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What are mean and variance of $W_i$, given that $Z_n=frac{sum{W_i}}{sqrt{n}sigma}sim N(0,1)$? [closed]

-1

$begingroup$

Let
$$Z_n=frac{sum{W_i}}{sqrt{n}sigma}sim N(0,1),$$
where $W_i=X_i-mu$. What are the mean and variance of $W_i$?

probability-theory statistics normal-distribution statistical-inference central-limit-theorem

share|cite|improve this question

edited Mar 21 at 23:42

Brian

1,508416

asked Mar 21 at 22:34

theQuestiontheQuestion

84

$endgroup$

closed as off-topic by NCh, Saad, Lee David Chung Lin, Leucippus, Eevee Trainer Mar 22 at 4:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – NCh, Saad, Lee David Chung Lin, Leucippus, Eevee Trainer

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know the distribution of $X_i$?
  $endgroup$
  – Minus One-Twelfth
  Mar 21 at 22:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It is a distribution $F_x$ with finite $E(X)=mu$ and $Var(X)=sigma^2>0$
  $endgroup$
  – theQuestion
  Mar 21 at 22:40
  

  
  
  
  

add a comment | 

-1

$begingroup$

Let
$$Z_n=frac{sum{W_i}}{sqrt{n}sigma}sim N(0,1),$$
where $W_i=X_i-mu$. What are the mean and variance of $W_i$?

probability-theory statistics normal-distribution statistical-inference central-limit-theorem

share|cite|improve this question

edited Mar 21 at 23:42

Brian

1,508416

asked Mar 21 at 22:34

theQuestiontheQuestion

84

$endgroup$

closed as off-topic by NCh, Saad, Lee David Chung Lin, Leucippus, Eevee Trainer Mar 22 at 4:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – NCh, Saad, Lee David Chung Lin, Leucippus, Eevee Trainer

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know the distribution of $X_i$?
  $endgroup$
  – Minus One-Twelfth
  Mar 21 at 22:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It is a distribution $F_x$ with finite $E(X)=mu$ and $Var(X)=sigma^2>0$
  $endgroup$
  – theQuestion
  Mar 21 at 22:40
  

  
  
  
  

add a comment | 

$begingroup$

Let
$$Z_n=frac{sum{W_i}}{sqrt{n}sigma}sim N(0,1),$$
where $W_i=X_i-mu$. What are the mean and variance of $W_i$?

probability-theory statistics normal-distribution statistical-inference central-limit-theorem

share|cite|improve this question

edited Mar 21 at 23:42

Brian

1,508416

asked Mar 21 at 22:34

theQuestiontheQuestion

84

$endgroup$

share|cite|improve this question

edited Mar 21 at 23:42

Brian

1,508416

asked Mar 21 at 22:34

theQuestiontheQuestion

84

share|cite|improve this question

edited Mar 21 at 23:42

Brian

1,508416

asked Mar 21 at 22:34

theQuestiontheQuestion

84

edited Mar 21 at 23:42

Brian

1,508416

edited Mar 21 at 23:42

Brian

1,508416

asked Mar 21 at 22:34

theQuestiontheQuestion

84

asked Mar 21 at 22:34

theQuestiontheQuestion

84

1 Answer
1

active

oldest

votes

0

$begingroup$

$newcommand{Var}{operatorname{Var}}newcommand{E}{mathbb{E}}$The mean of $W_i$ is $0$ and variance of $W_i$ is same as that of $X_i$, namely $sigma^2$. This follows from standard properties of mean and variance ($E(X-c) = E(X)-c$ and $Var(X-c)= Var(X)$, for any constant $c$).

share|cite|improve this answer

answered Mar 21 at 22:45

Minus One-TwelfthMinus One-Twelfth

3,398413

$endgroup$

add a comment | 

0

$begingroup$

$newcommand{Var}{operatorname{Var}}newcommand{E}{mathbb{E}}$The mean of $W_i$ is $0$ and variance of $W_i$ is same as that of $X_i$, namely $sigma^2$. This follows from standard properties of mean and variance ($E(X-c) = E(X)-c$ and $Var(X-c)= Var(X)$, for any constant $c$).

share|cite|improve this answer

answered Mar 21 at 22:45

Minus One-TwelfthMinus One-Twelfth

3,398413

$endgroup$

add a comment | 

0

$begingroup$

$newcommand{Var}{operatorname{Var}}newcommand{E}{mathbb{E}}$The mean of $W_i$ is $0$ and variance of $W_i$ is same as that of $X_i$, namely $sigma^2$. This follows from standard properties of mean and variance ($E(X-c) = E(X)-c$ and $Var(X-c)= Var(X)$, for any constant $c$).

share|cite|improve this answer

answered Mar 21 at 22:45

Minus One-TwelfthMinus One-Twelfth

3,398413

$endgroup$

add a comment | 

$begingroup$

$newcommand{Var}{operatorname{Var}}newcommand{E}{mathbb{E}}$The mean of $W_i$ is $0$ and variance of $W_i$ is same as that of $X_i$, namely $sigma^2$. This follows from standard properties of mean and variance ($E(X-c) = E(X)-c$ and $Var(X-c)= Var(X)$, for any constant $c$).

share|cite|improve this answer

answered Mar 21 at 22:45

Minus One-TwelfthMinus One-Twelfth

3,398413

$endgroup$

share|cite|improve this answer

answered Mar 21 at 22:45

Minus One-TwelfthMinus One-Twelfth

3,398413

answered Mar 21 at 22:45

Minus One-TwelfthMinus One-Twelfth

3,398413

answered Mar 21 at 22:45

Minus One-TwelfthMinus One-Twelfth

3,398413