A problem in probability theory about expected value and Fubini's theorem
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Let ${X_{n}:ngeq1}$ be a seqeunce of square-integrable independent random variables on $(Omega,mathcal{F},mathbb{P})$ with $mathbb{E}[X_{n}]=0$ for every $ngeq1$. Set $S_{n}:=sum_{j=1}^{n}X_{j}$ for each $ngeq1$. Assume that $sum_{ngeq1}mathbb{E}[X_{n}^{2}]<infty$. Prove that, for every $p>1$,
i)$$big(mathbb{E}big[sup_{ngeq1}|S_{n}|^{2p}big]big)^{frac{1}{p}}leqfrac{p}{p-1}big(mathbb{E}big[|S|^{2p}big]big)^{frac{1}{p}}$$
Without loss of generality assume that $|S|^2 in L^p$ and you might first consider $min{big( sup_n|S_n|^2big),K}$ for any fixed $K>0.$
ii) If $Sin L^q$ for some $qin (2,infty)$, then $S_n to S$ also in $L^q$.
Using the said assumptions, it's known that that $S_{n}to S$ a.s. for some random variable S and due to completeness of $L^2(Omega,mathcal{F},mathbb{P})$, I know that $S in L^2$ and $S to S_n$ also in $L^2$. I also can prove that, for every $t>0$,
$$mathbb{P}big(sup_{ngeq1} |S_{n}|^2>tbig)leqfrac{1}{t}mathbb{E}big[S^2;sup_{ngeq1}|S_{n}|^2>tbig].$$
I think in order to prove the claim, I can combine the above statement with the following result
$$mathbb{E}[X^p]=pint_{0}^{infty}t^{p-1}mathbb{P}(X> t)dt = pint_{0}^{infty}t^{p-1}mathbb{P}(Xgeq t)dt.$$
My idea is to use Fubini's theorem to get the result, but I don't know how to connect them together.
I'd appreciate any help.
probability-theory convergence expected-value
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Let ${X_{n}:ngeq1}$ be a seqeunce of square-integrable independent random variables on $(Omega,mathcal{F},mathbb{P})$ with $mathbb{E}[X_{n}]=0$ for every $ngeq1$. Set $S_{n}:=sum_{j=1}^{n}X_{j}$ for each $ngeq1$. Assume that $sum_{ngeq1}mathbb{E}[X_{n}^{2}]<infty$. Prove that, for every $p>1$,
i)$$big(mathbb{E}big[sup_{ngeq1}|S_{n}|^{2p}big]big)^{frac{1}{p}}leqfrac{p}{p-1}big(mathbb{E}big[|S|^{2p}big]big)^{frac{1}{p}}$$
Without loss of generality assume that $|S|^2 in L^p$ and you might first consider $min{big( sup_n|S_n|^2big),K}$ for any fixed $K>0.$
ii) If $Sin L^q$ for some $qin (2,infty)$, then $S_n to S$ also in $L^q$.
Using the said assumptions, it's known that that $S_{n}to S$ a.s. for some random variable S and due to completeness of $L^2(Omega,mathcal{F},mathbb{P})$, I know that $S in L^2$ and $S to S_n$ also in $L^2$. I also can prove that, for every $t>0$,
$$mathbb{P}big(sup_{ngeq1} |S_{n}|^2>tbig)leqfrac{1}{t}mathbb{E}big[S^2;sup_{ngeq1}|S_{n}|^2>tbig].$$
I think in order to prove the claim, I can combine the above statement with the following result
$$mathbb{E}[X^p]=pint_{0}^{infty}t^{p-1}mathbb{P}(X> t)dt = pint_{0}^{infty}t^{p-1}mathbb{P}(Xgeq t)dt.$$
My idea is to use Fubini's theorem to get the result, but I don't know how to connect them together.
I'd appreciate any help.
probability-theory convergence expected-value
add a comment |
up vote
1
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favorite
up vote
1
down vote
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Let ${X_{n}:ngeq1}$ be a seqeunce of square-integrable independent random variables on $(Omega,mathcal{F},mathbb{P})$ with $mathbb{E}[X_{n}]=0$ for every $ngeq1$. Set $S_{n}:=sum_{j=1}^{n}X_{j}$ for each $ngeq1$. Assume that $sum_{ngeq1}mathbb{E}[X_{n}^{2}]<infty$. Prove that, for every $p>1$,
i)$$big(mathbb{E}big[sup_{ngeq1}|S_{n}|^{2p}big]big)^{frac{1}{p}}leqfrac{p}{p-1}big(mathbb{E}big[|S|^{2p}big]big)^{frac{1}{p}}$$
Without loss of generality assume that $|S|^2 in L^p$ and you might first consider $min{big( sup_n|S_n|^2big),K}$ for any fixed $K>0.$
ii) If $Sin L^q$ for some $qin (2,infty)$, then $S_n to S$ also in $L^q$.
Using the said assumptions, it's known that that $S_{n}to S$ a.s. for some random variable S and due to completeness of $L^2(Omega,mathcal{F},mathbb{P})$, I know that $S in L^2$ and $S to S_n$ also in $L^2$. I also can prove that, for every $t>0$,
$$mathbb{P}big(sup_{ngeq1} |S_{n}|^2>tbig)leqfrac{1}{t}mathbb{E}big[S^2;sup_{ngeq1}|S_{n}|^2>tbig].$$
I think in order to prove the claim, I can combine the above statement with the following result
$$mathbb{E}[X^p]=pint_{0}^{infty}t^{p-1}mathbb{P}(X> t)dt = pint_{0}^{infty}t^{p-1}mathbb{P}(Xgeq t)dt.$$
My idea is to use Fubini's theorem to get the result, but I don't know how to connect them together.
I'd appreciate any help.
probability-theory convergence expected-value
Let ${X_{n}:ngeq1}$ be a seqeunce of square-integrable independent random variables on $(Omega,mathcal{F},mathbb{P})$ with $mathbb{E}[X_{n}]=0$ for every $ngeq1$. Set $S_{n}:=sum_{j=1}^{n}X_{j}$ for each $ngeq1$. Assume that $sum_{ngeq1}mathbb{E}[X_{n}^{2}]<infty$. Prove that, for every $p>1$,
i)$$big(mathbb{E}big[sup_{ngeq1}|S_{n}|^{2p}big]big)^{frac{1}{p}}leqfrac{p}{p-1}big(mathbb{E}big[|S|^{2p}big]big)^{frac{1}{p}}$$
Without loss of generality assume that $|S|^2 in L^p$ and you might first consider $min{big( sup_n|S_n|^2big),K}$ for any fixed $K>0.$
ii) If $Sin L^q$ for some $qin (2,infty)$, then $S_n to S$ also in $L^q$.
Using the said assumptions, it's known that that $S_{n}to S$ a.s. for some random variable S and due to completeness of $L^2(Omega,mathcal{F},mathbb{P})$, I know that $S in L^2$ and $S to S_n$ also in $L^2$. I also can prove that, for every $t>0$,
$$mathbb{P}big(sup_{ngeq1} |S_{n}|^2>tbig)leqfrac{1}{t}mathbb{E}big[S^2;sup_{ngeq1}|S_{n}|^2>tbig].$$
I think in order to prove the claim, I can combine the above statement with the following result
$$mathbb{E}[X^p]=pint_{0}^{infty}t^{p-1}mathbb{P}(X> t)dt = pint_{0}^{infty}t^{p-1}mathbb{P}(Xgeq t)dt.$$
My idea is to use Fubini's theorem to get the result, but I don't know how to connect them together.
I'd appreciate any help.
probability-theory convergence expected-value
probability-theory convergence expected-value
asked Nov 15 at 16:27
Weak Nullstellensatz
163
163
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1 Answer
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Observe that
$$
mathbb Eleft[S^2;sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright]
=int_{0}^{+infty}Prleft(left{S^2gt sright}cap left{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}right)mathrm ds=int_0^{t/2}+int_{t/2}^{+infty}.
$$
Bound the first part by $t/2 Prleft{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}$ and the second one by $int_{t/2}^{+infty}Pr left{S^2gt sright}mathrm ds$. We get
$$
tPrleft{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}leqslant 2int_{t/2}^{+infty}Pr left{S^2gt sright}mathrm ds.
$$
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Observe that
$$
mathbb Eleft[S^2;sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright]
=int_{0}^{+infty}Prleft(left{S^2gt sright}cap left{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}right)mathrm ds=int_0^{t/2}+int_{t/2}^{+infty}.
$$
Bound the first part by $t/2 Prleft{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}$ and the second one by $int_{t/2}^{+infty}Pr left{S^2gt sright}mathrm ds$. We get
$$
tPrleft{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}leqslant 2int_{t/2}^{+infty}Pr left{S^2gt sright}mathrm ds.
$$
add a comment |
up vote
0
down vote
Observe that
$$
mathbb Eleft[S^2;sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright]
=int_{0}^{+infty}Prleft(left{S^2gt sright}cap left{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}right)mathrm ds=int_0^{t/2}+int_{t/2}^{+infty}.
$$
Bound the first part by $t/2 Prleft{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}$ and the second one by $int_{t/2}^{+infty}Pr left{S^2gt sright}mathrm ds$. We get
$$
tPrleft{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}leqslant 2int_{t/2}^{+infty}Pr left{S^2gt sright}mathrm ds.
$$
add a comment |
up vote
0
down vote
up vote
0
down vote
Observe that
$$
mathbb Eleft[S^2;sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright]
=int_{0}^{+infty}Prleft(left{S^2gt sright}cap left{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}right)mathrm ds=int_0^{t/2}+int_{t/2}^{+infty}.
$$
Bound the first part by $t/2 Prleft{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}$ and the second one by $int_{t/2}^{+infty}Pr left{S^2gt sright}mathrm ds$. We get
$$
tPrleft{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}leqslant 2int_{t/2}^{+infty}Pr left{S^2gt sright}mathrm ds.
$$
Observe that
$$
mathbb Eleft[S^2;sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright]
=int_{0}^{+infty}Prleft(left{S^2gt sright}cap left{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}right)mathrm ds=int_0^{t/2}+int_{t/2}^{+infty}.
$$
Bound the first part by $t/2 Prleft{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}$ and the second one by $int_{t/2}^{+infty}Pr left{S^2gt sright}mathrm ds$. We get
$$
tPrleft{sup_{ngeqslant 1}leftlvert S_nrightrvert^2gt tright}leqslant 2int_{t/2}^{+infty}Pr left{S^2gt sright}mathrm ds.
$$
answered Nov 16 at 16:12
Davide Giraudo
124k16149254
124k16149254
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