Existence of nonnegative and nonconstant martingales
Are there martingales that are nonnegative and nonconstant? If so, are their any intuitive examples for such?
probability-theory examples-counterexamples martingales
edited Nov 27 '18 at 10:08
Davide Giraudo
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asked Nov 26 '18 at 20:05
Tesla
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Are there martingales that are nonnegative and nonconstant? If so, are their any intuitive examples for such?
probability-theory examples-counterexamples martingales
edited Nov 27 '18 at 10:08
Davide Giraudo
125k16150259
asked Nov 26 '18 at 20:05
Tesla
928426
add a comment |
Are there martingales that are nonnegative and nonconstant? If so, are their any intuitive examples for such?
probability-theory examples-counterexamples martingales
edited Nov 27 '18 at 10:08
Davide Giraudo
125k16150259
asked Nov 26 '18 at 20:05
Tesla
928426
Are there martingales that are nonnegative and nonconstant? If so, are their any intuitive examples for such?
probability-theory examples-counterexamples martingales
probability-theory examples-counterexamples martingales
edited Nov 27 '18 at 10:08
Davide Giraudo
125k16150259
asked Nov 26 '18 at 20:05
Tesla
928426
edited Nov 27 '18 at 10:08
Davide Giraudo
125k16150259
asked Nov 26 '18 at 20:05
Tesla
928426
edited Nov 27 '18 at 10:08
Davide Giraudo
125k16150259
edited Nov 27 '18 at 10:08
Davide Giraudo
125k16150259
edited Nov 27 '18 at 10:08
Davide Giraudo
125k16150259
125k16150259
asked Nov 26 '18 at 20:05
Tesla
928426
asked Nov 26 '18 at 20:05
Tesla
928426
asked Nov 26 '18 at 20:05
Tesla
928426
928426
add a comment |
add a comment |
2 Answers
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Example 1: Let $(X_j)_{j in mathbb{N}}$ be a sequence of independent random variables such that $$mathbb{P}(X_j = - 2^{-j})=frac{1}{2} quad text{and} quad mathbb{P}(X_j=2^{-j})=frac{1}{2}$$ and define $$M_n := 1 + sum_{j=1}^n X_j.$$ As $mathbb{E}(X_j)=0$ for all $j in mathbb{N}$, the independence of the random variables $(X_j)_{j in mathbb{N}}$ implies that $(M_n)_{n in mathbb{N}}$ is a martingale. Moreover, the process is clearly nonconstant. As $$sum_{j=1}^{infty} |X_j| = sum_{j=1}^{infty} 2^{-j} = 1$$ we have $$M_n = 1+ sum_{j=1}^n X_j geq 1- sum_{j=1}^{infty} |X_j| geq 0,$$ i.e. $M_n geq 0$ for all $n in mathbb{N}$.
Example 2: Let $(xi_j)_{j in mathbb{N}}$ be independent identically distributed random variables such that $$mathbb{P}(xi_j = 3/2) = frac{1}{2} quad text{and} quad mathbb{P}(xi_j = 1/2) = frac{1}{2}.$$ Since $mathbb{E}(xi_j)=1$ for all $j geq 1$, it follows from the independence of the random variables that $$M_n := prod_{j=1}^n xi_j$$ is a martingale. Clearly, $(M_n)_{n in mathbb{N}}$ is non-constant and non-negative.
Remark: It is not difficult to show that that the only non-negative martingale $(M_n)_{n in mathbb{N}}$ with $M_0=0$ is the trivial martingale.
answered Nov 26 '18 at 20:32
saz
78.1k758122
add a comment |
Another standard example is to take simple random walk started at $1$ and stopped when it hits 0. That is, let $S_n = 1 + xi_1 + xi_2 + dots + xi_n$ where $xi_i$ are iid $pm1$, $T = min{n : S_n = 0}$, and $X_n = S_{n wedge T}$.
This is nonconstant in the sense that none of the $X_n$, $n ge 1$, is a constant random variable. But it is "eventually constant" in the sense that for almost every $omega$, there is an $N(omega)$ such that $X_n(omega) = 0$ for all $n ge N(omega)$. In other words, almost surely, it is zero from some time onwards, but that time is random and unbounded.
answered Nov 26 '18 at 20:38
Nate Eldredge
62.2k681169
add a comment |
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2 Answers
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2 Answers
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Example 1: Let $(X_j)_{j in mathbb{N}}$ be a sequence of independent random variables such that $$mathbb{P}(X_j = - 2^{-j})=frac{1}{2} quad text{and} quad mathbb{P}(X_j=2^{-j})=frac{1}{2}$$ and define $$M_n := 1 + sum_{j=1}^n X_j.$$ As $mathbb{E}(X_j)=0$ for all $j in mathbb{N}$, the independence of the random variables $(X_j)_{j in mathbb{N}}$ implies that $(M_n)_{n in mathbb{N}}$ is a martingale. Moreover, the process is clearly nonconstant. As $$sum_{j=1}^{infty} |X_j| = sum_{j=1}^{infty} 2^{-j} = 1$$ we have $$M_n = 1+ sum_{j=1}^n X_j geq 1- sum_{j=1}^{infty} |X_j| geq 0,$$ i.e. $M_n geq 0$ for all $n in mathbb{N}$.
Example 2: Let $(xi_j)_{j in mathbb{N}}$ be independent identically distributed random variables such that $$mathbb{P}(xi_j = 3/2) = frac{1}{2} quad text{and} quad mathbb{P}(xi_j = 1/2) = frac{1}{2}.$$ Since $mathbb{E}(xi_j)=1$ for all $j geq 1$, it follows from the independence of the random variables that $$M_n := prod_{j=1}^n xi_j$$ is a martingale. Clearly, $(M_n)_{n in mathbb{N}}$ is non-constant and non-negative.
Remark: It is not difficult to show that that the only non-negative martingale $(M_n)_{n in mathbb{N}}$ with $M_0=0$ is the trivial martingale.
answered Nov 26 '18 at 20:32
saz
78.1k758122
add a comment |
Example 1: Let $(X_j)_{j in mathbb{N}}$ be a sequence of independent random variables such that $$mathbb{P}(X_j = - 2^{-j})=frac{1}{2} quad text{and} quad mathbb{P}(X_j=2^{-j})=frac{1}{2}$$ and define $$M_n := 1 + sum_{j=1}^n X_j.$$ As $mathbb{E}(X_j)=0$ for all $j in mathbb{N}$, the independence of the random variables $(X_j)_{j in mathbb{N}}$ implies that $(M_n)_{n in mathbb{N}}$ is a martingale. Moreover, the process is clearly nonconstant. As $$sum_{j=1}^{infty} |X_j| = sum_{j=1}^{infty} 2^{-j} = 1$$ we have $$M_n = 1+ sum_{j=1}^n X_j geq 1- sum_{j=1}^{infty} |X_j| geq 0,$$ i.e. $M_n geq 0$ for all $n in mathbb{N}$.
Example 2: Let $(xi_j)_{j in mathbb{N}}$ be independent identically distributed random variables such that $$mathbb{P}(xi_j = 3/2) = frac{1}{2} quad text{and} quad mathbb{P}(xi_j = 1/2) = frac{1}{2}.$$ Since $mathbb{E}(xi_j)=1$ for all $j geq 1$, it follows from the independence of the random variables that $$M_n := prod_{j=1}^n xi_j$$ is a martingale. Clearly, $(M_n)_{n in mathbb{N}}$ is non-constant and non-negative.
Remark: It is not difficult to show that that the only non-negative martingale $(M_n)_{n in mathbb{N}}$ with $M_0=0$ is the trivial martingale.
answered Nov 26 '18 at 20:32
saz
78.1k758122
add a comment |
Example 1: Let $(X_j)_{j in mathbb{N}}$ be a sequence of independent random variables such that $$mathbb{P}(X_j = - 2^{-j})=frac{1}{2} quad text{and} quad mathbb{P}(X_j=2^{-j})=frac{1}{2}$$ and define $$M_n := 1 + sum_{j=1}^n X_j.$$ As $mathbb{E}(X_j)=0$ for all $j in mathbb{N}$, the independence of the random variables $(X_j)_{j in mathbb{N}}$ implies that $(M_n)_{n in mathbb{N}}$ is a martingale. Moreover, the process is clearly nonconstant. As $$sum_{j=1}^{infty} |X_j| = sum_{j=1}^{infty} 2^{-j} = 1$$ we have $$M_n = 1+ sum_{j=1}^n X_j geq 1- sum_{j=1}^{infty} |X_j| geq 0,$$ i.e. $M_n geq 0$ for all $n in mathbb{N}$.
Example 2: Let $(xi_j)_{j in mathbb{N}}$ be independent identically distributed random variables such that $$mathbb{P}(xi_j = 3/2) = frac{1}{2} quad text{and} quad mathbb{P}(xi_j = 1/2) = frac{1}{2}.$$ Since $mathbb{E}(xi_j)=1$ for all $j geq 1$, it follows from the independence of the random variables that $$M_n := prod_{j=1}^n xi_j$$ is a martingale. Clearly, $(M_n)_{n in mathbb{N}}$ is non-constant and non-negative.
Remark: It is not difficult to show that that the only non-negative martingale $(M_n)_{n in mathbb{N}}$ with $M_0=0$ is the trivial martingale.
answered Nov 26 '18 at 20:32
saz
78.1k758122
Example 1: Let $(X_j)_{j in mathbb{N}}$ be a sequence of independent random variables such that $$mathbb{P}(X_j = - 2^{-j})=frac{1}{2} quad text{and} quad mathbb{P}(X_j=2^{-j})=frac{1}{2}$$ and define $$M_n := 1 + sum_{j=1}^n X_j.$$ As $mathbb{E}(X_j)=0$ for all $j in mathbb{N}$, the independence of the random variables $(X_j)_{j in mathbb{N}}$ implies that $(M_n)_{n in mathbb{N}}$ is a martingale. Moreover, the process is clearly nonconstant. As $$sum_{j=1}^{infty} |X_j| = sum_{j=1}^{infty} 2^{-j} = 1$$ we have $$M_n = 1+ sum_{j=1}^n X_j geq 1- sum_{j=1}^{infty} |X_j| geq 0,$$ i.e. $M_n geq 0$ for all $n in mathbb{N}$.
Example 2: Let $(xi_j)_{j in mathbb{N}}$ be independent identically distributed random variables such that $$mathbb{P}(xi_j = 3/2) = frac{1}{2} quad text{and} quad mathbb{P}(xi_j = 1/2) = frac{1}{2}.$$ Since $mathbb{E}(xi_j)=1$ for all $j geq 1$, it follows from the independence of the random variables that $$M_n := prod_{j=1}^n xi_j$$ is a martingale. Clearly, $(M_n)_{n in mathbb{N}}$ is non-constant and non-negative.
Remark: It is not difficult to show that that the only non-negative martingale $(M_n)_{n in mathbb{N}}$ with $M_0=0$ is the trivial martingale.
answered Nov 26 '18 at 20:32
saz
78.1k758122
edited Nov 27 '18 at 10:48
answered Nov 26 '18 at 20:32
saz
78.1k758122
answered Nov 26 '18 at 20:32
saz
78.1k758122
answered Nov 26 '18 at 20:32
saz
78.1k758122
78.1k758122
add a comment |
add a comment |
Another standard example is to take simple random walk started at $1$ and stopped when it hits 0. That is, let $S_n = 1 + xi_1 + xi_2 + dots + xi_n$ where $xi_i$ are iid $pm1$, $T = min{n : S_n = 0}$, and $X_n = S_{n wedge T}$.
This is nonconstant in the sense that none of the $X_n$, $n ge 1$, is a constant random variable. But it is "eventually constant" in the sense that for almost every $omega$, there is an $N(omega)$ such that $X_n(omega) = 0$ for all $n ge N(omega)$. In other words, almost surely, it is zero from some time onwards, but that time is random and unbounded.
answered Nov 26 '18 at 20:38
Nate Eldredge
62.2k681169
add a comment |
Another standard example is to take simple random walk started at $1$ and stopped when it hits 0. That is, let $S_n = 1 + xi_1 + xi_2 + dots + xi_n$ where $xi_i$ are iid $pm1$, $T = min{n : S_n = 0}$, and $X_n = S_{n wedge T}$.
This is nonconstant in the sense that none of the $X_n$, $n ge 1$, is a constant random variable. But it is "eventually constant" in the sense that for almost every $omega$, there is an $N(omega)$ such that $X_n(omega) = 0$ for all $n ge N(omega)$. In other words, almost surely, it is zero from some time onwards, but that time is random and unbounded.
answered Nov 26 '18 at 20:38
Nate Eldredge
62.2k681169
add a comment |
Another standard example is to take simple random walk started at $1$ and stopped when it hits 0. That is, let $S_n = 1 + xi_1 + xi_2 + dots + xi_n$ where $xi_i$ are iid $pm1$, $T = min{n : S_n = 0}$, and $X_n = S_{n wedge T}$.
This is nonconstant in the sense that none of the $X_n$, $n ge 1$, is a constant random variable. But it is "eventually constant" in the sense that for almost every $omega$, there is an $N(omega)$ such that $X_n(omega) = 0$ for all $n ge N(omega)$. In other words, almost surely, it is zero from some time onwards, but that time is random and unbounded.
answered Nov 26 '18 at 20:38
Nate Eldredge
62.2k681169
Another standard example is to take simple random walk started at $1$ and stopped when it hits 0. That is, let $S_n = 1 + xi_1 + xi_2 + dots + xi_n$ where $xi_i$ are iid $pm1$, $T = min{n : S_n = 0}$, and $X_n = S_{n wedge T}$.
This is nonconstant in the sense that none of the $X_n$, $n ge 1$, is a constant random variable. But it is "eventually constant" in the sense that for almost every $omega$, there is an $N(omega)$ such that $X_n(omega) = 0$ for all $n ge N(omega)$. In other words, almost surely, it is zero from some time onwards, but that time is random and unbounded.
answered Nov 26 '18 at 20:38
Nate Eldredge
62.2k681169
answered Nov 26 '18 at 20:38
Nate Eldredge
62.2k681169
answered Nov 26 '18 at 20:38
Nate Eldredge
62.2k681169
answered Nov 26 '18 at 20:38
Nate Eldredge
62.2k681169
62.2k681169
add a comment |
add a comment |
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