an adapted process. Also note that $X_{n-1}ge 1 implies X_nge 1$, hence $X_nge 1$ a.s. We find that $Bbb EX_1 =2<infty$ as a base case and
$$begin{eqnarray}
Bbb Eleft[X_{n+1}1_{{Z_{n+1}le N}}|F_nright] &=&Bbb Eleft[left(log X_n+1right)^{Z_{n+1}}1_{{Z_{n+1}le N}}|X_nright]\&= &sum_{k=0}^N frac{e^{-1}left(log X_n +1right)^k}{k!}\
&stackrel{Ntoinfty}longrightarrow&e^{-1}e^{log X_n +1}=X_n.
end{eqnarray}$$ By induction and monotone convergence theorem, it follows $Bbb E[X_{n+1}|F_n]$ is integrable and $Bbb E[X_{n+1}|F_n]= X_n$
- Part (b):
First note that since $(X_n, F_n)$ is a $L^1$-bounded martingale, there exists
$$
X =lim_{ntoinfty} X_nin [1,infty) quadtext{a.s.}
$$ by martingale convergence theorem. Let $A ={X>1}$.
By taking logarithms, we find that
$$
log X_{n} =Z_n log left(log X_{n-1}+1right).
$$ From this we obtain
$$
Z_n stackrel{ntoinfty}longrightarrow frac{log X}{log left(log X+1right)} quadtext{a.s. on } A.
$$ On the other hand, by strong law of large numbers
$$
frac{Z_1+Z_2+cdots +Z_n}{n}stackrel{ntoinfty}longrightarrow 1 quadtext{a.s.}
$$ Since the Cesaro limit is equal to the limit if both exist, we have
$$
X=log X+1 quadtext{a.s. on } A.
$$ Since $log t < t-1$ for all $tne 1$, this is impossible on $A={X> 1}$. Hence $Asubset A^c$ a.s. and $Bbb P(A)=0$ as desired.
- Part (c):
Note that if $(X_n,F_n)$ is $L^p$-bounded for $p>1$, martingale convergence theorem implies that
$$
X_n to X=1quadtext{both a.s. and in} L^p.
$$ But this implies that
$$
X_1 =Bbb E[X_n |F_1] stackrel{ntoinfty}longrightarrow Bbb E [X|F_1] =1 text{ in } L^p
$$ leading to the contradiction.
answered Jan 28 at 13:18
SongSong
18.5k21651
$endgroup$
$begingroup$
Why is $X_n$ bounded in $L^1$?
$endgroup$
– Joachim Doyle
Jan 28 at 13:37
$begingroup$
And could you explain why we have a contradiction in (c)?
$endgroup$
– Joachim Doyle
Jan 28 at 13:41
$begingroup$
1. Since $X_n$ is non-negative and $X_n$ is a martingale. 2. Since $X_1=1+Z_1$ is not constant $1$. Is it convincing?
$endgroup$
– Song
Jan 28 at 13:42
$begingroup$
Ah, I see 1. now. Because of non-negative and martingale we have $mathbb{E}|X_n|=mathbb{E}(X_n)=mathbb{E}(X_1)=2$. Thank you
$endgroup$
– Joachim Doyle
Jan 28 at 13:55
add a comment |
$begingroup$
Observe that we can prove by induction that $X_n$ is $mathcal F_n$-measurable and that $X_ngeqslant 1$ almost surely. Since $Z_n$ is independent of $X_{n-1}$, it follows that
$$mathbb Eleftlvert X_nrightrvert=mathbb Eleft[sum_{k=0}^{+infty}e^{-1}frac{1}{k!}left(1+log X_{n-1}right)^kright]=e^{-1}mathbb Eleft[ expleft(1+log X_{n-1}right) right]=mathbb Eleftlvert X_{n-1}rightrvert.$$
answered Jan 28 at 11:30
Davide GiraudoDavide Giraudo
128k17154268
$endgroup$
$begingroup$
Could you take a look at my edit, would that be correct for the conditional expectation? Also how would I start for b) and c)?
$endgroup$
– Joachim Doyle
Jan 28 at 12:13
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3090731%2flogarithm-of-martingale-to-the-power-of-a-poisson-random-variable-is-a-martingal%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
- Part (a):
As @Davide pointed out, we can see $X_{n-1}in F_{n-1} implies X_nin F_n$ easily. By induction, $(X_n,F_n)$ is an adapted process. Also note that $X_{n-1}ge 1 implies X_nge 1$, hence $X_nge 1$ a.s. We find that $Bbb EX_1 =2<infty$ as a base case and
$$begin{eqnarray}
Bbb Eleft[X_{n+1}1_{{Z_{n+1}le N}}|F_nright] &=&Bbb Eleft[left(log X_n+1right)^{Z_{n+1}}1_{{Z_{n+1}le N}}|X_nright]\&= &sum_{k=0}^N frac{e^{-1}left(log X_n +1right)^k}{k!}\
&stackrel{Ntoinfty}longrightarrow&e^{-1}e^{log X_n +1}=X_n.
end{eqnarray}$$ By induction and monotone convergence theorem, it follows $Bbb E[X_{n+1}|F_n]$ is integrable and $Bbb E[X_{n+1}|F_n]= X_n$
- Part (b):
First note that since $(X_n, F_n)$ is a $L^1$-bounded martingale, there exists
$$
X =lim_{ntoinfty} X_nin [1,infty) quadtext{a.s.}
$$ by martingale convergence theorem. Let $A ={X>1}$.
By taking logarithms, we find that
$$
log X_{n} =Z_n log left(log X_{n-1}+1right).
$$ From this we obtain
$$
Z_n stackrel{ntoinfty}longrightarrow frac{log X}{log left(log X+1right)} quadtext{a.s. on } A.
$$ On the other hand, by strong law of large numbers
$$
frac{Z_1+Z_2+cdots +Z_n}{n}stackrel{ntoinfty}longrightarrow 1 quadtext{a.s.}
$$ Since the Cesaro limit is equal to the limit if both exist, we have
$$
X=log X+1 quadtext{a.s. on } A.
$$ Since $log t < t-1$ for all $tne 1$, this is impossible on $A={X> 1}$. Hence $Asubset A^c$ a.s. and $Bbb P(A)=0$ as desired.
- Part (c):
Note that if $(X_n,F_n)$ is $L^p$-bounded for $p>1$, martingale convergence theorem implies that
$$
X_n to X=1quadtext{both a.s. and in} L^p.
$$ But this implies that
$$
X_1 =Bbb E[X_n |F_1] stackrel{ntoinfty}longrightarrow Bbb E [X|F_1] =1 text{ in } L^p
$$ leading to the contradiction.
answered Jan 28 at 13:18
SongSong
18.5k21651
$endgroup$
$begingroup$
Why is $X_n$ bounded in $L^1$?
$endgroup$
– Joachim Doyle
Jan 28 at 13:37
$begingroup$
And could you explain why we have a contradiction in (c)?
$endgroup$
– Joachim Doyle
Jan 28 at 13:41
$begingroup$
1. Since $X_n$ is non-negative and $X_n$ is a martingale. 2. Since $X_1=1+Z_1$ is not constant $1$. Is it convincing?
$endgroup$
– Song
Jan 28 at 13:42
$begingroup$
Ah, I see 1. now. Because of non-negative and martingale we have $mathbb{E}|X_n|=mathbb{E}(X_n)=mathbb{E}(X_1)=2$. Thank you
$endgroup$
– Joachim Doyle
Jan 28 at 13:55
add a comment |
$begingroup$
- Part (a):
As @Davide pointed out, we can see $X_{n-1}in F_{n-1} implies X_nin F_n$ easily. By induction, $(X_n,F_n)$ is an adapted process. Also note that $X_{n-1}ge 1 implies X_nge 1$, hence $X_nge 1$ a.s. We find that $Bbb EX_1 =2<infty$ as a base case and
$$begin{eqnarray}
Bbb Eleft[X_{n+1}1_{{Z_{n+1}le N}}|F_nright] &=&Bbb Eleft[left(log X_n+1right)^{Z_{n+1}}1_{{Z_{n+1}le N}}|X_nright]\&= &sum_{k=0}^N frac{e^{-1}left(log X_n +1right)^k}{k!}\
&stackrel{Ntoinfty}longrightarrow&e^{-1}e^{log X_n +1}=X_n.
end{eqnarray}$$ By induction and monotone convergence theorem, it follows $Bbb E[X_{n+1}|F_n]$ is integrable and $Bbb E[X_{n+1}|F_n]= X_n$
- Part (b):
First note that since $(X_n, F_n)$ is a $L^1$-bounded martingale, there exists
$$
X =lim_{ntoinfty} X_nin [1,infty) quadtext{a.s.}
$$ by martingale convergence theorem. Let $A ={X>1}$.
By taking logarithms, we find that
$$
log X_{n} =Z_n log left(log X_{n-1}+1right).
$$ From this we obtain
$$
Z_n stackrel{ntoinfty}longrightarrow frac{log X}{log left(log X+1right)} quadtext{a.s. on } A.
$$ On the other hand, by strong law of large numbers
$$
frac{Z_1+Z_2+cdots +Z_n}{n}stackrel{ntoinfty}longrightarrow 1 quadtext{a.s.}
$$ Since the Cesaro limit is equal to the limit if both exist, we have
$$
X=log X+1 quadtext{a.s. on } A.
$$ Since $log t < t-1$ for all $tne 1$, this is impossible on $A={X> 1}$. Hence $Asubset A^c$ a.s. and $Bbb P(A)=0$ as desired.
- Part (c):
Note that if $(X_n,F_n)$ is $L^p$-bounded for $p>1$, martingale convergence theorem implies that
$$
X_n to X=1quadtext{both a.s. and in} L^p.
$$ But this implies that
$$
X_1 =Bbb E[X_n |F_1] stackrel{ntoinfty}longrightarrow Bbb E [X|F_1] =1 text{ in } L^p
$$ leading to the contradiction.
answered Jan 28 at 13:18
SongSong
18.5k21651
$endgroup$
$begingroup$
Why is $X_n$ bounded in $L^1$?
$endgroup$
– Joachim Doyle
Jan 28 at 13:37
$begingroup$
And could you explain why we have a contradiction in (c)?
$endgroup$
– Joachim Doyle
Jan 28 at 13:41
$begingroup$
1. Since $X_n$ is non-negative and $X_n$ is a martingale. 2. Since $X_1=1+Z_1$ is not constant $1$. Is it convincing?
$endgroup$
– Song
Jan 28 at 13:42
$begingroup$
Ah, I see 1. now. Because of non-negative and martingale we have $mathbb{E}|X_n|=mathbb{E}(X_n)=mathbb{E}(X_1)=2$. Thank you
$endgroup$
– Joachim Doyle
Jan 28 at 13:55
add a comment |
$begingroup$
- Part (a):
As @Davide pointed out, we can see $X_{n-1}in F_{n-1} implies X_nin F_n$ easily. By induction, $(X_n,F_n)$ is an adapted process. Also note that $X_{n-1}ge 1 implies X_nge 1$, hence $X_nge 1$ a.s. We find that $Bbb EX_1 =2<infty$ as a base case and
$$begin{eqnarray}
Bbb Eleft[X_{n+1}1_{{Z_{n+1}le N}}|F_nright] &=&Bbb Eleft[left(log X_n+1right)^{Z_{n+1}}1_{{Z_{n+1}le N}}|X_nright]\&= &sum_{k=0}^N frac{e^{-1}left(log X_n +1right)^k}{k!}\
&stackrel{Ntoinfty}longrightarrow&e^{-1}e^{log X_n +1}=X_n.
end{eqnarray}$$ By induction and monotone convergence theorem, it follows $Bbb E[X_{n+1}|F_n]$ is integrable and $Bbb E[X_{n+1}|F_n]= X_n$
- Part (b):
First note that since $(X_n, F_n)$ is a $L^1$-bounded martingale, there exists
$$
X =lim_{ntoinfty} X_nin [1,infty) quadtext{a.s.}
$$ by martingale convergence theorem. Let $A ={X>1}$.
By taking logarithms, we find that
$$
log X_{n} =Z_n log left(log X_{n-1}+1right).
$$ From this we obtain
$$
Z_n stackrel{ntoinfty}longrightarrow frac{log X}{log left(log X+1right)} quadtext{a.s. on } A.
$$ On the other hand, by strong law of large numbers
$$
frac{Z_1+Z_2+cdots +Z_n}{n}stackrel{ntoinfty}longrightarrow 1 quadtext{a.s.}
$$ Since the Cesaro limit is equal to the limit if both exist, we have
$$
X=log X+1 quadtext{a.s. on } A.
$$ Since $log t < t-1$ for all $tne 1$, this is impossible on $A={X> 1}$. Hence $Asubset A^c$ a.s. and $Bbb P(A)=0$ as desired.
- Part (c):
Note that if $(X_n,F_n)$ is $L^p$-bounded for $p>1$, martingale convergence theorem implies that
$$
X_n to X=1quadtext{both a.s. and in} L^p.
$$ But this implies that
$$
X_1 =Bbb E[X_n |F_1] stackrel{ntoinfty}longrightarrow Bbb E [X|F_1] =1 text{ in } L^p
$$ leading to the contradiction.
answered Jan 28 at 13:18
SongSong
18.5k21651
$endgroup$
- Part (a):
As @Davide pointed out, we can see $X_{n-1}in F_{n-1} implies X_nin F_n$ easily. By induction, $(X_n,F_n)$ is an adapted process. Also note that $X_{n-1}ge 1 implies X_nge 1$, hence $X_nge 1$ a.s. We find that $Bbb EX_1 =2<infty$ as a base case and
$$begin{eqnarray}
Bbb Eleft[X_{n+1}1_{{Z_{n+1}le N}}|F_nright] &=&Bbb Eleft[left(log X_n+1right)^{Z_{n+1}}1_{{Z_{n+1}le N}}|X_nright]\&= &sum_{k=0}^N frac{e^{-1}left(log X_n +1right)^k}{k!}\
&stackrel{Ntoinfty}longrightarrow&e^{-1}e^{log X_n +1}=X_n.
end{eqnarray}$$ By induction and monotone convergence theorem, it follows $Bbb E[X_{n+1}|F_n]$ is integrable and $Bbb E[X_{n+1}|F_n]= X_n$
- Part (b):
First note that since $(X_n, F_n)$ is a $L^1$-bounded martingale, there exists
$$
X =lim_{ntoinfty} X_nin [1,infty) quadtext{a.s.}
$$ by martingale convergence theorem. Let $A ={X>1}$.
By taking logarithms, we find that
$$
log X_{n} =Z_n log left(log X_{n-1}+1right).
$$ From this we obtain
$$
Z_n stackrel{ntoinfty}longrightarrow frac{log X}{log left(log X+1right)} quadtext{a.s. on } A.
$$ On the other hand, by strong law of large numbers
$$
frac{Z_1+Z_2+cdots +Z_n}{n}stackrel{ntoinfty}longrightarrow 1 quadtext{a.s.}
$$ Since the Cesaro limit is equal to the limit if both exist, we have
$$
X=log X+1 quadtext{a.s. on } A.
$$ Since $log t < t-1$ for all $tne 1$, this is impossible on $A={X> 1}$. Hence $Asubset A^c$ a.s. and $Bbb P(A)=0$ as desired.
- Part (c):
Note that if $(X_n,F_n)$ is $L^p$-bounded for $p>1$, martingale convergence theorem implies that
$$
X_n to X=1quadtext{both a.s. and in} L^p.
$$ But this implies that
$$
X_1 =Bbb E[X_n |F_1] stackrel{ntoinfty}longrightarrow Bbb E [X|F_1] =1 text{ in } L^p
$$ leading to the contradiction.
answered Jan 28 at 13:18
SongSong
18.5k21651
answered Jan 28 at 13:18
SongSong
18.5k21651
answered Jan 28 at 13:18
SongSong
18.5k21651
answered Jan 28 at 13:18
SongSong
18.5k21651
18.5k21651
$begingroup$
Why is $X_n$ bounded in $L^1$?
$endgroup$
– Joachim Doyle
Jan 28 at 13:37
$begingroup$
And could you explain why we have a contradiction in (c)?
$endgroup$
– Joachim Doyle
Jan 28 at 13:41
$begingroup$
1. Since $X_n$ is non-negative and $X_n$ is a martingale. 2. Since $X_1=1+Z_1$ is not constant $1$. Is it convincing?
$endgroup$
– Song
Jan 28 at 13:42
$begingroup$
Ah, I see 1. now. Because of non-negative and martingale we have $mathbb{E}|X_n|=mathbb{E}(X_n)=mathbb{E}(X_1)=2$. Thank you
$endgroup$
– Joachim Doyle
Jan 28 at 13:55
add a comment |
$begingroup$
Why is $X_n$ bounded in $L^1$?
$endgroup$
– Joachim Doyle
Jan 28 at 13:37
$begingroup$
And could you explain why we have a contradiction in (c)?
$endgroup$
– Joachim Doyle
Jan 28 at 13:41
$begingroup$
1. Since $X_n$ is non-negative and $X_n$ is a martingale. 2. Since $X_1=1+Z_1$ is not constant $1$. Is it convincing?
$endgroup$
– Song
Jan 28 at 13:42
$begingroup$
Ah, I see 1. now. Because of non-negative and martingale we have $mathbb{E}|X_n|=mathbb{E}(X_n)=mathbb{E}(X_1)=2$. Thank you
$endgroup$
– Joachim Doyle
Jan 28 at 13:55
$begingroup$
Why is $X_n$ bounded in $L^1$?
$endgroup$
– Joachim Doyle
Jan 28 at 13:37
$begingroup$
Why is $X_n$ bounded in $L^1$?
$endgroup$
– Joachim Doyle
Jan 28 at 13:37
$begingroup$
And could you explain why we have a contradiction in (c)?
$endgroup$
– Joachim Doyle
Jan 28 at 13:41
$begingroup$
And could you explain why we have a contradiction in (c)?
$endgroup$
– Joachim Doyle
Jan 28 at 13:41
$begingroup$
1. Since $X_n$ is non-negative and $X_n$ is a martingale. 2. Since $X_1=1+Z_1$ is not constant $1$. Is it convincing?
$endgroup$
– Song
Jan 28 at 13:42
$begingroup$
1. Since $X_n$ is non-negative and $X_n$ is a martingale. 2. Since $X_1=1+Z_1$ is not constant $1$. Is it convincing?
$endgroup$
– Song
Jan 28 at 13:42
$begingroup$
Ah, I see 1. now. Because of non-negative and martingale we have $mathbb{E}|X_n|=mathbb{E}(X_n)=mathbb{E}(X_1)=2$. Thank you
$endgroup$
– Joachim Doyle
Jan 28 at 13:55
$begingroup$
Ah, I see 1. now. Because of non-negative and martingale we have $mathbb{E}|X_n|=mathbb{E}(X_n)=mathbb{E}(X_1)=2$. Thank you
$endgroup$
– Joachim Doyle
Jan 28 at 13:55
add a comment |
$begingroup$
Observe that we can prove by induction that $X_n$ is $mathcal F_n$-measurable and that $X_ngeqslant 1$ almost surely. Since $Z_n$ is independent of $X_{n-1}$, it follows that
$$mathbb Eleftlvert X_nrightrvert=mathbb Eleft[sum_{k=0}^{+infty}e^{-1}frac{1}{k!}left(1+log X_{n-1}right)^kright]=e^{-1}mathbb Eleft[ expleft(1+log X_{n-1}right) right]=mathbb Eleftlvert X_{n-1}rightrvert.$$
answered Jan 28 at 11:30
Davide GiraudoDavide Giraudo
128k17154268
$endgroup$
$begingroup$
Could you take a look at my edit, would that be correct for the conditional expectation? Also how would I start for b) and c)?
$endgroup$
– Joachim Doyle
Jan 28 at 12:13
add a comment |
$begingroup$
Observe that we can prove by induction that $X_n$ is $mathcal F_n$-measurable and that $X_ngeqslant 1$ almost surely. Since $Z_n$ is independent of $X_{n-1}$, it follows that
$$mathbb Eleftlvert X_nrightrvert=mathbb Eleft[sum_{k=0}^{+infty}e^{-1}frac{1}{k!}left(1+log X_{n-1}right)^kright]=e^{-1}mathbb Eleft[ expleft(1+log X_{n-1}right) right]=mathbb Eleftlvert X_{n-1}rightrvert.$$
answered Jan 28 at 11:30
Davide GiraudoDavide Giraudo
128k17154268
$endgroup$
$begingroup$
Could you take a look at my edit, would that be correct for the conditional expectation? Also how would I start for b) and c)?
$endgroup$
– Joachim Doyle
Jan 28 at 12:13
add a comment |
$begingroup$
Observe that we can prove by induction that $X_n$ is $mathcal F_n$-measurable and that $X_ngeqslant 1$ almost surely. Since $Z_n$ is independent of $X_{n-1}$, it follows that
$$mathbb Eleftlvert X_nrightrvert=mathbb Eleft[sum_{k=0}^{+infty}e^{-1}frac{1}{k!}left(1+log X_{n-1}right)^kright]=e^{-1}mathbb Eleft[ expleft(1+log X_{n-1}right) right]=mathbb Eleftlvert X_{n-1}rightrvert.$$
answered Jan 28 at 11:30
Davide GiraudoDavide Giraudo
128k17154268
$endgroup$
Observe that we can prove by induction that $X_n$ is $mathcal F_n$-measurable and that $X_ngeqslant 1$ almost surely. Since $Z_n$ is independent of $X_{n-1}$, it follows that
$$mathbb Eleftlvert X_nrightrvert=mathbb Eleft[sum_{k=0}^{+infty}e^{-1}frac{1}{k!}left(1+log X_{n-1}right)^kright]=e^{-1}mathbb Eleft[ expleft(1+log X_{n-1}right) right]=mathbb Eleftlvert X_{n-1}rightrvert.$$
answered Jan 28 at 11:30
Davide GiraudoDavide Giraudo
128k17154268
answered Jan 28 at 11:30
Davide GiraudoDavide Giraudo
128k17154268
answered Jan 28 at 11:30
Davide GiraudoDavide Giraudo
128k17154268
answered Jan 28 at 11:30
Davide GiraudoDavide Giraudo
128k17154268
128k17154268
$begingroup$
Could you take a look at my edit, would that be correct for the conditional expectation? Also how would I start for b) and c)?
$endgroup$
– Joachim Doyle
Jan 28 at 12:13
add a comment |
$begingroup$
Could you take a look at my edit, would that be correct for the conditional expectation? Also how would I start for b) and c)?
$endgroup$
– Joachim Doyle
Jan 28 at 12:13
$begingroup$
Could you take a look at my edit, would that be correct for the conditional expectation? Also how would I start for b) and c)?
$endgroup$
– Joachim Doyle
Jan 28 at 12:13
$begingroup$
Could you take a look at my edit, would that be correct for the conditional expectation? Also how would I start for b) and c)?
$endgroup$
– Joachim Doyle
Jan 28 at 12:13
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3090731%2flogarithm-of-martingale-to-the-power-of-a-poisson-random-variable-is-a-martingal%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
