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Deriving the equation for variance

1

$begingroup$

I am trying to understand variance, and when I look at the fully derived equation on wikipedia, I do not understand how we are able to make the jump from B to C.

When simplifying, 2(XE[X]) where does the 2X go? I understand how we get 2E[X], but I'm not sure why or how the 2X is able to be removed.

Sorry for this stupid question. I haven't taken a math class in about 7 years, and I'm trying to get back in to it... Thanks for your help :)

self-study variance

share|cite|improve this question

edited 5 hours ago

snarkshark

asked 6 hours ago

snarksharksnarkshark

82

New contributor

snarkshark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is not a stupid question. It's not just simplifying... notice that B has an expectation of the entire expression, while C does not. (And welcome to CV, snarkshark!)
  $endgroup$
  – Alexis
  6 hours ago
  
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hi, @Alexis :) I noticed that, but I'm still confused. Does E[-2XE[X]] somehow become 1? I dont understand how its eliminated from the equation still :'(
  $endgroup$
  – snarkshark
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As an aside, you might read about the self-study tag, and edit your question (the "edit" link at lower left) to inlcude it.
  $endgroup$
  – Alexis
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well simplifying does enter thereafter. What is $E[(-2)]$? What is $E[(X)]$? What is $E[(E[X])]$? And so, what is $E[(-2XE[X])]$?
  $endgroup$
  – Alexis
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  E[(-2)] = -2 (averaging itself is just itself); E[X] = X; and E[(E[X])] = E[X] = X; so... 𝐸[(−2𝑋𝐸[𝑋])] just ends up being -2X and its removed from the equation because its a constant? Thanks for your patience with my dumb questions ^^;;
  $endgroup$
  – snarkshark
  5 hours ago
  
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

I am trying to understand variance, and when I look at the fully derived equation on wikipedia, I do not understand how we are able to make the jump from B to C.

When simplifying, 2(XE[X]) where does the 2X go? I understand how we get 2E[X], but I'm not sure why or how the 2X is able to be removed.

Sorry for this stupid question. I haven't taken a math class in about 7 years, and I'm trying to get back in to it... Thanks for your help :)

self-study variance

share|cite|improve this question

edited 5 hours ago

snarkshark

asked 6 hours ago

snarksharksnarkshark

82

New contributor

snarkshark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is not a stupid question. It's not just simplifying... notice that B has an expectation of the entire expression, while C does not. (And welcome to CV, snarkshark!)
  $endgroup$
  – Alexis
  6 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hi, @Alexis :) I noticed that, but I'm still confused. Does E[-2XE[X]] somehow become 1? I dont understand how its eliminated from the equation still :'(
  $endgroup$
  – snarkshark
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As an aside, you might read about the self-study tag, and edit your question (the "edit" link at lower left) to inlcude it.
  $endgroup$
  – Alexis
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well simplifying does enter thereafter. What is $E[(-2)]$? What is $E[(X)]$? What is $E[(E[X])]$? And so, what is $E[(-2XE[X])]$?
  $endgroup$
  – Alexis
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  E[(-2)] = -2 (averaging itself is just itself); E[X] = X; and E[(E[X])] = E[X] = X; so... 𝐸[(−2𝑋𝐸[𝑋])] just ends up being -2X and its removed from the equation because its a constant? Thanks for your patience with my dumb questions ^^;;
  $endgroup$
  – snarkshark
  5 hours ago
  
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

I am trying to understand variance, and when I look at the fully derived equation on wikipedia, I do not understand how we are able to make the jump from B to C.

When simplifying, 2(XE[X]) where does the 2X go? I understand how we get 2E[X], but I'm not sure why or how the 2X is able to be removed.

Sorry for this stupid question. I haven't taken a math class in about 7 years, and I'm trying to get back in to it... Thanks for your help :)

self-study variance

share|cite|improve this question

edited 5 hours ago

snarkshark

asked 6 hours ago

snarksharksnarkshark

82

New contributor

snarkshark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 5 hours ago

snarkshark

asked 6 hours ago

snarksharksnarkshark

82

New contributor

snarkshark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 5 hours ago

snarkshark

asked 6 hours ago

snarksharksnarkshark

82

New contributor

snarkshark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 6 hours ago

snarksharksnarkshark

82

New contributor

snarkshark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 6 hours ago

snarksharksnarkshark

82

2 Answers
2

active

oldest

votes

1

$begingroup$

You distribute $E[.]$ operator into each summand, since it is a linear operator by definition. So, the middle term is $E[2XE[X]]=2E[XE[X]]$. When calculated $E[X]$ is constant, and therefore removed from the expectation (let $E[X]=mu$):

$$2E[Xmu]=2mu E[X]=2E[X]E[X]=2E[X]^2$$

share|cite|improve this answer

edited 5 hours ago

answered 6 hours ago

gunesgunes

6,6551215

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hi! Thanks so much for taking the time to answer. A few questions: 1) Why do you put a dot in E[.] operator? Is this some kind of common notation? 2) To clarify, since E[] is essentially just an averaging equation, is that why you can move the 2 outside of the inner E[]? I don't really have the intuition for it being linear and such yet ^^;;
  $endgroup$
  – snarkshark
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @snarkshark, 1. it doesn't have a special meaning. Notationally, dot generally refers to some expression. 2. Intuitionally yes. If average of $X$ is $E[X]$, then average of $aX$ is $aE[X]$. Mathematically, Think about $E[f(X)]=sum f(x)p_X(x)$ for discrete RVs. If $f(X)=aX$, where $a$ is a constant, then $E[aX]=sum axp_X(x)=asum x p_X(x)$. Situation is similar if $sum$ is replaced by $int$ when the variable is continuous.
  $endgroup$
  – gunes
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I didn't realize that summation symbols always implied discrete :) Thanks! That makes a lot more sense now. Thank you
  $endgroup$
  – snarkshark
  5 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

The 2X doesn't disappear. When you distribute the E[] operator, it turns into 2E[X]. Thus E[... -2XE[X] + ...] becomes ... - 2E[X]E[X] + ... (because E[E[X]]=E[X]) and then the term simplifies with the other E[X]² term.

share|cite|improve this answer

answered 2 hours ago

Anony-MousseAnony-Mousse

30.7k54180

$endgroup$

add a comment | 

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1

$begingroup$

You distribute $E[.]$ operator into each summand, since it is a linear operator by definition. So, the middle term is $E[2XE[X]]=2E[XE[X]]$. When calculated $E[X]$ is constant, and therefore removed from the expectation (let $E[X]=mu$):

$$2E[Xmu]=2mu E[X]=2E[X]E[X]=2E[X]^2$$

share|cite|improve this answer

edited 5 hours ago

answered 6 hours ago

gunesgunes

6,6551215

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hi! Thanks so much for taking the time to answer. A few questions: 1) Why do you put a dot in E[.] operator? Is this some kind of common notation? 2) To clarify, since E[] is essentially just an averaging equation, is that why you can move the 2 outside of the inner E[]? I don't really have the intuition for it being linear and such yet ^^;;
  $endgroup$
  – snarkshark
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @snarkshark, 1. it doesn't have a special meaning. Notationally, dot generally refers to some expression. 2. Intuitionally yes. If average of $X$ is $E[X]$, then average of $aX$ is $aE[X]$. Mathematically, Think about $E[f(X)]=sum f(x)p_X(x)$ for discrete RVs. If $f(X)=aX$, where $a$ is a constant, then $E[aX]=sum axp_X(x)=asum x p_X(x)$. Situation is similar if $sum$ is replaced by $int$ when the variable is continuous.
  $endgroup$
  – gunes
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I didn't realize that summation symbols always implied discrete :) Thanks! That makes a lot more sense now. Thank you
  $endgroup$
  – snarkshark
  5 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

You distribute $E[.]$ operator into each summand, since it is a linear operator by definition. So, the middle term is $E[2XE[X]]=2E[XE[X]]$. When calculated $E[X]$ is constant, and therefore removed from the expectation (let $E[X]=mu$):

$$2E[Xmu]=2mu E[X]=2E[X]E[X]=2E[X]^2$$

share|cite|improve this answer

edited 5 hours ago

answered 6 hours ago

gunesgunes

6,6551215

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hi! Thanks so much for taking the time to answer. A few questions: 1) Why do you put a dot in E[.] operator? Is this some kind of common notation? 2) To clarify, since E[] is essentially just an averaging equation, is that why you can move the 2 outside of the inner E[]? I don't really have the intuition for it being linear and such yet ^^;;
  $endgroup$
  – snarkshark
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @snarkshark, 1. it doesn't have a special meaning. Notationally, dot generally refers to some expression. 2. Intuitionally yes. If average of $X$ is $E[X]$, then average of $aX$ is $aE[X]$. Mathematically, Think about $E[f(X)]=sum f(x)p_X(x)$ for discrete RVs. If $f(X)=aX$, where $a$ is a constant, then $E[aX]=sum axp_X(x)=asum x p_X(x)$. Situation is similar if $sum$ is replaced by $int$ when the variable is continuous.
  $endgroup$
  – gunes
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I didn't realize that summation symbols always implied discrete :) Thanks! That makes a lot more sense now. Thank you
  $endgroup$
  – snarkshark
  5 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

You distribute $E[.]$ operator into each summand, since it is a linear operator by definition. So, the middle term is $E[2XE[X]]=2E[XE[X]]$. When calculated $E[X]$ is constant, and therefore removed from the expectation (let $E[X]=mu$):

$$2E[Xmu]=2mu E[X]=2E[X]E[X]=2E[X]^2$$

share|cite|improve this answer

edited 5 hours ago

answered 6 hours ago

gunesgunes

6,6551215

$endgroup$

share|cite|improve this answer

edited 5 hours ago

answered 6 hours ago

gunesgunes

6,6551215

answered 6 hours ago

gunesgunes

6,6551215

answered 6 hours ago

gunesgunes

6,6551215

1

$begingroup$

The 2X doesn't disappear. When you distribute the E[] operator, it turns into 2E[X]. Thus E[... -2XE[X] + ...] becomes ... - 2E[X]E[X] + ... (because E[E[X]]=E[X]) and then the term simplifies with the other E[X]² term.

share|cite|improve this answer

answered 2 hours ago

Anony-MousseAnony-Mousse

30.7k54180

$endgroup$

add a comment | 

1

$begingroup$

The 2X doesn't disappear. When you distribute the E[] operator, it turns into 2E[X]. Thus E[... -2XE[X] + ...] becomes ... - 2E[X]E[X] + ... (because E[E[X]]=E[X]) and then the term simplifies with the other E[X]² term.

share|cite|improve this answer

answered 2 hours ago

Anony-MousseAnony-Mousse

30.7k54180

$endgroup$

add a comment | 

$begingroup$

The 2X doesn't disappear. When you distribute the E[] operator, it turns into 2E[X]. Thus E[... -2XE[X] + ...] becomes ... - 2E[X]E[X] + ... (because E[E[X]]=E[X]) and then the term simplifies with the other E[X]² term.

share|cite|improve this answer

answered 2 hours ago

Anony-MousseAnony-Mousse

30.7k54180

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Anony-MousseAnony-Mousse

30.7k54180

answered 2 hours ago

Anony-MousseAnony-Mousse

30.7k54180

answered 2 hours ago

Anony-MousseAnony-Mousse

30.7k54180