Convergence in probability implies convergence in expectation.

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I would like to see a full solution to the following problem. I have tried approaching it in a number of different ways, however I don't seem to get to the desired result. No point posting what I did, as they just lead to dead end roads.

Show that if $X_nto 0$ in probability then:
$$
Eleft[frac1+right]to 0, mbox as ntoinfty
$$

The converse implication holds as well, but that one is easy to prove using Chebyshev's Inequality.

probability expectation

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asked Jul 20 at 10:30

Andrei Crisan

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up vote
0
down vote

favorite

I would like to see a full solution to the following problem. I have tried approaching it in a number of different ways, however I don't seem to get to the desired result. No point posting what I did, as they just lead to dead end roads.

Show that if $X_nto 0$ in probability then:
$$
Eleft[frac1+right]to 0, mbox as ntoinfty
$$

The converse implication holds as well, but that one is easy to prove using Chebyshev's Inequality.

probability expectation

share|cite|improve this question

asked Jul 20 at 10:30

Andrei Crisan

3219

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I would like to see a full solution to the following problem. I have tried approaching it in a number of different ways, however I don't seem to get to the desired result. No point posting what I did, as they just lead to dead end roads.

Show that if $X_nto 0$ in probability then:
$$
Eleft[frac1+right]to 0, mbox as ntoinfty
$$

The converse implication holds as well, but that one is easy to prove using Chebyshev's Inequality.

probability expectation

share|cite|improve this question

asked Jul 20 at 10:30

Andrei Crisan

3219

I would like to see a full solution to the following problem. I have tried approaching it in a number of different ways, however I don't seem to get to the desired result. No point posting what I did, as they just lead to dead end roads.

Show that if $X_nto 0$ in probability then:
$$
Eleft[frac1+right]to 0, mbox as ntoinfty
$$

The converse implication holds as well, but that one is easy to prove using Chebyshev's Inequality.

probability expectation

share|cite|improve this question

asked Jul 20 at 10:30

Andrei Crisan

3219

share|cite|improve this question

share|cite|improve this question

asked Jul 20 at 10:30

Andrei Crisan

3219

asked Jul 20 at 10:30

Andrei Crisan

3219

asked Jul 20 at 10:30

Andrei Crisan

3219

3219

add a comment | 

add a comment | 

1 Answer
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accepted

Use Dominated Convergence Theorem. $frac 1+leq 1$. To use the usual version of DCT you have to go to a.s. convergent subsequences, but DCT is valid for convergence in probability too. Alternatively, let $epsilon in (0,1)$ and $delta =frac epsilon 1-epsilon $ Then $Efrac 1+ =Efrac 1+I_X_n+Efrac 1+I_ <delta leq EI_X_n+frac delta 1+delta =PX_n +epsilon $. The result follows from this.

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edited Jul 20 at 10:41

answered Jul 20 at 10:34

Kavi Rama Murthy

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Use Dominated Convergence Theorem. $frac 1+leq 1$. To use the usual version of DCT you have to go to a.s. convergent subsequences, but DCT is valid for convergence in probability too. Alternatively, let $epsilon in (0,1)$ and $delta =frac epsilon 1-epsilon $ Then $Efrac 1+ =Efrac 1+I_X_n+Efrac 1+I_ <delta leq EI_X_n+frac delta 1+delta =PX_n +epsilon $. The result follows from this.

share|cite|improve this answer

edited Jul 20 at 10:41

answered Jul 20 at 10:34

Kavi Rama Murthy

20.6k2830

add a comment | 

up vote
1
down vote



accepted

Use Dominated Convergence Theorem. $frac 1+leq 1$. To use the usual version of DCT you have to go to a.s. convergent subsequences, but DCT is valid for convergence in probability too. Alternatively, let $epsilon in (0,1)$ and $delta =frac epsilon 1-epsilon $ Then $Efrac 1+ =Efrac 1+I_X_n+Efrac 1+I_ <delta leq EI_X_n+frac delta 1+delta =PX_n +epsilon $. The result follows from this.

share|cite|improve this answer

edited Jul 20 at 10:41

answered Jul 20 at 10:34

Kavi Rama Murthy

20.6k2830

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Use Dominated Convergence Theorem. $frac 1+leq 1$. To use the usual version of DCT you have to go to a.s. convergent subsequences, but DCT is valid for convergence in probability too. Alternatively, let $epsilon in (0,1)$ and $delta =frac epsilon 1-epsilon $ Then $Efrac 1+ =Efrac 1+I_X_n+Efrac 1+I_ <delta leq EI_X_n+frac delta 1+delta =PX_n +epsilon $. The result follows from this.

share|cite|improve this answer

edited Jul 20 at 10:41

answered Jul 20 at 10:34

Kavi Rama Murthy

20.6k2830

Use Dominated Convergence Theorem. $frac 1+leq 1$. To use the usual version of DCT you have to go to a.s. convergent subsequences, but DCT is valid for convergence in probability too. Alternatively, let $epsilon in (0,1)$ and $delta =frac epsilon 1-epsilon $ Then $Efrac 1+ =Efrac 1+I_X_n+Efrac 1+I_ <delta leq EI_X_n+frac delta 1+delta =PX_n +epsilon $. The result follows from this.

share|cite|improve this answer

edited Jul 20 at 10:41

answered Jul 20 at 10:34

Kavi Rama Murthy

20.6k2830

share|cite|improve this answer

share|cite|improve this answer

edited Jul 20 at 10:41

edited Jul 20 at 10:41

edited Jul 20 at 10:41

answered Jul 20 at 10:34

Kavi Rama Murthy

20.6k2830

answered Jul 20 at 10:34

Kavi Rama Murthy

20.6k2830

answered Jul 20 at 10:34

Kavi Rama Murthy

20.6k2830

20.6k2830

add a comment | 

add a comment | 

 

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