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0

$begingroup$

I think my textbook has an error in the worked solutions, but I want to double check I just don't misunderstand.

Let $X_i$ be a random variable such that $P(X_i geq x ) = e/xlog x$ for $x geq e$

We know that if $X geq 0$ then $E[X] = int_0^infty P(Xgeq x)dx$

Now, the textbook gives the answer as

$$E[X] = int_e^infty e/xlog x space dx$$

However, surely using the above formula it should actually be

$$E[X] = e+ int_e^infty e/xlog x space dx$$

Since $P(X > y) = 1$ for $y in [0,e]$?

Or am I misunderstanding something basic here?

Edit:

The fact $E[X]$ is infinite doesn't matter, as the question does the same for the truncated variables $E[X 1_(]$

probability

share|cite|improve this question

edited Mar 22 at 1:01

Xiaomi

asked Mar 21 at 14:48

XiaomiXiaomi

1,081115

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Where does the summand $e$ in your calculation come from?
  $endgroup$
  – J. Wang
  Mar 21 at 15:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the name of the survival function?
  $endgroup$
  – callculus
  Mar 21 at 16:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $E(X)$ does not converge. See here
  $endgroup$
  – callculus
  Mar 21 at 16:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know it does not converge, But supposing for a moment it did.... Since this working does the same for the truncated variable (which does converge)
  $endgroup$
  – Xiaomi
  Mar 22 at 1:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.Wang from integrating $P(X > y)=1$ from $0$ to $e$
  $endgroup$
  – Xiaomi
  Mar 22 at 1:00
  
  

  
  
  
  

add a comment | 

0

$begingroup$

I think my textbook has an error in the worked solutions, but I want to double check I just don't misunderstand.

Let $X_i$ be a random variable such that $P(X_i geq x ) = e/xlog x$ for $x geq e$

We know that if $X geq 0$ then $E[X] = int_0^infty P(Xgeq x)dx$

Now, the textbook gives the answer as

$$E[X] = int_e^infty e/xlog x space dx$$

However, surely using the above formula it should actually be

$$E[X] = e+ int_e^infty e/xlog x space dx$$

Since $P(X > y) = 1$ for $y in [0,e]$?

Or am I misunderstanding something basic here?

Edit:

The fact $E[X]$ is infinite doesn't matter, as the question does the same for the truncated variables $E[X 1_(]$

probability

share|cite|improve this question

edited Mar 22 at 1:01

Xiaomi

asked Mar 21 at 14:48

XiaomiXiaomi

1,081115

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Where does the summand $e$ in your calculation come from?
  $endgroup$
  – J. Wang
  Mar 21 at 15:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the name of the survival function?
  $endgroup$
  – callculus
  Mar 21 at 16:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $E(X)$ does not converge. See here
  $endgroup$
  – callculus
  Mar 21 at 16:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know it does not converge, But supposing for a moment it did.... Since this working does the same for the truncated variable (which does converge)
  $endgroup$
  – Xiaomi
  Mar 22 at 1:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.Wang from integrating $P(X > y)=1$ from $0$ to $e$
  $endgroup$
  – Xiaomi
  Mar 22 at 1:00
  
  

  
  
  
  

add a comment | 

$begingroup$

I think my textbook has an error in the worked solutions, but I want to double check I just don't misunderstand.

Let $X_i$ be a random variable such that $P(X_i geq x ) = e/xlog x$ for $x geq e$

We know that if $X geq 0$ then $E[X] = int_0^infty P(Xgeq x)dx$

Now, the textbook gives the answer as

$$E[X] = int_e^infty e/xlog x space dx$$

However, surely using the above formula it should actually be

$$E[X] = e+ int_e^infty e/xlog x space dx$$

Since $P(X > y) = 1$ for $y in [0,e]$?

Or am I misunderstanding something basic here?

Edit:

The fact $E[X]$ is infinite doesn't matter, as the question does the same for the truncated variables $E[X 1_(]$

probability

share|cite|improve this question

edited Mar 22 at 1:01

Xiaomi

asked Mar 21 at 14:48

XiaomiXiaomi

1,081115

$endgroup$

share|cite|improve this question

edited Mar 22 at 1:01

Xiaomi

asked Mar 21 at 14:48

XiaomiXiaomi

1,081115

share|cite|improve this question

edited Mar 22 at 1:01

Xiaomi

asked Mar 21 at 14:48

XiaomiXiaomi

1,081115

asked Mar 21 at 14:48

XiaomiXiaomi

1,081115

asked Mar 21 at 14:48

XiaomiXiaomi

1,081115