Third Absolute Moment of a Simple Random Walk
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This might be a very basic question. Let $S_n$ denote a simple random walk at time $n$, i.e. $S_n = sum_{i=1}^n X_i$ , where $X_1,ldots,X_n$ are i.i.d. Rademacher with $mathbb{P}(X_1=1) = mathbb{P}(X_1=-1) = frac{1}{2}$. Is there any upper bound for the quantity $mathbb{E}left(|S_n|^3right)$ in terms of powers of $n$? My interest is in the power of $n$ in the tightest such upper bound.
Any help will be greatly appreciated
probability random-walk
asked Dec 6 '18 at 19:06
UsermathUsermath
1306
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add a comment |
$begingroup$
This might be a very basic question. Let $S_n$ denote a simple random walk at time $n$, i.e. $S_n = sum_{i=1}^n X_i$ , where $X_1,ldots,X_n$ are i.i.d. Rademacher with $mathbb{P}(X_1=1) = mathbb{P}(X_1=-1) = frac{1}{2}$. Is there any upper bound for the quantity $mathbb{E}left(|S_n|^3right)$ in terms of powers of $n$? My interest is in the power of $n$ in the tightest such upper bound.
Any help will be greatly appreciated
probability random-walk
asked Dec 6 '18 at 19:06
UsermathUsermath
1306
$endgroup$
1
$begingroup$
Think I got it. A simple moment calculation will give $mathbb{E}S_n^4 = O(n^2)$. Then, from Holder's inequality, it will follow that $mathbb{E}|S_n|^3 leq (mathbb{E}S_n^4)^{3/4} = O(n^{3/2})$. Also, this is the tightest power of $n$ one can get, since $mathbb{E}|S_n|^3 geq (mathbb{E}S_n^2)^{3/2}=n^{3/2}$.
$endgroup$
– Usermath
Dec 6 '18 at 19:34
add a comment |
$begingroup$
This might be a very basic question. Let $S_n$ denote a simple random walk at time $n$, i.e. $S_n = sum_{i=1}^n X_i$ , where $X_1,ldots,X_n$ are i.i.d. Rademacher with $mathbb{P}(X_1=1) = mathbb{P}(X_1=-1) = frac{1}{2}$. Is there any upper bound for the quantity $mathbb{E}left(|S_n|^3right)$ in terms of powers of $n$? My interest is in the power of $n$ in the tightest such upper bound.
Any help will be greatly appreciated
probability random-walk
asked Dec 6 '18 at 19:06
UsermathUsermath
1306
$endgroup$
This might be a very basic question. Let $S_n$ denote a simple random walk at time $n$, i.e. $S_n = sum_{i=1}^n X_i$ , where $X_1,ldots,X_n$ are i.i.d. Rademacher with $mathbb{P}(X_1=1) = mathbb{P}(X_1=-1) = frac{1}{2}$. Is there any upper bound for the quantity $mathbb{E}left(|S_n|^3right)$ in terms of powers of $n$? My interest is in the power of $n$ in the tightest such upper bound.
Any help will be greatly appreciated
probability random-walk
probability random-walk
asked Dec 6 '18 at 19:06
UsermathUsermath
1306
asked Dec 6 '18 at 19:06
UsermathUsermath
1306
asked Dec 6 '18 at 19:06
UsermathUsermath
1306
asked Dec 6 '18 at 19:06
UsermathUsermath
1306
asked Dec 6 '18 at 19:06
UsermathUsermath
1306
1306
1
$begingroup$
Think I got it. A simple moment calculation will give $mathbb{E}S_n^4 = O(n^2)$. Then, from Holder's inequality, it will follow that $mathbb{E}|S_n|^3 leq (mathbb{E}S_n^4)^{3/4} = O(n^{3/2})$. Also, this is the tightest power of $n$ one can get, since $mathbb{E}|S_n|^3 geq (mathbb{E}S_n^2)^{3/2}=n^{3/2}$.
$endgroup$
– Usermath
Dec 6 '18 at 19:34
add a comment |
1
$begingroup$
Think I got it. A simple moment calculation will give $mathbb{E}S_n^4 = O(n^2)$. Then, from Holder's inequality, it will follow that $mathbb{E}|S_n|^3 leq (mathbb{E}S_n^4)^{3/4} = O(n^{3/2})$. Also, this is the tightest power of $n$ one can get, since $mathbb{E}|S_n|^3 geq (mathbb{E}S_n^2)^{3/2}=n^{3/2}$.
$endgroup$
– Usermath
Dec 6 '18 at 19:34
1
1
$begingroup$
Think I got it. A simple moment calculation will give $mathbb{E}S_n^4 = O(n^2)$. Then, from Holder's inequality, it will follow that $mathbb{E}|S_n|^3 leq (mathbb{E}S_n^4)^{3/4} = O(n^{3/2})$. Also, this is the tightest power of $n$ one can get, since $mathbb{E}|S_n|^3 geq (mathbb{E}S_n^2)^{3/2}=n^{3/2}$.
$endgroup$
– Usermath
Dec 6 '18 at 19:34
$begingroup$
Think I got it. A simple moment calculation will give $mathbb{E}S_n^4 = O(n^2)$. Then, from Holder's inequality, it will follow that $mathbb{E}|S_n|^3 leq (mathbb{E}S_n^4)^{3/4} = O(n^{3/2})$. Also, this is the tightest power of $n$ one can get, since $mathbb{E}|S_n|^3 geq (mathbb{E}S_n^2)^{3/2}=n^{3/2}$.
$endgroup$
– Usermath
Dec 6 '18 at 19:34
add a comment |
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$begingroup$
Think I got it. A simple moment calculation will give $mathbb{E}S_n^4 = O(n^2)$. Then, from Holder's inequality, it will follow that $mathbb{E}|S_n|^3 leq (mathbb{E}S_n^4)^{3/4} = O(n^{3/2})$. Also, this is the tightest power of $n$ one can get, since $mathbb{E}|S_n|^3 geq (mathbb{E}S_n^2)^{3/2}=n^{3/2}$.
$endgroup$
– Usermath
Dec 6 '18 at 19:34