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Limit of $Y_n = left(prodlimits_{i=1}^{n} X_iright)^{1/n}$ for $ntoinfty$.

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1

$begingroup$

Let $(X_n)_{ngeq1}$ be a sequence of i.i.d. random variables such that $P(X_1 = 1) = P(X_1 = 2) = frac{1}{2}$. Let $(Y_n)_{ngeq1}$ be defined as

$$Y_n = left(prod_{i=1}^{n} X_iright)^{1/n}$$ for all $n geq 1$.

Show that there exists (and determine it) a real number $a$ such that $Y_n to a$ almost surely as $n to infty$.

probability-theory

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edited Mar 10 at 15:07

Mars Plastic

1,451121

asked Mar 10 at 14:13

Jasper Jasper

607

$endgroup$

add a comment | 

1

$begingroup$

Let $(X_n)_{ngeq1}$ be a sequence of i.i.d. random variables such that $P(X_1 = 1) = P(X_1 = 2) = frac{1}{2}$. Let $(Y_n)_{ngeq1}$ be defined as

$$Y_n = left(prod_{i=1}^{n} X_iright)^{1/n}$$ for all $n geq 1$.

Show that there exists (and determine it) a real number $a$ such that $Y_n to a$ almost surely as $n to infty$.

probability-theory

share|cite|improve this question

edited Mar 10 at 15:07

Mars Plastic

1,451121

asked Mar 10 at 14:13

Jasper Jasper

607

$endgroup$

add a comment | 

$begingroup$

Let $(X_n)_{ngeq1}$ be a sequence of i.i.d. random variables such that $P(X_1 = 1) = P(X_1 = 2) = frac{1}{2}$. Let $(Y_n)_{ngeq1}$ be defined as

$$Y_n = left(prod_{i=1}^{n} X_iright)^{1/n}$$ for all $n geq 1$.

Show that there exists (and determine it) a real number $a$ such that $Y_n to a$ almost surely as $n to infty$.

probability-theory

share|cite|improve this question

edited Mar 10 at 15:07

Mars Plastic

1,451121

asked Mar 10 at 14:13

Jasper Jasper

607

$endgroup$

share|cite|improve this question

edited Mar 10 at 15:07

Mars Plastic

1,451121

asked Mar 10 at 14:13

Jasper Jasper

607

share|cite|improve this question

edited Mar 10 at 15:07

Mars Plastic

1,451121

asked Mar 10 at 14:13

Jasper Jasper

607

edited Mar 10 at 15:07

Mars Plastic

1,451121

edited Mar 10 at 15:07

Mars Plastic

1,451121

asked Mar 10 at 14:13

Jasper Jasper

607

asked Mar 10 at 14:13

Jasper Jasper

607

3 Answers
3

active

oldest

votes

1

$begingroup$

$$log(Y_n) = frac{sum_{i=1}^{n} log(X_i)}{n}$$

With strong law of large numbers $log(Y_n)$ converges almost surely to $E(log(X_1))$.
I let you do the computation of the expected value.

share|cite|improve this answer

answered Mar 10 at 14:32

JenniferJennifer

8,60521837

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $mathbb{E}$(log($X_1$)) = $frac{1}{2}$(log(1) + log(2), but I don't see why log($Y_n$) converges almost surely to $mathbb{E}$(log($X_1$)).
  $endgroup$
  – Jasper
  Mar 10 at 20:59
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The random varaibles $log(X_1), dots ,log(X_n)$ are iid so you can apply strong law of large numbers thus $log(Y_n) = frac{sum_{i=1}^{n} log(X_i)}{n}$ convers almost surely to $E(log(X_1))$
  $endgroup$
  – Jennifer
  Mar 10 at 21:14
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Hints:

• Observe that: $$Y_{n}=2^{frac{1}{n}sum_{i=1}^{n}mathbf{1}_{left{ X_{i}=2right} }}$$
  where the $mathbf{1}_{left{ X_{i}=2right} }$ are iid with Bernoulli
  distribution that has parameter $frac{1}{2}$.




• If $a_n$ converges then so does $2^{a_n}$.



• SLLN

share|cite|improve this answer

edited Mar 10 at 14:35

answered Mar 10 at 14:27

drhabdrhab

103k545136

$endgroup$

add a comment | 

0

$begingroup$

Hint: Look at $log (Y_n)$ and use the law of large numbers for $(log(X_n))_{nge 1}$.

share|cite|improve this answer

answered Mar 10 at 14:32

Mars PlasticMars Plastic

1,451121

$endgroup$

add a comment | 

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1

$begingroup$

$$log(Y_n) = frac{sum_{i=1}^{n} log(X_i)}{n}$$

With strong law of large numbers $log(Y_n)$ converges almost surely to $E(log(X_1))$.
I let you do the computation of the expected value.

share|cite|improve this answer

answered Mar 10 at 14:32

JenniferJennifer

8,60521837

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $mathbb{E}$(log($X_1$)) = $frac{1}{2}$(log(1) + log(2), but I don't see why log($Y_n$) converges almost surely to $mathbb{E}$(log($X_1$)).
  $endgroup$
  – Jasper
  Mar 10 at 20:59
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The random varaibles $log(X_1), dots ,log(X_n)$ are iid so you can apply strong law of large numbers thus $log(Y_n) = frac{sum_{i=1}^{n} log(X_i)}{n}$ convers almost surely to $E(log(X_1))$
  $endgroup$
  – Jennifer
  Mar 10 at 21:14
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

$$log(Y_n) = frac{sum_{i=1}^{n} log(X_i)}{n}$$

With strong law of large numbers $log(Y_n)$ converges almost surely to $E(log(X_1))$.
I let you do the computation of the expected value.

share|cite|improve this answer

answered Mar 10 at 14:32

JenniferJennifer

8,60521837

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $mathbb{E}$(log($X_1$)) = $frac{1}{2}$(log(1) + log(2), but I don't see why log($Y_n$) converges almost surely to $mathbb{E}$(log($X_1$)).
  $endgroup$
  – Jasper
  Mar 10 at 20:59
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The random varaibles $log(X_1), dots ,log(X_n)$ are iid so you can apply strong law of large numbers thus $log(Y_n) = frac{sum_{i=1}^{n} log(X_i)}{n}$ convers almost surely to $E(log(X_1))$
  $endgroup$
  – Jennifer
  Mar 10 at 21:14
  
  
  

  
  
  
  

add a comment | 

$begingroup$

$$log(Y_n) = frac{sum_{i=1}^{n} log(X_i)}{n}$$

With strong law of large numbers $log(Y_n)$ converges almost surely to $E(log(X_1))$.
I let you do the computation of the expected value.

share|cite|improve this answer

answered Mar 10 at 14:32

JenniferJennifer

8,60521837

$endgroup$

share|cite|improve this answer

answered Mar 10 at 14:32

JenniferJennifer

8,60521837

answered Mar 10 at 14:32

JenniferJennifer

8,60521837

answered Mar 10 at 14:32

JenniferJennifer

8,60521837

1

$begingroup$

Hints:

• Observe that: $$Y_{n}=2^{frac{1}{n}sum_{i=1}^{n}mathbf{1}_{left{ X_{i}=2right} }}$$
  where the $mathbf{1}_{left{ X_{i}=2right} }$ are iid with Bernoulli
  distribution that has parameter $frac{1}{2}$.




• If $a_n$ converges then so does $2^{a_n}$.



• SLLN

share|cite|improve this answer

edited Mar 10 at 14:35

answered Mar 10 at 14:27

drhabdrhab

103k545136

$endgroup$

add a comment | 

1

$begingroup$

Hints:

• Observe that: $$Y_{n}=2^{frac{1}{n}sum_{i=1}^{n}mathbf{1}_{left{ X_{i}=2right} }}$$
  where the $mathbf{1}_{left{ X_{i}=2right} }$ are iid with Bernoulli
  distribution that has parameter $frac{1}{2}$.




• If $a_n$ converges then so does $2^{a_n}$.



• SLLN

share|cite|improve this answer

edited Mar 10 at 14:35

answered Mar 10 at 14:27

drhabdrhab

103k545136

$endgroup$

add a comment | 

$begingroup$

Hints:

• Observe that: $$Y_{n}=2^{frac{1}{n}sum_{i=1}^{n}mathbf{1}_{left{ X_{i}=2right} }}$$
  where the $mathbf{1}_{left{ X_{i}=2right} }$ are iid with Bernoulli
  distribution that has parameter $frac{1}{2}$.




• If $a_n$ converges then so does $2^{a_n}$.



• SLLN

share|cite|improve this answer

edited Mar 10 at 14:35

answered Mar 10 at 14:27

drhabdrhab

103k545136

$endgroup$

share|cite|improve this answer

edited Mar 10 at 14:35

answered Mar 10 at 14:27

drhabdrhab

103k545136

answered Mar 10 at 14:27

drhabdrhab

103k545136

answered Mar 10 at 14:27

drhabdrhab

103k545136

0

$begingroup$

Hint: Look at $log (Y_n)$ and use the law of large numbers for $(log(X_n))_{nge 1}$.

share|cite|improve this answer

answered Mar 10 at 14:32

Mars PlasticMars Plastic

1,451121

$endgroup$

add a comment | 

0

$begingroup$

Hint: Look at $log (Y_n)$ and use the law of large numbers for $(log(X_n))_{nge 1}$.

share|cite|improve this answer

answered Mar 10 at 14:32

Mars PlasticMars Plastic

1,451121

$endgroup$

add a comment | 

$begingroup$

Hint: Look at $log (Y_n)$ and use the law of large numbers for $(log(X_n))_{nge 1}$.

share|cite|improve this answer

answered Mar 10 at 14:32

Mars PlasticMars Plastic

1,451121

$endgroup$

share|cite|improve this answer

answered Mar 10 at 14:32

Mars PlasticMars Plastic

1,451121

answered Mar 10 at 14:32

Mars PlasticMars Plastic

1,451121

answered Mar 10 at 14:32

Mars PlasticMars Plastic

1,451121