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Proving formula for math expectation

0

1

$begingroup$

In a book I met a formula for math. expectation of a random variable $xi$ with distribution function $F(x)$:

$$Mxi=-int_-infty^0F(x)dx+int_0^infty(1-F(x))dx$$

I wonder how do I prove it?

My attempt follows:

$Mxiequivint_-infty^inftyxdF(x)=lim_ato-infty^btoinftyint_a^bxdF(x)$

Integrating in parts, I obtain

$int_a^bxdF(x)=(xF(x))rvert_a^b-int_a^bF(x)dx=bF(b)-aF(a)-int_a^0F(x)dx-int_0^bF(x)dx=bF(b)-aF(a)-int_a^0F(x)dx+int_0^b(1-F(x))dx-b=[-int_a^0F(x)dx+int_0^b(1-F(x))dx]+b(F(b)-1)-aF(a).$

Passing to the limit, I get

$Mxi=-int_-infty^0F(x)dx+int_0^infty(1-F(x))dx-lim_ato-inftyaF(a)+lim_btoinftyb(F(b)-1)$

So in order to prove the initial statement, I need to prove that for arbitrary distribution function $F$

$lim_ato-inftyaF(a)=0$

and

$lim_btoinftyb(F(b)-1)=0$

however I have no idea how to prove it and moreover I doubt that it's true.

probability probability-distributions

share|cite|improve this question

asked Mar 24 at 17:14

NickNick

1514

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I guess the last 2 assumptions should be true if the expected value is finite.
  $endgroup$
  – kludg
  Mar 24 at 17:34
  

  
  
  
  

add a comment | 

0

1

$begingroup$

In a book I met a formula for math. expectation of a random variable $xi$ with distribution function $F(x)$:

$$Mxi=-int_-infty^0F(x)dx+int_0^infty(1-F(x))dx$$

I wonder how do I prove it?

My attempt follows:

$Mxiequivint_-infty^inftyxdF(x)=lim_ato-infty^btoinftyint_a^bxdF(x)$

Integrating in parts, I obtain

$int_a^bxdF(x)=(xF(x))rvert_a^b-int_a^bF(x)dx=bF(b)-aF(a)-int_a^0F(x)dx-int_0^bF(x)dx=bF(b)-aF(a)-int_a^0F(x)dx+int_0^b(1-F(x))dx-b=[-int_a^0F(x)dx+int_0^b(1-F(x))dx]+b(F(b)-1)-aF(a).$

Passing to the limit, I get

$Mxi=-int_-infty^0F(x)dx+int_0^infty(1-F(x))dx-lim_ato-inftyaF(a)+lim_btoinftyb(F(b)-1)$

So in order to prove the initial statement, I need to prove that for arbitrary distribution function $F$

$lim_ato-inftyaF(a)=0$

and

$lim_btoinftyb(F(b)-1)=0$

however I have no idea how to prove it and moreover I doubt that it's true.

probability probability-distributions

share|cite|improve this question

asked Mar 24 at 17:14

NickNick

1514

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I guess the last 2 assumptions should be true if the expected value is finite.
  $endgroup$
  – kludg
  Mar 24 at 17:34
  

  
  
  
  

add a comment | 

$begingroup$

In a book I met a formula for math. expectation of a random variable $xi$ with distribution function $F(x)$:

$$Mxi=-int_-infty^0F(x)dx+int_0^infty(1-F(x))dx$$

I wonder how do I prove it?

My attempt follows:

$Mxiequivint_-infty^inftyxdF(x)=lim_ato-infty^btoinftyint_a^bxdF(x)$

Integrating in parts, I obtain

$int_a^bxdF(x)=(xF(x))rvert_a^b-int_a^bF(x)dx=bF(b)-aF(a)-int_a^0F(x)dx-int_0^bF(x)dx=bF(b)-aF(a)-int_a^0F(x)dx+int_0^b(1-F(x))dx-b=[-int_a^0F(x)dx+int_0^b(1-F(x))dx]+b(F(b)-1)-aF(a).$

Passing to the limit, I get

$Mxi=-int_-infty^0F(x)dx+int_0^infty(1-F(x))dx-lim_ato-inftyaF(a)+lim_btoinftyb(F(b)-1)$

So in order to prove the initial statement, I need to prove that for arbitrary distribution function $F$

$lim_ato-inftyaF(a)=0$

and

$lim_btoinftyb(F(b)-1)=0$

however I have no idea how to prove it and moreover I doubt that it's true.

probability probability-distributions

share|cite|improve this question

asked Mar 24 at 17:14

NickNick

1514

$endgroup$

share|cite|improve this question

asked Mar 24 at 17:14

NickNick

1514

share|cite|improve this question

asked Mar 24 at 17:14

NickNick

1514

asked Mar 24 at 17:14

NickNick

1514

asked Mar 24 at 17:14

NickNick

1514

1 Answer
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votes

1

$begingroup$

If the expectation is finite then both limits $$limlimits_ato-inftyint_-infty^a x,dF(x) text and limlimits_btoinftyint_b^infty x,dF(x)$$
are zero. Then
$$
0=limlimits_ato-inftyint_-infty^a x,dF(x) leq limlimits_ato-inftya int_-infty^a dF(x) =limlimits_ato-infty aF(a)leq 0.
$$
And
$$
0=limlimits_btoinftyint_b^infty x,dF(x) geq limlimits_btoinftyb int_b^infty dF(x) =limlimits_btoinfty b(1-F(b))geq 0.
$$

share|cite|improve this answer

answered Mar 25 at 0:59

NChNCh

7,1153825

$endgroup$

add a comment | 

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1

$begingroup$

If the expectation is finite then both limits $$limlimits_ato-inftyint_-infty^a x,dF(x) text and limlimits_btoinftyint_b^infty x,dF(x)$$
are zero. Then
$$
0=limlimits_ato-inftyint_-infty^a x,dF(x) leq limlimits_ato-inftya int_-infty^a dF(x) =limlimits_ato-infty aF(a)leq 0.
$$
And
$$
0=limlimits_btoinftyint_b^infty x,dF(x) geq limlimits_btoinftyb int_b^infty dF(x) =limlimits_btoinfty b(1-F(b))geq 0.
$$

share|cite|improve this answer

answered Mar 25 at 0:59

NChNCh

7,1153825

$endgroup$

add a comment | 

1

$begingroup$

If the expectation is finite then both limits $$limlimits_ato-inftyint_-infty^a x,dF(x) text and limlimits_btoinftyint_b^infty x,dF(x)$$
are zero. Then
$$
0=limlimits_ato-inftyint_-infty^a x,dF(x) leq limlimits_ato-inftya int_-infty^a dF(x) =limlimits_ato-infty aF(a)leq 0.
$$
And
$$
0=limlimits_btoinftyint_b^infty x,dF(x) geq limlimits_btoinftyb int_b^infty dF(x) =limlimits_btoinfty b(1-F(b))geq 0.
$$

share|cite|improve this answer

answered Mar 25 at 0:59

NChNCh

7,1153825

$endgroup$

add a comment | 

$begingroup$

If the expectation is finite then both limits $$limlimits_ato-inftyint_-infty^a x,dF(x) text and limlimits_btoinftyint_b^infty x,dF(x)$$
are zero. Then
$$
0=limlimits_ato-inftyint_-infty^a x,dF(x) leq limlimits_ato-inftya int_-infty^a dF(x) =limlimits_ato-infty aF(a)leq 0.
$$
And
$$
0=limlimits_btoinftyint_b^infty x,dF(x) geq limlimits_btoinftyb int_b^infty dF(x) =limlimits_btoinfty b(1-F(b))geq 0.
$$

share|cite|improve this answer

answered Mar 25 at 0:59

NChNCh

7,1153825

$endgroup$

share|cite|improve this answer

answered Mar 25 at 0:59

NChNCh

7,1153825

answered Mar 25 at 0:59

NChNCh

7,1153825

answered Mar 25 at 0:59

NChNCh

7,1153825