Mixing
Advice on correct usage of conditional expectation (proof verification)
$begingroup$
Assume $X_1, ldots, X_n$ are independent and identically distributed random variables where $X in {-1, 1}$ with equal probabilities, and $S_k$ represents their cumulative sum of up to $k$ variables.
The exercise is to show that $mathbb{E}S_{tau}^2 = mathbb{E}tau$, where $tau$ is a $(mathcal{F})_n$ measurable stopping time.
On one hand, it feels like
$$ mathbb{E}tau = sum_{j=1}^nj mathbb{P}(tau=j)=sum_{j=1}^n j mathbb{E}(mathbb{1}_{tau = j})$$
On the other hand,
$$ mathbb{E}(S^2_tau)=sum_{j=1}^nmathbb{E}(S_j^2 mathbb{1}_{tau = j})=sum_{j=1}^nmathbb{E}( mathbb{E}[S_j^2 mathbb{1}_{tau = j}|mathcal{F}_j]) = sum_{j=1}^nmathbb{E}(mathbb{1}_{tau = j} mathbb{E}[S_j^2 |mathcal{F}_j])$$
In essence, it is easy to check that $mathbb{E}(S_j^2)=j$ due to equal probabilities.
My question: is the last equation a proper way to pull out the $j$ from the expectation? That is, by stating that $mathbb{E}(S_j^2 | mathcal{F}_j)=j$? It feels like double expectation here is one way to pull out the $j$, but not entirely sure whether $mathcal{F}_j$ is the right choice here, since it makes the $S_j^2$ measurable.
On the other hand, is pulling out the $j$ necessary, or is it for some reason obvious that
$$ mathbb{E}S_j^2 mathbb{1}_{tau=j} = j mathbb{P}(tau = j)?$$
probability probability-theory martingales
asked Jan 16 at 20:54
sjoksjok
353
$endgroup$
add a comment |
$begingroup$
Assume $X_1, ldots, X_n$ are independent and identically distributed random variables where $X in {-1, 1}$ with equal probabilities, and $S_k$ represents their cumulative sum of up to $k$ variables.
The exercise is to show that $mathbb{E}S_{tau}^2 = mathbb{E}tau$, where $tau$ is a $(mathcal{F})_n$ measurable stopping time.
On one hand, it feels like
$$ mathbb{E}tau = sum_{j=1}^nj mathbb{P}(tau=j)=sum_{j=1}^n j mathbb{E}(mathbb{1}_{tau = j})$$
On the other hand,
$$ mathbb{E}(S^2_tau)=sum_{j=1}^nmathbb{E}(S_j^2 mathbb{1}_{tau = j})=sum_{j=1}^nmathbb{E}( mathbb{E}[S_j^2 mathbb{1}_{tau = j}|mathcal{F}_j]) = sum_{j=1}^nmathbb{E}(mathbb{1}_{tau = j} mathbb{E}[S_j^2 |mathcal{F}_j])$$
In essence, it is easy to check that $mathbb{E}(S_j^2)=j$ due to equal probabilities.
My question: is the last equation a proper way to pull out the $j$ from the expectation? That is, by stating that $mathbb{E}(S_j^2 | mathcal{F}_j)=j$? It feels like double expectation here is one way to pull out the $j$, but not entirely sure whether $mathcal{F}_j$ is the right choice here, since it makes the $S_j^2$ measurable.
On the other hand, is pulling out the $j$ necessary, or is it for some reason obvious that
$$ mathbb{E}S_j^2 mathbb{1}_{tau=j} = j mathbb{P}(tau = j)?$$
probability probability-theory martingales
asked Jan 16 at 20:54
sjoksjok
353
$endgroup$
$begingroup$
@d.k.o If I'm not mistaken, $mathbb{E}(X_i^2) = 1$ due to squares, and $mathbb{E}(X_i X_j) = 1cdot 1 cdot (frac{1}{2})^2 + (-1)cdot 1 cdot (frac{1}{2})^2 + 1cdot (-1) cdot (frac{1}{2})^2 + (-1)cdot (-1) cdot (frac{1}{2})^2 = 0$ for $i neq j$, so $mathbb{E}(S_j^2)$ should reduce to $sum mathbb{E}X_i^2 = j$
$endgroup$
– sjok
Jan 16 at 22:11
add a comment |
$begingroup$
Assume $X_1, ldots, X_n$ are independent and identically distributed random variables where $X in {-1, 1}$ with equal probabilities, and $S_k$ represents their cumulative sum of up to $k$ variables.
The exercise is to show that $mathbb{E}S_{tau}^2 = mathbb{E}tau$, where $tau$ is a $(mathcal{F})_n$ measurable stopping time.
On one hand, it feels like
$$ mathbb{E}tau = sum_{j=1}^nj mathbb{P}(tau=j)=sum_{j=1}^n j mathbb{E}(mathbb{1}_{tau = j})$$
On the other hand,
$$ mathbb{E}(S^2_tau)=sum_{j=1}^nmathbb{E}(S_j^2 mathbb{1}_{tau = j})=sum_{j=1}^nmathbb{E}( mathbb{E}[S_j^2 mathbb{1}_{tau = j}|mathcal{F}_j]) = sum_{j=1}^nmathbb{E}(mathbb{1}_{tau = j} mathbb{E}[S_j^2 |mathcal{F}_j])$$
In essence, it is easy to check that $mathbb{E}(S_j^2)=j$ due to equal probabilities.
My question: is the last equation a proper way to pull out the $j$ from the expectation? That is, by stating that $mathbb{E}(S_j^2 | mathcal{F}_j)=j$? It feels like double expectation here is one way to pull out the $j$, but not entirely sure whether $mathcal{F}_j$ is the right choice here, since it makes the $S_j^2$ measurable.
On the other hand, is pulling out the $j$ necessary, or is it for some reason obvious that
$$ mathbb{E}S_j^2 mathbb{1}_{tau=j} = j mathbb{P}(tau = j)?$$
probability probability-theory martingales
asked Jan 16 at 20:54
sjoksjok
353
$endgroup$
Assume $X_1, ldots, X_n$ are independent and identically distributed random variables where $X in {-1, 1}$ with equal probabilities, and $S_k$ represents their cumulative sum of up to $k$ variables.
The exercise is to show that $mathbb{E}S_{tau}^2 = mathbb{E}tau$, where $tau$ is a $(mathcal{F})_n$ measurable stopping time.
On one hand, it feels like
$$ mathbb{E}tau = sum_{j=1}^nj mathbb{P}(tau=j)=sum_{j=1}^n j mathbb{E}(mathbb{1}_{tau = j})$$
On the other hand,
$$ mathbb{E}(S^2_tau)=sum_{j=1}^nmathbb{E}(S_j^2 mathbb{1}_{tau = j})=sum_{j=1}^nmathbb{E}( mathbb{E}[S_j^2 mathbb{1}_{tau = j}|mathcal{F}_j]) = sum_{j=1}^nmathbb{E}(mathbb{1}_{tau = j} mathbb{E}[S_j^2 |mathcal{F}_j])$$
In essence, it is easy to check that $mathbb{E}(S_j^2)=j$ due to equal probabilities.
My question: is the last equation a proper way to pull out the $j$ from the expectation? That is, by stating that $mathbb{E}(S_j^2 | mathcal{F}_j)=j$? It feels like double expectation here is one way to pull out the $j$, but not entirely sure whether $mathcal{F}_j$ is the right choice here, since it makes the $S_j^2$ measurable.
On the other hand, is pulling out the $j$ necessary, or is it for some reason obvious that
$$ mathbb{E}S_j^2 mathbb{1}_{tau=j} = j mathbb{P}(tau = j)?$$
probability probability-theory martingales
probability probability-theory martingales
asked Jan 16 at 20:54
sjoksjok
353
asked Jan 16 at 20:54
sjoksjok
353
edited Jan 16 at 21:09
sjok
asked Jan 16 at 20:54
sjoksjok
353
asked Jan 16 at 20:54
sjoksjok
353
asked Jan 16 at 20:54
sjoksjok
353
353
$begingroup$
@d.k.o If I'm not mistaken, $mathbb{E}(X_i^2) = 1$ due to squares, and $mathbb{E}(X_i X_j) = 1cdot 1 cdot (frac{1}{2})^2 + (-1)cdot 1 cdot (frac{1}{2})^2 + 1cdot (-1) cdot (frac{1}{2})^2 + (-1)cdot (-1) cdot (frac{1}{2})^2 = 0$ for $i neq j$, so $mathbb{E}(S_j^2)$ should reduce to $sum mathbb{E}X_i^2 = j$
$endgroup$
– sjok
Jan 16 at 22:11
add a comment |
$begingroup$
@d.k.o If I'm not mistaken, $mathbb{E}(X_i^2) = 1$ due to squares, and $mathbb{E}(X_i X_j) = 1cdot 1 cdot (frac{1}{2})^2 + (-1)cdot 1 cdot (frac{1}{2})^2 + 1cdot (-1) cdot (frac{1}{2})^2 + (-1)cdot (-1) cdot (frac{1}{2})^2 = 0$ for $i neq j$, so $mathbb{E}(S_j^2)$ should reduce to $sum mathbb{E}X_i^2 = j$
$endgroup$
– sjok
Jan 16 at 22:11
$begingroup$
@d.k.o If I'm not mistaken, $mathbb{E}(X_i^2) = 1$ due to squares, and $mathbb{E}(X_i X_j) = 1cdot 1 cdot (frac{1}{2})^2 + (-1)cdot 1 cdot (frac{1}{2})^2 + 1cdot (-1) cdot (frac{1}{2})^2 + (-1)cdot (-1) cdot (frac{1}{2})^2 = 0$ for $i neq j$, so $mathbb{E}(S_j^2)$ should reduce to $sum mathbb{E}X_i^2 = j$
$endgroup$
– sjok
Jan 16 at 22:11
$begingroup$
@d.k.o If I'm not mistaken, $mathbb{E}(X_i^2) = 1$ due to squares, and $mathbb{E}(X_i X_j) = 1cdot 1 cdot (frac{1}{2})^2 + (-1)cdot 1 cdot (frac{1}{2})^2 + 1cdot (-1) cdot (frac{1}{2})^2 + (-1)cdot (-1) cdot (frac{1}{2})^2 = 0$ for $i neq j$, so $mathbb{E}(S_j^2)$ should reduce to $sum mathbb{E}X_i^2 = j$
$endgroup$
– sjok
Jan 16 at 22:11
add a comment |
1 Answer
1
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$begingroup$
First, ${S_k^2-k}$ is a martingale. Since $taule n$, the optional stopping theorem implies
$$
mathsf{E}[S_{tau}^2-tau]=mathsf{E}[S_1^2-1]=0.
$$
Regarding your question, let $n=4$, $tau:=minleft({kge 1:S_kge 1 }cup{n}right)$. Then
$$
3timesmathsf{P}(tau=3)=frac{3}{8},
$$
but
$$
mathsf{E}[S_tau^21{tau=3}]=mathsf{E}[S_tau^2mid tau=3]mathsf{P}(tau=3)=frac{1}{8}.
$$
Direct computation shows that
$$
mathsf{E}tau=sum_{j=1}^4 jmathsf{P}(tau=j)= 1timesfrac{1}{2}+2times 0+3timesfrac{1}{8}+4timesfrac{3}{8}=frac{19}{8}
$$
and
$$
mathsf{E}S_tau^2=sum_{j=1}^4 mathsf{E}[S_tau^21{tau=j}]=frac{1}{2}+0+frac{1}{8}+frac{7}{4}=frac{19}{8}.
$$
answered Jan 17 at 0:59
d.k.o.d.k.o.
9,685628
$endgroup$
$begingroup$
great counterexample, thanks. I still need to wrap my head around conditional expectations, apparently.
$endgroup$
– sjok
Jan 17 at 14:35
add a comment |
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1 Answer
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$begingroup$
First, ${S_k^2-k}$ is a martingale. Since $taule n$, the optional stopping theorem implies
$$
mathsf{E}[S_{tau}^2-tau]=mathsf{E}[S_1^2-1]=0.
$$
Regarding your question, let $n=4$, $tau:=minleft({kge 1:S_kge 1 }cup{n}right)$. Then
$$
3timesmathsf{P}(tau=3)=frac{3}{8},
$$
but
$$
mathsf{E}[S_tau^21{tau=3}]=mathsf{E}[S_tau^2mid tau=3]mathsf{P}(tau=3)=frac{1}{8}.
$$
Direct computation shows that
$$
mathsf{E}tau=sum_{j=1}^4 jmathsf{P}(tau=j)= 1timesfrac{1}{2}+2times 0+3timesfrac{1}{8}+4timesfrac{3}{8}=frac{19}{8}
$$
and
$$
mathsf{E}S_tau^2=sum_{j=1}^4 mathsf{E}[S_tau^21{tau=j}]=frac{1}{2}+0+frac{1}{8}+frac{7}{4}=frac{19}{8}.
$$
answered Jan 17 at 0:59
d.k.o.d.k.o.
9,685628
$endgroup$
$begingroup$
great counterexample, thanks. I still need to wrap my head around conditional expectations, apparently.
$endgroup$
– sjok
Jan 17 at 14:35
add a comment |
$begingroup$
First, ${S_k^2-k}$ is a martingale. Since $taule n$, the optional stopping theorem implies
$$
mathsf{E}[S_{tau}^2-tau]=mathsf{E}[S_1^2-1]=0.
$$
Regarding your question, let $n=4$, $tau:=minleft({kge 1:S_kge 1 }cup{n}right)$. Then
$$
3timesmathsf{P}(tau=3)=frac{3}{8},
$$
but
$$
mathsf{E}[S_tau^21{tau=3}]=mathsf{E}[S_tau^2mid tau=3]mathsf{P}(tau=3)=frac{1}{8}.
$$
Direct computation shows that
$$
mathsf{E}tau=sum_{j=1}^4 jmathsf{P}(tau=j)= 1timesfrac{1}{2}+2times 0+3timesfrac{1}{8}+4timesfrac{3}{8}=frac{19}{8}
$$
and
$$
mathsf{E}S_tau^2=sum_{j=1}^4 mathsf{E}[S_tau^21{tau=j}]=frac{1}{2}+0+frac{1}{8}+frac{7}{4}=frac{19}{8}.
$$
answered Jan 17 at 0:59
d.k.o.d.k.o.
9,685628
$endgroup$
$begingroup$
great counterexample, thanks. I still need to wrap my head around conditional expectations, apparently.
$endgroup$
– sjok
Jan 17 at 14:35
add a comment |
$begingroup$
First, ${S_k^2-k}$ is a martingale. Since $taule n$, the optional stopping theorem implies
$$
mathsf{E}[S_{tau}^2-tau]=mathsf{E}[S_1^2-1]=0.
$$
Regarding your question, let $n=4$, $tau:=minleft({kge 1:S_kge 1 }cup{n}right)$. Then
$$
3timesmathsf{P}(tau=3)=frac{3}{8},
$$
but
$$
mathsf{E}[S_tau^21{tau=3}]=mathsf{E}[S_tau^2mid tau=3]mathsf{P}(tau=3)=frac{1}{8}.
$$
Direct computation shows that
$$
mathsf{E}tau=sum_{j=1}^4 jmathsf{P}(tau=j)= 1timesfrac{1}{2}+2times 0+3timesfrac{1}{8}+4timesfrac{3}{8}=frac{19}{8}
$$
and
$$
mathsf{E}S_tau^2=sum_{j=1}^4 mathsf{E}[S_tau^21{tau=j}]=frac{1}{2}+0+frac{1}{8}+frac{7}{4}=frac{19}{8}.
$$
answered Jan 17 at 0:59
d.k.o.d.k.o.
9,685628
$endgroup$
First, ${S_k^2-k}$ is a martingale. Since $taule n$, the optional stopping theorem implies
$$
mathsf{E}[S_{tau}^2-tau]=mathsf{E}[S_1^2-1]=0.
$$
Regarding your question, let $n=4$, $tau:=minleft({kge 1:S_kge 1 }cup{n}right)$. Then
$$
3timesmathsf{P}(tau=3)=frac{3}{8},
$$
but
$$
mathsf{E}[S_tau^21{tau=3}]=mathsf{E}[S_tau^2mid tau=3]mathsf{P}(tau=3)=frac{1}{8}.
$$
Direct computation shows that
$$
mathsf{E}tau=sum_{j=1}^4 jmathsf{P}(tau=j)= 1timesfrac{1}{2}+2times 0+3timesfrac{1}{8}+4timesfrac{3}{8}=frac{19}{8}
$$
and
$$
mathsf{E}S_tau^2=sum_{j=1}^4 mathsf{E}[S_tau^21{tau=j}]=frac{1}{2}+0+frac{1}{8}+frac{7}{4}=frac{19}{8}.
$$
answered Jan 17 at 0:59
d.k.o.d.k.o.
9,685628
edited Jan 17 at 1:46
answered Jan 17 at 0:59
d.k.o.d.k.o.
9,685628
answered Jan 17 at 0:59
d.k.o.d.k.o.
9,685628
answered Jan 17 at 0:59
d.k.o.d.k.o.
9,685628
9,685628
$begingroup$
great counterexample, thanks. I still need to wrap my head around conditional expectations, apparently.
$endgroup$
– sjok
Jan 17 at 14:35
add a comment |
$begingroup$
great counterexample, thanks. I still need to wrap my head around conditional expectations, apparently.
$endgroup$
– sjok
Jan 17 at 14:35
$begingroup$
great counterexample, thanks. I still need to wrap my head around conditional expectations, apparently.
$endgroup$
– sjok
Jan 17 at 14:35
$begingroup$
great counterexample, thanks. I still need to wrap my head around conditional expectations, apparently.
$endgroup$
– sjok
Jan 17 at 14:35
add a comment |
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$begingroup$
@d.k.o If I'm not mistaken, $mathbb{E}(X_i^2) = 1$ due to squares, and $mathbb{E}(X_i X_j) = 1cdot 1 cdot (frac{1}{2})^2 + (-1)cdot 1 cdot (frac{1}{2})^2 + 1cdot (-1) cdot (frac{1}{2})^2 + (-1)cdot (-1) cdot (frac{1}{2})^2 = 0$ for $i neq j$, so $mathbb{E}(S_j^2)$ should reduce to $sum mathbb{E}X_i^2 = j$
$endgroup$
– sjok
Jan 16 at 22:11