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Expectation of PDF

0

$begingroup$

Let $X$ be a continuous random variable with density function

$$f(x)=begincases
fracx2 & textif; 0 leq x leq 2& \ \
0 &textotherwise&
endcases$$

What is $E[|X-E[X]|]$?

I've tried to evaluate E(X). And wouldn't the answer be 0?

But instead the answer given is $frac3281$

probability expected-value

share|cite|improve this question

edited Apr 2 at 15:39

Chinnapparaj R

6,91221030

asked Apr 2 at 15:36

MathxxMathxx

3,41211545

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note: the absolute value brackets. The expected absolute displacement from the mean will not be zero.
  $endgroup$
  – Graham Kemp
  Apr 3 at 0:09
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $X$ be a continuous random variable with density function

$$f(x)=begincases
fracx2 & textif; 0 leq x leq 2& \ \
0 &textotherwise&
endcases$$

What is $E[|X-E[X]|]$?

I've tried to evaluate E(X). And wouldn't the answer be 0?

But instead the answer given is $frac3281$

probability expected-value

share|cite|improve this question

edited Apr 2 at 15:39

Chinnapparaj R

6,91221030

asked Apr 2 at 15:36

MathxxMathxx

3,41211545

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note: the absolute value brackets. The expected absolute displacement from the mean will not be zero.
  $endgroup$
  – Graham Kemp
  Apr 3 at 0:09
  

  
  
  
  

add a comment | 

$begingroup$

Let $X$ be a continuous random variable with density function

$$f(x)=begincases
fracx2 & textif; 0 leq x leq 2& \ \
0 &textotherwise&
endcases$$

What is $E[|X-E[X]|]$?

I've tried to evaluate E(X). And wouldn't the answer be 0?

But instead the answer given is $frac3281$

probability expected-value

share|cite|improve this question

edited Apr 2 at 15:39

Chinnapparaj R

6,91221030

asked Apr 2 at 15:36

MathxxMathxx

3,41211545

$endgroup$

share|cite|improve this question

edited Apr 2 at 15:39

Chinnapparaj R

6,91221030

asked Apr 2 at 15:36

MathxxMathxx

3,41211545

share|cite|improve this question

edited Apr 2 at 15:39

Chinnapparaj R

6,91221030

asked Apr 2 at 15:36

MathxxMathxx

3,41211545

edited Apr 2 at 15:39

Chinnapparaj R

6,91221030

edited Apr 2 at 15:39

Chinnapparaj R

6,91221030

asked Apr 2 at 15:36

MathxxMathxx

3,41211545

asked Apr 2 at 15:36

MathxxMathxx

3,41211545

1 Answer
1

active

oldest

votes

1

$begingroup$

Let's evaluate $mathbbE(X)$ step by step. By definition, if $X$ is a continuous RV with PDF $f(x)$ then $mathbbE(X)=int_mathbbR x f(x) mathrmdx$. In your case since $f(x)=0$ outside of the interval $[0,2]$ we have
$$mathbbE(X)=int_0^2 x cdot fracx2 mathrmdx=frac12 int_0^2 x^2 mathrmdx$$
$$=frac12 cdotfrac13 x^3 |_0^2=frac16 (8-0)=frac86=frac43neq 0,$$
which is clearly non-zero. You should carefully check your integration, since this is but basic rules of integration—all that was used is the power rule $int x^n mathrmdx=frac1n+1 x^n+1$ and it is really not a matter of probability at this point once you know the definition of $mathbbE(X)$, it's just routine calculus).
Can you now compute $int |x-4/3| fracx2 mathrmdx$? Note you will have to split it up based of the definition of absolute value and find the appropriate limits of integration.

share|cite|improve this answer

answered Apr 2 at 23:19

LoveTooNap29LoveTooNap29

1,1981614

$endgroup$

add a comment | 

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1

$begingroup$

Let's evaluate $mathbbE(X)$ step by step. By definition, if $X$ is a continuous RV with PDF $f(x)$ then $mathbbE(X)=int_mathbbR x f(x) mathrmdx$. In your case since $f(x)=0$ outside of the interval $[0,2]$ we have
$$mathbbE(X)=int_0^2 x cdot fracx2 mathrmdx=frac12 int_0^2 x^2 mathrmdx$$
$$=frac12 cdotfrac13 x^3 |_0^2=frac16 (8-0)=frac86=frac43neq 0,$$
which is clearly non-zero. You should carefully check your integration, since this is but basic rules of integration—all that was used is the power rule $int x^n mathrmdx=frac1n+1 x^n+1$ and it is really not a matter of probability at this point once you know the definition of $mathbbE(X)$, it's just routine calculus).
Can you now compute $int |x-4/3| fracx2 mathrmdx$? Note you will have to split it up based of the definition of absolute value and find the appropriate limits of integration.

share|cite|improve this answer

answered Apr 2 at 23:19

LoveTooNap29LoveTooNap29

1,1981614

$endgroup$

add a comment | 

1

$begingroup$

Let's evaluate $mathbbE(X)$ step by step. By definition, if $X$ is a continuous RV with PDF $f(x)$ then $mathbbE(X)=int_mathbbR x f(x) mathrmdx$. In your case since $f(x)=0$ outside of the interval $[0,2]$ we have
$$mathbbE(X)=int_0^2 x cdot fracx2 mathrmdx=frac12 int_0^2 x^2 mathrmdx$$
$$=frac12 cdotfrac13 x^3 |_0^2=frac16 (8-0)=frac86=frac43neq 0,$$
which is clearly non-zero. You should carefully check your integration, since this is but basic rules of integration—all that was used is the power rule $int x^n mathrmdx=frac1n+1 x^n+1$ and it is really not a matter of probability at this point once you know the definition of $mathbbE(X)$, it's just routine calculus).
Can you now compute $int |x-4/3| fracx2 mathrmdx$? Note you will have to split it up based of the definition of absolute value and find the appropriate limits of integration.

share|cite|improve this answer

answered Apr 2 at 23:19

LoveTooNap29LoveTooNap29

1,1981614

$endgroup$

add a comment | 

$begingroup$

Let's evaluate $mathbbE(X)$ step by step. By definition, if $X$ is a continuous RV with PDF $f(x)$ then $mathbbE(X)=int_mathbbR x f(x) mathrmdx$. In your case since $f(x)=0$ outside of the interval $[0,2]$ we have
$$mathbbE(X)=int_0^2 x cdot fracx2 mathrmdx=frac12 int_0^2 x^2 mathrmdx$$
$$=frac12 cdotfrac13 x^3 |_0^2=frac16 (8-0)=frac86=frac43neq 0,$$
which is clearly non-zero. You should carefully check your integration, since this is but basic rules of integration—all that was used is the power rule $int x^n mathrmdx=frac1n+1 x^n+1$ and it is really not a matter of probability at this point once you know the definition of $mathbbE(X)$, it's just routine calculus).
Can you now compute $int |x-4/3| fracx2 mathrmdx$? Note you will have to split it up based of the definition of absolute value and find the appropriate limits of integration.

share|cite|improve this answer

answered Apr 2 at 23:19

LoveTooNap29LoveTooNap29

1,1981614

$endgroup$

share|cite|improve this answer

answered Apr 2 at 23:19

LoveTooNap29LoveTooNap29

1,1981614

answered Apr 2 at 23:19

LoveTooNap29LoveTooNap29

1,1981614

answered Apr 2 at 23:19

LoveTooNap29LoveTooNap29

1,1981614