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What does mean : “a.s. convergence doesn't come from a metric”?
In the Durrett : Probability theory and example, (4th edition), remark page 65 it says : Since there are sequence of r.v. that converge in probability but not a.s., it follow from theorem 2.3.3 that a.s. convergence does not come from a metric or even from a topology.
I'm not really sure to really understand what it mean. Could someone give explanation ? For information : theorem 2.3.3 say
Let $y_n$ a sequence of a topological space. If every subsequence $y_{m(n)}$ has a further subsequence $y_{m(n_k)}$ that converge to $y$, then $y_nto y$.
probability
asked Nov 20 '18 at 10:02
user617786
234
add a comment |
In the Durrett : Probability theory and example, (4th edition), remark page 65 it says : Since there are sequence of r.v. that converge in probability but not a.s., it follow from theorem 2.3.3 that a.s. convergence does not come from a metric or even from a topology.
I'm not really sure to really understand what it mean. Could someone give explanation ? For information : theorem 2.3.3 say
Let $y_n$ a sequence of a topological space. If every subsequence $y_{m(n)}$ has a further subsequence $y_{m(n_k)}$ that converge to $y$, then $y_nto y$.
probability
asked Nov 20 '18 at 10:02
user617786
234
add a comment |
In the Durrett : Probability theory and example, (4th edition), remark page 65 it says : Since there are sequence of r.v. that converge in probability but not a.s., it follow from theorem 2.3.3 that a.s. convergence does not come from a metric or even from a topology.
I'm not really sure to really understand what it mean. Could someone give explanation ? For information : theorem 2.3.3 say
Let $y_n$ a sequence of a topological space. If every subsequence $y_{m(n)}$ has a further subsequence $y_{m(n_k)}$ that converge to $y$, then $y_nto y$.
probability
asked Nov 20 '18 at 10:02
user617786
234
In the Durrett : Probability theory and example, (4th edition), remark page 65 it says : Since there are sequence of r.v. that converge in probability but not a.s., it follow from theorem 2.3.3 that a.s. convergence does not come from a metric or even from a topology.
I'm not really sure to really understand what it mean. Could someone give explanation ? For information : theorem 2.3.3 say
Let $y_n$ a sequence of a topological space. If every subsequence $y_{m(n)}$ has a further subsequence $y_{m(n_k)}$ that converge to $y$, then $y_nto y$.
probability
probability
asked Nov 20 '18 at 10:02
user617786
234
asked Nov 20 '18 at 10:02
user617786
234
asked Nov 20 '18 at 10:02
user617786
234
asked Nov 20 '18 at 10:02
user617786
234
asked Nov 20 '18 at 10:02
user617786
234
234
add a comment |
add a comment |
1 Answer
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Suppose ${X_n}$ converges in probability to $X$ but not almost surely. Take any subsequence ${X_{n_k}}$. Since this sequence converges in probability it has a subsequence ${X_{n_{k_j}}}$ which converges almost surely. Application of Theorem 2.3.3 shows that ${X_n}$ must converge almost surely, but it doesn't.
answered Nov 20 '18 at 10:12
Kavi Rama Murthy
50.4k31854
add a comment |
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1 Answer
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1 Answer
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Suppose ${X_n}$ converges in probability to $X$ but not almost surely. Take any subsequence ${X_{n_k}}$. Since this sequence converges in probability it has a subsequence ${X_{n_{k_j}}}$ which converges almost surely. Application of Theorem 2.3.3 shows that ${X_n}$ must converge almost surely, but it doesn't.
answered Nov 20 '18 at 10:12
Kavi Rama Murthy
50.4k31854
add a comment |
Suppose ${X_n}$ converges in probability to $X$ but not almost surely. Take any subsequence ${X_{n_k}}$. Since this sequence converges in probability it has a subsequence ${X_{n_{k_j}}}$ which converges almost surely. Application of Theorem 2.3.3 shows that ${X_n}$ must converge almost surely, but it doesn't.
answered Nov 20 '18 at 10:12
Kavi Rama Murthy
50.4k31854
add a comment |
Suppose ${X_n}$ converges in probability to $X$ but not almost surely. Take any subsequence ${X_{n_k}}$. Since this sequence converges in probability it has a subsequence ${X_{n_{k_j}}}$ which converges almost surely. Application of Theorem 2.3.3 shows that ${X_n}$ must converge almost surely, but it doesn't.
answered Nov 20 '18 at 10:12
Kavi Rama Murthy
50.4k31854
Suppose ${X_n}$ converges in probability to $X$ but not almost surely. Take any subsequence ${X_{n_k}}$. Since this sequence converges in probability it has a subsequence ${X_{n_{k_j}}}$ which converges almost surely. Application of Theorem 2.3.3 shows that ${X_n}$ must converge almost surely, but it doesn't.
answered Nov 20 '18 at 10:12
Kavi Rama Murthy
50.4k31854
answered Nov 20 '18 at 10:12
Kavi Rama Murthy
50.4k31854
answered Nov 20 '18 at 10:12
Kavi Rama Murthy
50.4k31854
answered Nov 20 '18 at 10:12
Kavi Rama Murthy
50.4k31854
50.4k31854
add a comment |
add a comment |
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