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Prove of $Eleft(|X+Y|^aright)ge Eleft(|Y|^aright)$?

Prove $Eleft[left(frac1nsum_i=1^nX_iright)^kright]leq Eleft[X_1left(fracX_1+(n-1)munright)^k-1right]$$Pleft(limsup left(X_n=0, X_ n+1=1,X_ n+2=0 right)right)$Expected value of function of minimum between two random variablesDoes $Pleft(X_1 < X_2 < X_3right) = Pleft(X_1 le X_2 le X_3right)$?An example where $Eleft[lim_n to inftyX_nright] neq lim_n to inftyEleft[X_nright]$Does $mathbbEleft[Xright]=infty$ imply $mathbbEleft[X^2right]=infty$?$lim_n to inftyPleft(left|1 over nsum_i=1^nleft(X_i-X_1+X_2+cdots + X_n over n right)^2 - sigma^2 right| >epsilonright)=0$Show that $Eleft[leftvertsum y_kx_krightvert^4right]le3Eleft[leftvertsum y_kx_krightvert^2right]^2$ for $(x_k)$ i.i.d. random signsProve that $E!left[sum_i=1^NX_iright]=mu E[N]$probability of $left[log_2n right]$ consecutive tails

1

$begingroup$

Let $E(V)$ be the expectation of $V$.
It is also known that $E(X)=0, a>1, Eleft(|X|^aright) < +infty$ and $Eleft(|Y|^aright)< +infty$. $X$ and $Y$ are independent.
How can I prove that $Eleft(|X+Y|^aright)ge Eleft(|Y|^aright)$?

probability random-variables independence expected-value

share|cite|improve this question

edited Mar 17 at 11:51

Ingix

5,077159

asked Mar 17 at 10:32

Andrey KomisarovAndrey Komisarov

675

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is $V$ here? You mean $Y$?
  $endgroup$
  – Jimmy R.
  Mar 17 at 11:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, it is only explanation what is E
  $endgroup$
  – Andrey Komisarov
  Mar 17 at 11:34
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $E(V)$ be the expectation of $V$.
It is also known that $E(X)=0, a>1, Eleft(|X|^aright) < +infty$ and $Eleft(|Y|^aright)< +infty$. $X$ and $Y$ are independent.
How can I prove that $Eleft(|X+Y|^aright)ge Eleft(|Y|^aright)$?

probability random-variables independence expected-value

share|cite|improve this question

edited Mar 17 at 11:51

Ingix

5,077159

asked Mar 17 at 10:32

Andrey KomisarovAndrey Komisarov

675

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is $V$ here? You mean $Y$?
  $endgroup$
  – Jimmy R.
  Mar 17 at 11:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, it is only explanation what is E
  $endgroup$
  – Andrey Komisarov
  Mar 17 at 11:34
  

  
  
  
  

add a comment | 

$begingroup$

Let $E(V)$ be the expectation of $V$.
It is also known that $E(X)=0, a>1, Eleft(|X|^aright) < +infty$ and $Eleft(|Y|^aright)< +infty$. $X$ and $Y$ are independent.
How can I prove that $Eleft(|X+Y|^aright)ge Eleft(|Y|^aright)$?

probability random-variables independence expected-value

share|cite|improve this question

edited Mar 17 at 11:51

Ingix

5,077159

asked Mar 17 at 10:32

Andrey KomisarovAndrey Komisarov

675

$endgroup$

share|cite|improve this question

edited Mar 17 at 11:51

Ingix

5,077159

asked Mar 17 at 10:32

Andrey KomisarovAndrey Komisarov

675

share|cite|improve this question

edited Mar 17 at 11:51

Ingix

5,077159

asked Mar 17 at 10:32

Andrey KomisarovAndrey Komisarov

675

edited Mar 17 at 11:51

Ingix

5,077159

edited Mar 17 at 11:51

Ingix

5,077159

asked Mar 17 at 10:32

Andrey KomisarovAndrey Komisarov

675

asked Mar 17 at 10:32

Andrey KomisarovAndrey Komisarov

675

1 Answer
1

active

oldest

votes

2

$begingroup$

Since $a>1$, the function $f: xmapsto |x|^a$ is convex. (Of course this is true for $age1$, too.)
Your result follows from Jensen's inequality: conditional on $Y$ we have $E(f(X+Y)|Y) ge f(E(X+Y|Y))=f(Y)$. And so on.

share|cite|improve this answer

edited Mar 17 at 12:45

answered Mar 17 at 12:40

kimchi loverkimchi lover

11.5k31229

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your Jensen is the wrong way around.
  $endgroup$
  – user159517
  Mar 17 at 12:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Whoops: so it is! Thanks for the catch.
  $endgroup$
  – kimchi lover
  Mar 17 at 12:46
  

  
  
  
  

add a comment | 

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2

$begingroup$

Since $a>1$, the function $f: xmapsto |x|^a$ is convex. (Of course this is true for $age1$, too.)
Your result follows from Jensen's inequality: conditional on $Y$ we have $E(f(X+Y)|Y) ge f(E(X+Y|Y))=f(Y)$. And so on.

share|cite|improve this answer

edited Mar 17 at 12:45

answered Mar 17 at 12:40

kimchi loverkimchi lover

11.5k31229

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your Jensen is the wrong way around.
  $endgroup$
  – user159517
  Mar 17 at 12:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Whoops: so it is! Thanks for the catch.
  $endgroup$
  – kimchi lover
  Mar 17 at 12:46
  

  
  
  
  

add a comment | 

2

$begingroup$

Since $a>1$, the function $f: xmapsto |x|^a$ is convex. (Of course this is true for $age1$, too.)
Your result follows from Jensen's inequality: conditional on $Y$ we have $E(f(X+Y)|Y) ge f(E(X+Y|Y))=f(Y)$. And so on.

share|cite|improve this answer

edited Mar 17 at 12:45

answered Mar 17 at 12:40

kimchi loverkimchi lover

11.5k31229

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your Jensen is the wrong way around.
  $endgroup$
  – user159517
  Mar 17 at 12:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Whoops: so it is! Thanks for the catch.
  $endgroup$
  – kimchi lover
  Mar 17 at 12:46
  

  
  
  
  

add a comment | 

$begingroup$

Since $a>1$, the function $f: xmapsto |x|^a$ is convex. (Of course this is true for $age1$, too.)
Your result follows from Jensen's inequality: conditional on $Y$ we have $E(f(X+Y)|Y) ge f(E(X+Y|Y))=f(Y)$. And so on.

share|cite|improve this answer

edited Mar 17 at 12:45

answered Mar 17 at 12:40

kimchi loverkimchi lover

11.5k31229

$endgroup$

share|cite|improve this answer

edited Mar 17 at 12:45

answered Mar 17 at 12:40

kimchi loverkimchi lover

11.5k31229

answered Mar 17 at 12:40

kimchi loverkimchi lover

11.5k31229

answered Mar 17 at 12:40

kimchi loverkimchi lover

11.5k31229