Proving a sequence forms a martingale
1
$begingroup$
Let $Omega = mathbb N = {1,2,3,cdots}$ and $mathscr F_n$ be the $sigma$-field generated by the sets ${1},{2},cdots,{[n+1,infty)}$
Define a probability on $mathbb N$ by setting $mathbb P([n,infty)) = frac 1 n$
Show that
1) $f_n =(n+1)mathbf1_{[n+1,infty)}$ is a martingale
2) $f_n to 0 quad a.s.$
I'm having difficulty dealing with the indicator r.v. How do we get it out of the conditional expectation for $mathbf E[f_{n+1}mid mathcal{F}_n]$?
I tried to graph it out and I think I can see why it forms a martingale but I'm not able to prove it explicitly.
$f_n$">
probability-theory martingales
edited Jan 5 at 20:06
saz
81.5k861127
asked Apr 22 '15 at 21:44
Srinivas EswarSrinivas Eswar
11911
$endgroup$
add a comment |
1
$begingroup$
Let $Omega = mathbb N = {1,2,3,cdots}$ and $mathscr F_n$ be the $sigma$-field generated by the sets ${1},{2},cdots,{[n+1,infty)}$
Define a probability on $mathbb N$ by setting $mathbb P([n,infty)) = frac 1 n$
Show that
1) $f_n =(n+1)mathbf1_{[n+1,infty)}$ is a martingale
2) $f_n to 0 quad a.s.$
I'm having difficulty dealing with the indicator r.v. How do we get it out of the conditional expectation for $mathbf E[f_{n+1}mid mathcal{F}_n]$?
I tried to graph it out and I think I can see why it forms a martingale but I'm not able to prove it explicitly.
$f_n$">
probability-theory martingales
edited Jan 5 at 20:06
saz
81.5k861127
asked Apr 22 '15 at 21:44
Srinivas EswarSrinivas Eswar
11911
$endgroup$
$begingroup$
@saz, Any luck with this question?
$endgroup$
– Srinivas Eswar
Apr 27 '15 at 15:08
$begingroup$
I know how to solve this question, yes. But since you don't show any effort, I don't feel motivated to answer it.
$endgroup$
– saz
Apr 27 '15 at 16:18
$begingroup$
@saz, I have tried two ways to show it but I don't know if they're the right paths. The first attempt was to take $mathbb E[f_{n+1} lvert mathscr F_n]$ and split $[n+1,infty)$ into ${n} cup [n+1,infty)$ but the indicator variable is still within the expectation. My next attempt was to graph it out but I still can't get to the answer. I am truly stuck at this point. A pointer in the right direction would help.
$endgroup$
– Srinivas Eswar
Apr 27 '15 at 17:16
$begingroup$
So, any progress?
$endgroup$
– saz
Apr 28 '15 at 18:10
$begingroup$
@saz I sort of follow what you're saying but we never used $mathscr G$ in our study. My course is "classical probability" and tends to avoid measure theory. I'm still working out step 1. I'll update once I wrap my head around it.
$endgroup$
– Srinivas Eswar
Apr 28 '15 at 22:19
add a comment |
1
1
1
$begingroup$
Let $Omega = mathbb N = {1,2,3,cdots}$ and $mathscr F_n$ be the $sigma$-field generated by the sets ${1},{2},cdots,{[n+1,infty)}$
Define a probability on $mathbb N$ by setting $mathbb P([n,infty)) = frac 1 n$
Show that
1) $f_n =(n+1)mathbf1_{[n+1,infty)}$ is a martingale
2) $f_n to 0 quad a.s.$
I'm having difficulty dealing with the indicator r.v. How do we get it out of the conditional expectation for $mathbf E[f_{n+1}mid mathcal{F}_n]$?
I tried to graph it out and I think I can see why it forms a martingale but I'm not able to prove it explicitly.
$f_n$">
probability-theory martingales
edited Jan 5 at 20:06
saz
81.5k861127
asked Apr 22 '15 at 21:44
Srinivas EswarSrinivas Eswar
11911
$endgroup$
Let $Omega = mathbb N = {1,2,3,cdots}$ and $mathscr F_n$ be the $sigma$-field generated by the sets ${1},{2},cdots,{[n+1,infty)}$
Define a probability on $mathbb N$ by setting $mathbb P([n,infty)) = frac 1 n$
Show that
1) $f_n =(n+1)mathbf1_{[n+1,infty)}$ is a martingale
2) $f_n to 0 quad a.s.$
I'm having difficulty dealing with the indicator r.v. How do we get it out of the conditional expectation for $mathbf E[f_{n+1}mid mathcal{F}_n]$?
I tried to graph it out and I think I can see why it forms a martingale but I'm not able to prove it explicitly.
$f_n$">
probability-theory martingales
probability-theory martingales
edited Jan 5 at 20:06
saz
81.5k861127
asked Apr 22 '15 at 21:44
Srinivas EswarSrinivas Eswar
11911
edited Jan 5 at 20:06
saz
81.5k861127
asked Apr 22 '15 at 21:44
Srinivas EswarSrinivas Eswar
11911
edited Jan 5 at 20:06
saz
81.5k861127
edited Jan 5 at 20:06
saz
81.5k861127
edited Jan 5 at 20:06
saz
81.5k861127
81.5k861127
asked Apr 22 '15 at 21:44
Srinivas EswarSrinivas Eswar
11911
asked Apr 22 '15 at 21:44
Srinivas EswarSrinivas Eswar
11911
asked Apr 22 '15 at 21:44
Srinivas EswarSrinivas Eswar
11911
11911
$begingroup$
@saz, Any luck with this question?
$endgroup$
– Srinivas Eswar
Apr 27 '15 at 15:08
$begingroup$
I know how to solve this question, yes. But since you don't show any effort, I don't feel motivated to answer it.
$endgroup$
– saz
Apr 27 '15 at 16:18
$begingroup$
@saz, I have tried two ways to show it but I don't know if they're the right paths. The first attempt was to take $mathbb E[f_{n+1} lvert mathscr F_n]$ and split $[n+1,infty)$ into ${n} cup [n+1,infty)$ but the indicator variable is still within the expectation. My next attempt was to graph it out but I still can't get to the answer. I am truly stuck at this point. A pointer in the right direction would help.
$endgroup$
– Srinivas Eswar
Apr 27 '15 at 17:16
$begingroup$
So, any progress?
$endgroup$
– saz
Apr 28 '15 at 18:10
$begingroup$
@saz I sort of follow what you're saying but we never used $mathscr G$ in our study. My course is "classical probability" and tends to avoid measure theory. I'm still working out step 1. I'll update once I wrap my head around it.
$endgroup$
– Srinivas Eswar
Apr 28 '15 at 22:19
add a comment |
$begingroup$
@saz, Any luck with this question?
$endgroup$
– Srinivas Eswar
Apr 27 '15 at 15:08
$begingroup$
I know how to solve this question, yes. But since you don't show any effort, I don't feel motivated to answer it.
$endgroup$
– saz
Apr 27 '15 at 16:18
$begingroup$
@saz, I have tried two ways to show it but I don't know if they're the right paths. The first attempt was to take $mathbb E[f_{n+1} lvert mathscr F_n]$ and split $[n+1,infty)$ into ${n} cup [n+1,infty)$ but the indicator variable is still within the expectation. My next attempt was to graph it out but I still can't get to the answer. I am truly stuck at this point. A pointer in the right direction would help.
$endgroup$
– Srinivas Eswar
Apr 27 '15 at 17:16
$begingroup$
So, any progress?
$endgroup$
– saz
Apr 28 '15 at 18:10
$begingroup$
@saz I sort of follow what you're saying but we never used $mathscr G$ in our study. My course is "classical probability" and tends to avoid measure theory. I'm still working out step 1. I'll update once I wrap my head around it.
$endgroup$
– Srinivas Eswar
Apr 28 '15 at 22:19
$begingroup$
@saz, Any luck with this question?
$endgroup$
– Srinivas Eswar
Apr 27 '15 at 15:08
$begingroup$
@saz, Any luck with this question?
$endgroup$
– Srinivas Eswar
Apr 27 '15 at 15:08
$begingroup$
I know how to solve this question, yes. But since you don't show any effort, I don't feel motivated to answer it.
$endgroup$
– saz
Apr 27 '15 at 16:18
$begingroup$
I know how to solve this question, yes. But since you don't show any effort, I don't feel motivated to answer it.
$endgroup$
– saz
Apr 27 '15 at 16:18
$begingroup$
@saz, I have tried two ways to show it but I don't know if they're the right paths. The first attempt was to take $mathbb E[f_{n+1} lvert mathscr F_n]$ and split $[n+1,infty)$ into ${n} cup [n+1,infty)$ but the indicator variable is still within the expectation. My next attempt was to graph it out but I still can't get to the answer. I am truly stuck at this point. A pointer in the right direction would help.
$endgroup$
– Srinivas Eswar
Apr 27 '15 at 17:16
$begingroup$
@saz, I have tried two ways to show it but I don't know if they're the right paths. The first attempt was to take $mathbb E[f_{n+1} lvert mathscr F_n]$ and split $[n+1,infty)$ into ${n} cup [n+1,infty)$ but the indicator variable is still within the expectation. My next attempt was to graph it out but I still can't get to the answer. I am truly stuck at this point. A pointer in the right direction would help.
$endgroup$
– Srinivas Eswar
Apr 27 '15 at 17:16
$begingroup$
So, any progress?
$endgroup$
– saz
Apr 28 '15 at 18:10
$begingroup$
So, any progress?
$endgroup$
– saz
Apr 28 '15 at 18:10
$begingroup$
@saz I sort of follow what you're saying but we never used $mathscr G$ in our study. My course is "classical probability" and tends to avoid measure theory. I'm still working out step 1. I'll update once I wrap my head around it.
$endgroup$
– Srinivas Eswar
Apr 28 '15 at 22:19
$begingroup$
@saz I sort of follow what you're saying but we never used $mathscr G$ in our study. My course is "classical probability" and tends to avoid measure theory. I'm still working out step 1. I'll update once I wrap my head around it.
$endgroup$
– Srinivas Eswar
Apr 28 '15 at 22:19
add a comment |
1 Answer
1
active
oldest
votes
2
$begingroup$
Hints:
- It is well-known that a random variable $Y in L^1$ equals $mathbb{E}(X mid mathcal{F})$ if and only if it is $mathcal{F}$-measurable and $$int_G Y , dmathbb{P} = int_G X , dmathbb{P} qquad text{for all $G in mathcal{G}$}$$ where $mathcal{G}$ denotes a $cap$-stable generator of $mathcal{F}$, i.e. $sigma(mathcal{G}) = mathcal{F}$ and $$G_1, G_2 in mathcal{G} Rightarrow G_1 cap G_2 in mathcal{G}.$$ Check that $mathcal{G}_n := {emptyset,{1},{2},ldots,[n+1,infty)}$ is a $cap$-stable generator of $mathcal{F}_n$.
- In order to show that $f_n$ is a martingale, it suffices to show $$mathbb{E}(f_n mid mathcal{F}_{n-1}) = f_{n-1};$$ by step 1 this is equivalent to $$int_G f_n , dmathbb{P} = int_G f_{n-1} , dmathbb{P} tag{1}$$ for all $G in mathcal{G}_{n-1}$.
- By the very definition of $f_n$, we have $$begin{align*} A&:= int_G f_n , dmathbb{P} = (n+1) mathbb{P}(G cap [n+1,infty)) \ B &:= int_G f_{n-1} , dmathbb{P} = n mathbb{P}(G cap [n,infty)). end{align*}$$ We have to prove $A=B$ for all $G in mathcal{G}_{n-1}$. Consider two cases separately:
- $G = {j}$ for some $j in {1,ldots,n-1}$: Since $G cap [ninfty) = G cap [n+1,infty)=emptyset$, we get $A=B=0$.
- $G = [n,infty)$: Conclude from $G cap [n,infty)= [n,infty)$ and $G cap [n+1,infty) = [n+1,infty)$ that $A=B=1$.
answered Apr 27 '15 at 18:24
sazsaz
81.5k861127
$endgroup$
$begingroup$
I finally got back to this guy. I'm still not clear on a lot of things here. It's basically step 1. If I accept that I'm able to follow steps 2-3. Can you point me to some sources where I can read up on step 1. I can see how the addition of $phi$ has made $mathscr{G_n}$ $cap$-stable but I don't know anything about $L^1$ spaces or why this result (i.e. step 1) holds.
$endgroup$
– Srinivas Eswar
Aug 28 '15 at 18:36
$begingroup$
@SrinivasEswar What $phi$ are you talking about? Do you know the result for $mathcal{G} := mathcal{F}$? In this case, it is a standard characterization of cond. expectations which you can find in any textbook on this topic.
$endgroup$
– saz
Aug 28 '15 at 22:15
$begingroup$
I was referring to the null set. Thanks for the help.
$endgroup$
– Srinivas Eswar
Aug 28 '15 at 23:45
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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2
$begingroup$
Hints:
- It is well-known that a random variable $Y in L^1$ equals $mathbb{E}(X mid mathcal{F})$ if and only if it is $mathcal{F}$-measurable and $$int_G Y , dmathbb{P} = int_G X , dmathbb{P} qquad text{for all $G in mathcal{G}$}$$ where $mathcal{G}$ denotes a $cap$-stable generator of $mathcal{F}$, i.e. $sigma(mathcal{G}) = mathcal{F}$ and $$G_1, G_2 in mathcal{G} Rightarrow G_1 cap G_2 in mathcal{G}.$$ Check that $mathcal{G}_n := {emptyset,{1},{2},ldots,[n+1,infty)}$ is a $cap$-stable generator of $mathcal{F}_n$.
- In order to show that $f_n$ is a martingale, it suffices to show $$mathbb{E}(f_n mid mathcal{F}_{n-1}) = f_{n-1};$$ by step 1 this is equivalent to $$int_G f_n , dmathbb{P} = int_G f_{n-1} , dmathbb{P} tag{1}$$ for all $G in mathcal{G}_{n-1}$.
- By the very definition of $f_n$, we have $$begin{align*} A&:= int_G f_n , dmathbb{P} = (n+1) mathbb{P}(G cap [n+1,infty)) \ B &:= int_G f_{n-1} , dmathbb{P} = n mathbb{P}(G cap [n,infty)). end{align*}$$ We have to prove $A=B$ for all $G in mathcal{G}_{n-1}$. Consider two cases separately:
- $G = {j}$ for some $j in {1,ldots,n-1}$: Since $G cap [ninfty) = G cap [n+1,infty)=emptyset$, we get $A=B=0$.
- $G = [n,infty)$: Conclude from $G cap [n,infty)= [n,infty)$ and $G cap [n+1,infty) = [n+1,infty)$ that $A=B=1$.
answered Apr 27 '15 at 18:24
sazsaz
81.5k861127
$endgroup$
$begingroup$
I finally got back to this guy. I'm still not clear on a lot of things here. It's basically step 1. If I accept that I'm able to follow steps 2-3. Can you point me to some sources where I can read up on step 1. I can see how the addition of $phi$ has made $mathscr{G_n}$ $cap$-stable but I don't know anything about $L^1$ spaces or why this result (i.e. step 1) holds.
$endgroup$
– Srinivas Eswar
Aug 28 '15 at 18:36
$begingroup$
@SrinivasEswar What $phi$ are you talking about? Do you know the result for $mathcal{G} := mathcal{F}$? In this case, it is a standard characterization of cond. expectations which you can find in any textbook on this topic.
$endgroup$
– saz
Aug 28 '15 at 22:15
$begingroup$
I was referring to the null set. Thanks for the help.
$endgroup$
– Srinivas Eswar
Aug 28 '15 at 23:45
add a comment |
2
$begingroup$
Hints:
- It is well-known that a random variable $Y in L^1$ equals $mathbb{E}(X mid mathcal{F})$ if and only if it is $mathcal{F}$-measurable and $$int_G Y , dmathbb{P} = int_G X , dmathbb{P} qquad text{for all $G in mathcal{G}$}$$ where $mathcal{G}$ denotes a $cap$-stable generator of $mathcal{F}$, i.e. $sigma(mathcal{G}) = mathcal{F}$ and $$G_1, G_2 in mathcal{G} Rightarrow G_1 cap G_2 in mathcal{G}.$$ Check that $mathcal{G}_n := {emptyset,{1},{2},ldots,[n+1,infty)}$ is a $cap$-stable generator of $mathcal{F}_n$.
- In order to show that $f_n$ is a martingale, it suffices to show $$mathbb{E}(f_n mid mathcal{F}_{n-1}) = f_{n-1};$$ by step 1 this is equivalent to $$int_G f_n , dmathbb{P} = int_G f_{n-1} , dmathbb{P} tag{1}$$ for all $G in mathcal{G}_{n-1}$.
- By the very definition of $f_n$, we have $$begin{align*} A&:= int_G f_n , dmathbb{P} = (n+1) mathbb{P}(G cap [n+1,infty)) \ B &:= int_G f_{n-1} , dmathbb{P} = n mathbb{P}(G cap [n,infty)). end{align*}$$ We have to prove $A=B$ for all $G in mathcal{G}_{n-1}$. Consider two cases separately:
- $G = {j}$ for some $j in {1,ldots,n-1}$: Since $G cap [ninfty) = G cap [n+1,infty)=emptyset$, we get $A=B=0$.
- $G = [n,infty)$: Conclude from $G cap [n,infty)= [n,infty)$ and $G cap [n+1,infty) = [n+1,infty)$ that $A=B=1$.
answered Apr 27 '15 at 18:24
sazsaz
81.5k861127
$endgroup$
$begingroup$
I finally got back to this guy. I'm still not clear on a lot of things here. It's basically step 1. If I accept that I'm able to follow steps 2-3. Can you point me to some sources where I can read up on step 1. I can see how the addition of $phi$ has made $mathscr{G_n}$ $cap$-stable but I don't know anything about $L^1$ spaces or why this result (i.e. step 1) holds.
$endgroup$
– Srinivas Eswar
Aug 28 '15 at 18:36
$begingroup$
@SrinivasEswar What $phi$ are you talking about? Do you know the result for $mathcal{G} := mathcal{F}$? In this case, it is a standard characterization of cond. expectations which you can find in any textbook on this topic.
$endgroup$
– saz
Aug 28 '15 at 22:15
$begingroup$
I was referring to the null set. Thanks for the help.
$endgroup$
– Srinivas Eswar
Aug 28 '15 at 23:45
add a comment |
2
2
2
$begingroup$
Hints:
- It is well-known that a random variable $Y in L^1$ equals $mathbb{E}(X mid mathcal{F})$ if and only if it is $mathcal{F}$-measurable and $$int_G Y , dmathbb{P} = int_G X , dmathbb{P} qquad text{for all $G in mathcal{G}$}$$ where $mathcal{G}$ denotes a $cap$-stable generator of $mathcal{F}$, i.e. $sigma(mathcal{G}) = mathcal{F}$ and $$G_1, G_2 in mathcal{G} Rightarrow G_1 cap G_2 in mathcal{G}.$$ Check that $mathcal{G}_n := {emptyset,{1},{2},ldots,[n+1,infty)}$ is a $cap$-stable generator of $mathcal{F}_n$.
- In order to show that $f_n$ is a martingale, it suffices to show $$mathbb{E}(f_n mid mathcal{F}_{n-1}) = f_{n-1};$$ by step 1 this is equivalent to $$int_G f_n , dmathbb{P} = int_G f_{n-1} , dmathbb{P} tag{1}$$ for all $G in mathcal{G}_{n-1}$.
- By the very definition of $f_n$, we have $$begin{align*} A&:= int_G f_n , dmathbb{P} = (n+1) mathbb{P}(G cap [n+1,infty)) \ B &:= int_G f_{n-1} , dmathbb{P} = n mathbb{P}(G cap [n,infty)). end{align*}$$ We have to prove $A=B$ for all $G in mathcal{G}_{n-1}$. Consider two cases separately:
- $G = {j}$ for some $j in {1,ldots,n-1}$: Since $G cap [ninfty) = G cap [n+1,infty)=emptyset$, we get $A=B=0$.
- $G = [n,infty)$: Conclude from $G cap [n,infty)= [n,infty)$ and $G cap [n+1,infty) = [n+1,infty)$ that $A=B=1$.
answered Apr 27 '15 at 18:24
sazsaz
81.5k861127
$endgroup$
Hints:
- It is well-known that a random variable $Y in L^1$ equals $mathbb{E}(X mid mathcal{F})$ if and only if it is $mathcal{F}$-measurable and $$int_G Y , dmathbb{P} = int_G X , dmathbb{P} qquad text{for all $G in mathcal{G}$}$$ where $mathcal{G}$ denotes a $cap$-stable generator of $mathcal{F}$, i.e. $sigma(mathcal{G}) = mathcal{F}$ and $$G_1, G_2 in mathcal{G} Rightarrow G_1 cap G_2 in mathcal{G}.$$ Check that $mathcal{G}_n := {emptyset,{1},{2},ldots,[n+1,infty)}$ is a $cap$-stable generator of $mathcal{F}_n$.
- In order to show that $f_n$ is a martingale, it suffices to show $$mathbb{E}(f_n mid mathcal{F}_{n-1}) = f_{n-1};$$ by step 1 this is equivalent to $$int_G f_n , dmathbb{P} = int_G f_{n-1} , dmathbb{P} tag{1}$$ for all $G in mathcal{G}_{n-1}$.
- By the very definition of $f_n$, we have $$begin{align*} A&:= int_G f_n , dmathbb{P} = (n+1) mathbb{P}(G cap [n+1,infty)) \ B &:= int_G f_{n-1} , dmathbb{P} = n mathbb{P}(G cap [n,infty)). end{align*}$$ We have to prove $A=B$ for all $G in mathcal{G}_{n-1}$. Consider two cases separately:
- $G = {j}$ for some $j in {1,ldots,n-1}$: Since $G cap [ninfty) = G cap [n+1,infty)=emptyset$, we get $A=B=0$.
- $G = [n,infty)$: Conclude from $G cap [n,infty)= [n,infty)$ and $G cap [n+1,infty) = [n+1,infty)$ that $A=B=1$.
answered Apr 27 '15 at 18:24
sazsaz
81.5k861127
edited Apr 28 '15 at 17:30
answered Apr 27 '15 at 18:24
sazsaz
81.5k861127
answered Apr 27 '15 at 18:24
sazsaz
81.5k861127
answered Apr 27 '15 at 18:24
sazsaz
81.5k861127
81.5k861127
$begingroup$
I finally got back to this guy. I'm still not clear on a lot of things here. It's basically step 1. If I accept that I'm able to follow steps 2-3. Can you point me to some sources where I can read up on step 1. I can see how the addition of $phi$ has made $mathscr{G_n}$ $cap$-stable but I don't know anything about $L^1$ spaces or why this result (i.e. step 1) holds.
$endgroup$
– Srinivas Eswar
Aug 28 '15 at 18:36
$begingroup$
@SrinivasEswar What $phi$ are you talking about? Do you know the result for $mathcal{G} := mathcal{F}$? In this case, it is a standard characterization of cond. expectations which you can find in any textbook on this topic.
$endgroup$
– saz
Aug 28 '15 at 22:15
$begingroup$
I was referring to the null set. Thanks for the help.
$endgroup$
– Srinivas Eswar
Aug 28 '15 at 23:45
add a comment |
$begingroup$
I finally got back to this guy. I'm still not clear on a lot of things here. It's basically step 1. If I accept that I'm able to follow steps 2-3. Can you point me to some sources where I can read up on step 1. I can see how the addition of $phi$ has made $mathscr{G_n}$ $cap$-stable but I don't know anything about $L^1$ spaces or why this result (i.e. step 1) holds.
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– Srinivas Eswar
Aug 28 '15 at 18:36
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@SrinivasEswar What $phi$ are you talking about? Do you know the result for $mathcal{G} := mathcal{F}$? In this case, it is a standard characterization of cond. expectations which you can find in any textbook on this topic.
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– saz
Aug 28 '15 at 22:15
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I was referring to the null set. Thanks for the help.
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– Srinivas Eswar
Aug 28 '15 at 23:45
$begingroup$
I finally got back to this guy. I'm still not clear on a lot of things here. It's basically step 1. If I accept that I'm able to follow steps 2-3. Can you point me to some sources where I can read up on step 1. I can see how the addition of $phi$ has made $mathscr{G_n}$ $cap$-stable but I don't know anything about $L^1$ spaces or why this result (i.e. step 1) holds.
$endgroup$
– Srinivas Eswar
Aug 28 '15 at 18:36
$begingroup$
I finally got back to this guy. I'm still not clear on a lot of things here. It's basically step 1. If I accept that I'm able to follow steps 2-3. Can you point me to some sources where I can read up on step 1. I can see how the addition of $phi$ has made $mathscr{G_n}$ $cap$-stable but I don't know anything about $L^1$ spaces or why this result (i.e. step 1) holds.
$endgroup$
– Srinivas Eswar
Aug 28 '15 at 18:36
$begingroup$
@SrinivasEswar What $phi$ are you talking about? Do you know the result for $mathcal{G} := mathcal{F}$? In this case, it is a standard characterization of cond. expectations which you can find in any textbook on this topic.
$endgroup$
– saz
Aug 28 '15 at 22:15
$begingroup$
@SrinivasEswar What $phi$ are you talking about? Do you know the result for $mathcal{G} := mathcal{F}$? In this case, it is a standard characterization of cond. expectations which you can find in any textbook on this topic.
$endgroup$
– saz
Aug 28 '15 at 22:15
$begingroup$
I was referring to the null set. Thanks for the help.
$endgroup$
– Srinivas Eswar
Aug 28 '15 at 23:45
$begingroup$
I was referring to the null set. Thanks for the help.
$endgroup$
– Srinivas Eswar
Aug 28 '15 at 23:45
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@saz, Any luck with this question?
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– Srinivas Eswar
Apr 27 '15 at 15:08
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I know how to solve this question, yes. But since you don't show any effort, I don't feel motivated to answer it.
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– saz
Apr 27 '15 at 16:18
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@saz, I have tried two ways to show it but I don't know if they're the right paths. The first attempt was to take $mathbb E[f_{n+1} lvert mathscr F_n]$ and split $[n+1,infty)$ into ${n} cup [n+1,infty)$ but the indicator variable is still within the expectation. My next attempt was to graph it out but I still can't get to the answer. I am truly stuck at this point. A pointer in the right direction would help.
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– Srinivas Eswar
Apr 27 '15 at 17:16
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So, any progress?
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– saz
Apr 28 '15 at 18:10
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@saz I sort of follow what you're saying but we never used $mathscr G$ in our study. My course is "classical probability" and tends to avoid measure theory. I'm still working out step 1. I'll update once I wrap my head around it.
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– Srinivas Eswar
Apr 28 '15 at 22:19