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Convergence of series of normally distributed random variables

Convergence of series of normally distributed random variables

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3

1

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$X_1, X_2,...$ are independent and each $X_n$ has normal distribution $N(0,a^n)$ for $a>0$.
I need to find probability that the series $$sum_n=1^infty X_n$$ converges.
How do I approach this problem?

probability-theory normal-distribution probability-limit-theorems

share|cite|improve this question

asked 8 hours ago

Maja BlumensteinMaja Blumenstein

4581010 bronze badges

$endgroup$

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  It looks like an exercise in both Borel Cantelli Lemmas. It's probably easier to think of it as $sum_n=1 sqrta^n Z_n$ where each $Z_n$ is standard normal.
  $endgroup$
  – WoolierThanThou
  8 hours ago
  

  
  
  
  


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  For instance, for $ageq 1,$ you should be able to prove that there is some $varepsilon>0$ such that $Z_n>varepsilon$ infinitely often with probability $1$.
  $endgroup$
  – WoolierThanThou
  8 hours ago
  

  
  
  
  


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  @WoolierThanThou Care to clarify? B-C lemma involves sum $sum P(X_n)$, how do I get from this to convergence of series of rv's?
  $endgroup$
  – Maja Blumenstein
  7 hours ago
  

  
  
  
  


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  @WoolierThanThou From your second comment I see how I can use it to prove divergence, but how to prove convergence?
  $endgroup$
  – Maja Blumenstein
  7 hours ago
  

  
  
  
  


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  Well, you can just use Kavi's proof, but you can also end up estimating, say, $mathbbP(|Z_n|geq a^-n/4)$ and apply, say, the Chebyshev inequality, to get that these probabilities are summable. Hence, $sqrta^n |Z_n|leq a^n/4$ eventually.
  $endgroup$
  – WoolierThanThou
  7 hours ago
  
  

  
  
  
  

add a comment | 

3

1

$begingroup$

$X_1, X_2,...$ are independent and each $X_n$ has normal distribution $N(0,a^n)$ for $a>0$.
I need to find probability that the series $$sum_n=1^infty X_n$$ converges.
How do I approach this problem?

probability-theory normal-distribution probability-limit-theorems

share|cite|improve this question

asked 8 hours ago

Maja BlumensteinMaja Blumenstein

4581010 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It looks like an exercise in both Borel Cantelli Lemmas. It's probably easier to think of it as $sum_n=1 sqrta^n Z_n$ where each $Z_n$ is standard normal.
  $endgroup$
  – WoolierThanThou
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For instance, for $ageq 1,$ you should be able to prove that there is some $varepsilon>0$ such that $Z_n>varepsilon$ infinitely often with probability $1$.
  $endgroup$
  – WoolierThanThou
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @WoolierThanThou Care to clarify? B-C lemma involves sum $sum P(X_n)$, how do I get from this to convergence of series of rv's?
  $endgroup$
  – Maja Blumenstein
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @WoolierThanThou From your second comment I see how I can use it to prove divergence, but how to prove convergence?
  $endgroup$
  – Maja Blumenstein
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, you can just use Kavi's proof, but you can also end up estimating, say, $mathbbP(|Z_n|geq a^-n/4)$ and apply, say, the Chebyshev inequality, to get that these probabilities are summable. Hence, $sqrta^n |Z_n|leq a^n/4$ eventually.
  $endgroup$
  – WoolierThanThou
  7 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

$X_1, X_2,...$ are independent and each $X_n$ has normal distribution $N(0,a^n)$ for $a>0$.
I need to find probability that the series $$sum_n=1^infty X_n$$ converges.
How do I approach this problem?

probability-theory normal-distribution probability-limit-theorems

share|cite|improve this question

asked 8 hours ago

Maja BlumensteinMaja Blumenstein

4581010 bronze badges

$endgroup$

share|cite|improve this question

asked 8 hours ago

Maja BlumensteinMaja Blumenstein

4581010 bronze badges

share|cite|improve this question

asked 8 hours ago

Maja BlumensteinMaja Blumenstein

4581010 bronze badges

asked 8 hours ago

Maja BlumensteinMaja Blumenstein

4581010 bronze badges

asked 8 hours ago

Maja BlumensteinMaja Blumenstein

4581010 bronze badges

2 Answers
2

active

oldest

votes

5

$begingroup$

Hints: the series converges with probability $0$ or $1$ by $0-1$ law.

If $a <1$ then $sum E|X_n| <infty$ so the series converges with probability $1$.

If $a=1$ the series converges with probability $0$. In fact no i.i.d. series converges except when the terms are all $0$ with probability $1$.

When $a>1$ it is easy to see that $|X_n|$ tends to $infty$ in probability which implies that the series cannot converges with probability $1$ and hence it converges with probability $0$.

You can also solve this problem using Kolmogorov's 3-series theorem.

share|cite|improve this answer

answered 8 hours ago

Kavi Rama MurthyKavi Rama Murthy

111k66 gold badges4646 silver badges8888 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "In fact no i.i.d. series converges except when the terms are all 0 with probability 1." Is that "In fact no i.i.d. series converges (except when the terms are all 0) with probability 1." or "In fact no i.i.d. series converges (except when the terms are all 0 with probability 1)."?
  $endgroup$
  – Acccumulation
  1 min ago
  

  
  
  
  

add a comment | 

2

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The event $(sum_ngeq 1 X_n text converges)$ is a tail event. By Kolmogorov's $0-1$ law, it has probability $0$ or $1$.

When $a<1$, $sum_ngeq 1 E(X_n^2)<infty$ and by Kolmogorov's two-series theorem, $sum_ngeq 1 X_n$ converges almost surely.

When $ageq 1$, $S_n:=sum_k= 1^n X_k$ has distribution $mathcal N(0,n)$ or $mathcal N(0,afraca^n-1a-1$). Suppose for the sake of contradiction that $S_n$ converges a.s to some $S$. Then $S_nto S$ in $L^2$ (and $Sin L^2$), hence $E(S_n^2)to E(S^2)$, but $E(S_n^2)$ diverges, a contradiction. So $$P(sum_ngeq 1 X_n text converges) = 0$$

share|cite|improve this answer

answered 7 hours ago

Gabriel RomonGabriel Romon

19.5k55 gold badges3737 silver badges8888 bronze badges

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add a comment | 

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5

$begingroup$

Hints: the series converges with probability $0$ or $1$ by $0-1$ law.

If $a <1$ then $sum E|X_n| <infty$ so the series converges with probability $1$.

If $a=1$ the series converges with probability $0$. In fact no i.i.d. series converges except when the terms are all $0$ with probability $1$.

When $a>1$ it is easy to see that $|X_n|$ tends to $infty$ in probability which implies that the series cannot converges with probability $1$ and hence it converges with probability $0$.

You can also solve this problem using Kolmogorov's 3-series theorem.

share|cite|improve this answer

answered 8 hours ago

Kavi Rama MurthyKavi Rama Murthy

111k66 gold badges4646 silver badges8888 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "In fact no i.i.d. series converges except when the terms are all 0 with probability 1." Is that "In fact no i.i.d. series converges (except when the terms are all 0) with probability 1." or "In fact no i.i.d. series converges (except when the terms are all 0 with probability 1)."?
  $endgroup$
  – Acccumulation
  1 min ago
  

  
  
  
  

add a comment | 

5

$begingroup$

Hints: the series converges with probability $0$ or $1$ by $0-1$ law.

If $a <1$ then $sum E|X_n| <infty$ so the series converges with probability $1$.

If $a=1$ the series converges with probability $0$. In fact no i.i.d. series converges except when the terms are all $0$ with probability $1$.

When $a>1$ it is easy to see that $|X_n|$ tends to $infty$ in probability which implies that the series cannot converges with probability $1$ and hence it converges with probability $0$.

You can also solve this problem using Kolmogorov's 3-series theorem.

share|cite|improve this answer

answered 8 hours ago

Kavi Rama MurthyKavi Rama Murthy

111k66 gold badges4646 silver badges8888 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "In fact no i.i.d. series converges except when the terms are all 0 with probability 1." Is that "In fact no i.i.d. series converges (except when the terms are all 0) with probability 1." or "In fact no i.i.d. series converges (except when the terms are all 0 with probability 1)."?
  $endgroup$
  – Acccumulation
  1 min ago
  

  
  
  
  

add a comment | 

$begingroup$

Hints: the series converges with probability $0$ or $1$ by $0-1$ law.

If $a <1$ then $sum E|X_n| <infty$ so the series converges with probability $1$.

If $a=1$ the series converges with probability $0$. In fact no i.i.d. series converges except when the terms are all $0$ with probability $1$.

When $a>1$ it is easy to see that $|X_n|$ tends to $infty$ in probability which implies that the series cannot converges with probability $1$ and hence it converges with probability $0$.

You can also solve this problem using Kolmogorov's 3-series theorem.

share|cite|improve this answer

answered 8 hours ago

Kavi Rama MurthyKavi Rama Murthy

111k66 gold badges4646 silver badges8888 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Kavi Rama MurthyKavi Rama Murthy

111k66 gold badges4646 silver badges8888 bronze badges

answered 8 hours ago

Kavi Rama MurthyKavi Rama Murthy

111k66 gold badges4646 silver badges8888 bronze badges

answered 8 hours ago

Kavi Rama MurthyKavi Rama Murthy

111k66 gold badges4646 silver badges8888 bronze badges

2

$begingroup$

The event $(sum_ngeq 1 X_n text converges)$ is a tail event. By Kolmogorov's $0-1$ law, it has probability $0$ or $1$.

When $a<1$, $sum_ngeq 1 E(X_n^2)<infty$ and by Kolmogorov's two-series theorem, $sum_ngeq 1 X_n$ converges almost surely.

When $ageq 1$, $S_n:=sum_k= 1^n X_k$ has distribution $mathcal N(0,n)$ or $mathcal N(0,afraca^n-1a-1$). Suppose for the sake of contradiction that $S_n$ converges a.s to some $S$. Then $S_nto S$ in $L^2$ (and $Sin L^2$), hence $E(S_n^2)to E(S^2)$, but $E(S_n^2)$ diverges, a contradiction. So $$P(sum_ngeq 1 X_n text converges) = 0$$

share|cite|improve this answer

answered 7 hours ago

Gabriel RomonGabriel Romon

19.5k55 gold badges3737 silver badges8888 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

The event $(sum_ngeq 1 X_n text converges)$ is a tail event. By Kolmogorov's $0-1$ law, it has probability $0$ or $1$.

When $a<1$, $sum_ngeq 1 E(X_n^2)<infty$ and by Kolmogorov's two-series theorem, $sum_ngeq 1 X_n$ converges almost surely.

When $ageq 1$, $S_n:=sum_k= 1^n X_k$ has distribution $mathcal N(0,n)$ or $mathcal N(0,afraca^n-1a-1$). Suppose for the sake of contradiction that $S_n$ converges a.s to some $S$. Then $S_nto S$ in $L^2$ (and $Sin L^2$), hence $E(S_n^2)to E(S^2)$, but $E(S_n^2)$ diverges, a contradiction. So $$P(sum_ngeq 1 X_n text converges) = 0$$

share|cite|improve this answer

answered 7 hours ago

Gabriel RomonGabriel Romon

19.5k55 gold badges3737 silver badges8888 bronze badges

$endgroup$

add a comment | 

$begingroup$

The event $(sum_ngeq 1 X_n text converges)$ is a tail event. By Kolmogorov's $0-1$ law, it has probability $0$ or $1$.

When $a<1$, $sum_ngeq 1 E(X_n^2)<infty$ and by Kolmogorov's two-series theorem, $sum_ngeq 1 X_n$ converges almost surely.

When $ageq 1$, $S_n:=sum_k= 1^n X_k$ has distribution $mathcal N(0,n)$ or $mathcal N(0,afraca^n-1a-1$). Suppose for the sake of contradiction that $S_n$ converges a.s to some $S$. Then $S_nto S$ in $L^2$ (and $Sin L^2$), hence $E(S_n^2)to E(S^2)$, but $E(S_n^2)$ diverges, a contradiction. So $$P(sum_ngeq 1 X_n text converges) = 0$$

share|cite|improve this answer

answered 7 hours ago

Gabriel RomonGabriel Romon

19.5k55 gold badges3737 silver badges8888 bronze badges

$endgroup$

share|cite|improve this answer

answered 7 hours ago

Gabriel RomonGabriel Romon

19.5k55 gold badges3737 silver badges8888 bronze badges

answered 7 hours ago

Gabriel RomonGabriel Romon

19.5k55 gold badges3737 silver badges8888 bronze badges

answered 7 hours ago

Gabriel RomonGabriel Romon

19.5k55 gold badges3737 silver badges8888 bronze badges