LLN and CLT on a sum with a random number of terms

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Consider a random sequence $ (N_n)_n geq 1 subset mathbbN^* $ such that $ mathbbP( lim N_n = infty ) = 1 $

Let $ (X_n)_n geq 1 $ be an iid sequence with $ m = mathbbE[X_1] $ and $ sigma^2 = mathbbV[X_1] $

We define $ Z_n = sqrtfracnsigma^2( overlineX_n - m ) $

$ bullet $ Show that $ ( overlineX_N_n)_n geq 1 rightarrow m : $ almost surely

$ bullet $ Show that $(Z_N_n) $ converges in distribution towards $ mathcalN(0,1) $

---

How could I justify neatly this exercise?

I've tried writing a few things ,

Let $ A = lim N_n = infty $ , $ forall omega in A $, we get : $ frac1N_n( omega) sum_k=1^N_n( omega) X_k (omega) rightarrow m $ by LLN ?

With the CLT we get $ Z_n rightarrow mathcalN(0,1) $ and by convergence of measures it follows that

For $ epsilon > 0 $

So $ exists K_1 geq 1, forall n geq K_1, | mathbbE[f(Z_n)] - mathbbE[f(Y)] | leq epsilon/2 : : : $ with f continuous and bounded

And as $ N_n rightarrow infty $ in probability this means that

$ exists K_2 > 0 geq 1, forall n geq K_2, mathbbP(N_n < K_1) leq epsilon/ f(2M) $

So, $ forall n leq K_2, $ $ | mathbbE[f(Z_N_n) mathbb1_ (N_n < K_1) cup ( (N_n geq K_1) ] - mathbbE[f(Y)] | leq epsilon /2 + | mathbbE[f(Z_N_n) mathbb1( N_n < K_1) ] $ ..... ?

What would be a correct way of writing all this? Thanks in advance

probability-theory stochastic-processes probability-limit-theorems

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edited Jul 30 at 6:55

asked Jul 29 at 20:39

Psylex

568

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean exactly by a "random sequence $N_n$"?
  – Math1000
  Aug 1 at 8:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Math1000 $ N_n $ is a sequence of integer valued random variables (measurable functions). A discrete time stochastic process.
  – Psylex
  Aug 6 at 9:47
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

Consider a random sequence $ (N_n)_n geq 1 subset mathbbN^* $ such that $ mathbbP( lim N_n = infty ) = 1 $

Let $ (X_n)_n geq 1 $ be an iid sequence with $ m = mathbbE[X_1] $ and $ sigma^2 = mathbbV[X_1] $

We define $ Z_n = sqrtfracnsigma^2( overlineX_n - m ) $

$ bullet $ Show that $ ( overlineX_N_n)_n geq 1 rightarrow m : $ almost surely

$ bullet $ Show that $(Z_N_n) $ converges in distribution towards $ mathcalN(0,1) $

---

How could I justify neatly this exercise?

I've tried writing a few things ,

Let $ A = lim N_n = infty $ , $ forall omega in A $, we get : $ frac1N_n( omega) sum_k=1^N_n( omega) X_k (omega) rightarrow m $ by LLN ?

With the CLT we get $ Z_n rightarrow mathcalN(0,1) $ and by convergence of measures it follows that

For $ epsilon > 0 $

So $ exists K_1 geq 1, forall n geq K_1, | mathbbE[f(Z_n)] - mathbbE[f(Y)] | leq epsilon/2 : : : $ with f continuous and bounded

And as $ N_n rightarrow infty $ in probability this means that

$ exists K_2 > 0 geq 1, forall n geq K_2, mathbbP(N_n < K_1) leq epsilon/ f(2M) $

So, $ forall n leq K_2, $ $ | mathbbE[f(Z_N_n) mathbb1_ (N_n < K_1) cup ( (N_n geq K_1) ] - mathbbE[f(Y)] | leq epsilon /2 + | mathbbE[f(Z_N_n) mathbb1( N_n < K_1) ] $ ..... ?

What would be a correct way of writing all this? Thanks in advance

probability-theory stochastic-processes probability-limit-theorems

share|cite|improve this question

edited Jul 30 at 6:55

asked Jul 29 at 20:39

Psylex

568

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean exactly by a "random sequence $N_n$"?
  – Math1000
  Aug 1 at 8:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Math1000 $ N_n $ is a sequence of integer valued random variables (measurable functions). A discrete time stochastic process.
  – Psylex
  Aug 6 at 9:47
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Consider a random sequence $ (N_n)_n geq 1 subset mathbbN^* $ such that $ mathbbP( lim N_n = infty ) = 1 $

Let $ (X_n)_n geq 1 $ be an iid sequence with $ m = mathbbE[X_1] $ and $ sigma^2 = mathbbV[X_1] $

We define $ Z_n = sqrtfracnsigma^2( overlineX_n - m ) $

$ bullet $ Show that $ ( overlineX_N_n)_n geq 1 rightarrow m : $ almost surely

$ bullet $ Show that $(Z_N_n) $ converges in distribution towards $ mathcalN(0,1) $

---

How could I justify neatly this exercise?

I've tried writing a few things ,

Let $ A = lim N_n = infty $ , $ forall omega in A $, we get : $ frac1N_n( omega) sum_k=1^N_n( omega) X_k (omega) rightarrow m $ by LLN ?

With the CLT we get $ Z_n rightarrow mathcalN(0,1) $ and by convergence of measures it follows that

For $ epsilon > 0 $

So $ exists K_1 geq 1, forall n geq K_1, | mathbbE[f(Z_n)] - mathbbE[f(Y)] | leq epsilon/2 : : : $ with f continuous and bounded

And as $ N_n rightarrow infty $ in probability this means that

$ exists K_2 > 0 geq 1, forall n geq K_2, mathbbP(N_n < K_1) leq epsilon/ f(2M) $

So, $ forall n leq K_2, $ $ | mathbbE[f(Z_N_n) mathbb1_ (N_n < K_1) cup ( (N_n geq K_1) ] - mathbbE[f(Y)] | leq epsilon /2 + | mathbbE[f(Z_N_n) mathbb1( N_n < K_1) ] $ ..... ?

What would be a correct way of writing all this? Thanks in advance

probability-theory stochastic-processes probability-limit-theorems

share|cite|improve this question

edited Jul 30 at 6:55

asked Jul 29 at 20:39

Psylex

568

Consider a random sequence $ (N_n)_n geq 1 subset mathbbN^* $ such that $ mathbbP( lim N_n = infty ) = 1 $

Let $ (X_n)_n geq 1 $ be an iid sequence with $ m = mathbbE[X_1] $ and $ sigma^2 = mathbbV[X_1] $

We define $ Z_n = sqrtfracnsigma^2( overlineX_n - m ) $

$ bullet $ Show that $ ( overlineX_N_n)_n geq 1 rightarrow m : $ almost surely

$ bullet $ Show that $(Z_N_n) $ converges in distribution towards $ mathcalN(0,1) $

---

How could I justify neatly this exercise?

I've tried writing a few things ,

Let $ A = lim N_n = infty $ , $ forall omega in A $, we get : $ frac1N_n( omega) sum_k=1^N_n( omega) X_k (omega) rightarrow m $ by LLN ?

With the CLT we get $ Z_n rightarrow mathcalN(0,1) $ and by convergence of measures it follows that

For $ epsilon > 0 $

So $ exists K_1 geq 1, forall n geq K_1, | mathbbE[f(Z_n)] - mathbbE[f(Y)] | leq epsilon/2 : : : $ with f continuous and bounded

And as $ N_n rightarrow infty $ in probability this means that

$ exists K_2 > 0 geq 1, forall n geq K_2, mathbbP(N_n < K_1) leq epsilon/ f(2M) $

So, $ forall n leq K_2, $ $ | mathbbE[f(Z_N_n) mathbb1_ (N_n < K_1) cup ( (N_n geq K_1) ] - mathbbE[f(Y)] | leq epsilon /2 + | mathbbE[f(Z_N_n) mathbb1( N_n < K_1) ] $ ..... ?

What would be a correct way of writing all this? Thanks in advance

probability-theory stochastic-processes probability-limit-theorems

share|cite|improve this question

edited Jul 30 at 6:55

asked Jul 29 at 20:39

Psylex

568

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 6:55

edited Jul 30 at 6:55

edited Jul 30 at 6:55

asked Jul 29 at 20:39

Psylex

568

asked Jul 29 at 20:39

Psylex

568

asked Jul 29 at 20:39

Psylex

568

568

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean exactly by a "random sequence $N_n$"?
  – Math1000
  Aug 1 at 8:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Math1000 $ N_n $ is a sequence of integer valued random variables (measurable functions). A discrete time stochastic process.
  – Psylex
  Aug 6 at 9:47
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean exactly by a "random sequence $N_n$"?
  – Math1000
  Aug 1 at 8:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Math1000 $ N_n $ is a sequence of integer valued random variables (measurable functions). A discrete time stochastic process.
  – Psylex
  Aug 6 at 9:47
  
  

  
  
  
  

What do you mean exactly by a "random sequence $N_n$"?
– Math1000
Aug 1 at 8:51

What do you mean exactly by a "random sequence $N_n$"?
– Math1000
Aug 1 at 8:51

@Math1000 $ N_n $ is a sequence of integer valued random variables (measurable functions). A discrete time stochastic process.
– Psylex
Aug 6 at 9:47

@Math1000 $ N_n $ is a sequence of integer valued random variables (measurable functions). A discrete time stochastic process.
– Psylex
Aug 6 at 9:47

add a comment | 

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