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Positive Sub-Martingale with Limit = 0?

0

$begingroup$

We have $u(t+1)=u(t)r_t+1$, with $u(0) = 1$, and $r_t$ is a random variable with 2 possible values:

$2/9$ appears with the possibility of 80%,

$8$ appears with the possibility of 20%.

Then we have
$$
lim_ttoinftyE[u(t)]=lim_ttoinftyleft[left(frac29right)^0.8ttimes (8)^0.2tright] = limlimits_ttoinfty(0.300times1.516)^t = 0 dotsdots(1)
$$

$$
lim_ttoinftyE[u(t)]= lim_ttoinftyleft(E[r_1]E[r_2]dots E[r_t]right)=(frac29times0.8+8times0.2)^t=infty. dotsdots(2)
$$

Question 1: (1) and (2) seems to conflict with each other.

At the same time, we have

beginalignmathbbE[u(t+1)|mathcalF_t]-u(t)&=u(t)(mathbbE[r_t+1|mathcalF_t]-1) \ &=u(t)left(0.8timesfrac29+0.2times8 - 1right) \
&= 0.778u(t)>0.
endalign

It seems that $u(t)$ is a sub-martingale.

Question 2: $u(t)$ is a sub-martingale seems to conflict with (1).

import numpy as np
import random as random

u = 1
n = 100

for n in range(n):
 r = random.random()
 if r < 0.8:
 u = u * 2 / 9
 else:
 u = u * 8

print(u) # the outputs are always like 2.9307827938878015e-44

Question 3:
Can we say $u(t)$ converge in distribution/probability to zero, how to prove it?

I'm a beginner with stupid questions, these questions mess me up, don't know where are the mistakes or misunderstandings.

real-analysis statistics martingales

share|cite|improve this question

edited Mar 28 at 8:26

ZhongHua Yan

asked Mar 28 at 2:42

ZhongHua YanZhongHua Yan

12

New contributor

ZhongHua Yan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have no idea how you got $lim u(t+1)=0$. You are pretending that $r_t$ and $u(t)$ are all just numbers and not random. The argument is completely invalid.
  $endgroup$
  – Kavi Rama Murthy
  Mar 28 at 6:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi, thanks for your reply, Kavi Rama Murthy, I have updated my questions, can you give me a hint?
  $endgroup$
  – ZhongHua Yan
  Mar 28 at 8:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The first calculation is still wrong . the second one is right. I have no idea what kind of argument leads to (1).
  $endgroup$
  – Kavi Rama Murthy
  Mar 28 at 8:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you again, then how to explain the numerical simulation and what knowledge should I learn to solve question 3?
  $endgroup$
  – ZhongHua Yan
  Mar 28 at 8:31
  

  
  
  
  

add a comment | 

0

$begingroup$

We have $u(t+1)=u(t)r_t+1$, with $u(0) = 1$, and $r_t$ is a random variable with 2 possible values:

$2/9$ appears with the possibility of 80%,

$8$ appears with the possibility of 20%.

Then we have
$$
lim_ttoinftyE[u(t)]=lim_ttoinftyleft[left(frac29right)^0.8ttimes (8)^0.2tright] = limlimits_ttoinfty(0.300times1.516)^t = 0 dotsdots(1)
$$

$$
lim_ttoinftyE[u(t)]= lim_ttoinftyleft(E[r_1]E[r_2]dots E[r_t]right)=(frac29times0.8+8times0.2)^t=infty. dotsdots(2)
$$

Question 1: (1) and (2) seems to conflict with each other.

At the same time, we have

beginalignmathbbE[u(t+1)|mathcalF_t]-u(t)&=u(t)(mathbbE[r_t+1|mathcalF_t]-1) \ &=u(t)left(0.8timesfrac29+0.2times8 - 1right) \
&= 0.778u(t)>0.
endalign

It seems that $u(t)$ is a sub-martingale.

Question 2: $u(t)$ is a sub-martingale seems to conflict with (1).

import numpy as np
import random as random

u = 1
n = 100

for n in range(n):
 r = random.random()
 if r < 0.8:
 u = u * 2 / 9
 else:
 u = u * 8

print(u) # the outputs are always like 2.9307827938878015e-44

Question 3:
Can we say $u(t)$ converge in distribution/probability to zero, how to prove it?

I'm a beginner with stupid questions, these questions mess me up, don't know where are the mistakes or misunderstandings.

real-analysis statistics martingales

share|cite|improve this question

edited Mar 28 at 8:26

ZhongHua Yan

asked Mar 28 at 2:42

ZhongHua YanZhongHua Yan

12

New contributor

ZhongHua Yan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have no idea how you got $lim u(t+1)=0$. You are pretending that $r_t$ and $u(t)$ are all just numbers and not random. The argument is completely invalid.
  $endgroup$
  – Kavi Rama Murthy
  Mar 28 at 6:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi, thanks for your reply, Kavi Rama Murthy, I have updated my questions, can you give me a hint?
  $endgroup$
  – ZhongHua Yan
  Mar 28 at 8:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The first calculation is still wrong . the second one is right. I have no idea what kind of argument leads to (1).
  $endgroup$
  – Kavi Rama Murthy
  Mar 28 at 8:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you again, then how to explain the numerical simulation and what knowledge should I learn to solve question 3?
  $endgroup$
  – ZhongHua Yan
  Mar 28 at 8:31
  

  
  
  
  

add a comment | 

$begingroup$

We have $u(t+1)=u(t)r_t+1$, with $u(0) = 1$, and $r_t$ is a random variable with 2 possible values:

$2/9$ appears with the possibility of 80%,

$8$ appears with the possibility of 20%.

Then we have
$$
lim_ttoinftyE[u(t)]=lim_ttoinftyleft[left(frac29right)^0.8ttimes (8)^0.2tright] = limlimits_ttoinfty(0.300times1.516)^t = 0 dotsdots(1)
$$

$$
lim_ttoinftyE[u(t)]= lim_ttoinftyleft(E[r_1]E[r_2]dots E[r_t]right)=(frac29times0.8+8times0.2)^t=infty. dotsdots(2)
$$

Question 1: (1) and (2) seems to conflict with each other.

At the same time, we have

beginalignmathbbE[u(t+1)|mathcalF_t]-u(t)&=u(t)(mathbbE[r_t+1|mathcalF_t]-1) \ &=u(t)left(0.8timesfrac29+0.2times8 - 1right) \
&= 0.778u(t)>0.
endalign

It seems that $u(t)$ is a sub-martingale.

Question 2: $u(t)$ is a sub-martingale seems to conflict with (1).

import numpy as np
import random as random

u = 1
n = 100

for n in range(n):
 r = random.random()
 if r < 0.8:
 u = u * 2 / 9
 else:
 u = u * 8

print(u) # the outputs are always like 2.9307827938878015e-44

Question 3:
Can we say $u(t)$ converge in distribution/probability to zero, how to prove it?

I'm a beginner with stupid questions, these questions mess me up, don't know where are the mistakes or misunderstandings.

real-analysis statistics martingales

share|cite|improve this question

edited Mar 28 at 8:26

ZhongHua Yan

asked Mar 28 at 2:42

ZhongHua YanZhongHua Yan

12

New contributor

ZhongHua Yan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

import numpy as np
import random as random

u = 1
n = 100

for n in range(n):
 r = random.random()
 if r < 0.8:
 u = u * 2 / 9
 else:
 u = u * 8

print(u) # the outputs are always like 2.9307827938878015e-44

share|cite|improve this question

edited Mar 28 at 8:26

ZhongHua Yan

asked Mar 28 at 2:42

ZhongHua YanZhongHua Yan

12

New contributor

ZhongHua Yan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 28 at 8:26

ZhongHua Yan

asked Mar 28 at 2:42

ZhongHua YanZhongHua Yan

12

New contributor

ZhongHua Yan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 28 at 2:42

ZhongHua YanZhongHua Yan

12

New contributor

ZhongHua Yan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 28 at 2:42

ZhongHua YanZhongHua Yan

12