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
asked Mar 31 at 16:07
snarksharksnarkshark
82
New contributor
snarkshark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
|
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
asked Mar 31 at 16:07
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
Mar 31 at 16:18
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
Mar 31 at 16:21
$begingroup$
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
Mar 31 at 16:21
$begingroup$
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
Mar 31 at 16:23
$begingroup$
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
Mar 31 at 16:37
|
show 1 more comment
1
1
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
asked Mar 31 at 16:07
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.
$endgroup$
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
self-study variance
asked Mar 31 at 16:07
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 Mar 31 at 16:07
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.
edited Mar 31 at 16:26
snarkshark
asked Mar 31 at 16:07
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 Mar 31 at 16:07
snarksharksnarkshark
82
asked Mar 31 at 16:07
snarksharksnarkshark
82
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.
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.
snarkshark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
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
Mar 31 at 16:18
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
Mar 31 at 16:21
$begingroup$
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
Mar 31 at 16:21
$begingroup$
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
Mar 31 at 16:23
$begingroup$
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
Mar 31 at 16:37
|
show 1 more comment
$begingroup$
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
Mar 31 at 16:18
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
Mar 31 at 16:21
$begingroup$
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
Mar 31 at 16:21
$begingroup$
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
Mar 31 at 16:23
$begingroup$
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
Mar 31 at 16:37
$begingroup$
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
Mar 31 at 16:18
$begingroup$
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
Mar 31 at 16:18
1
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
Mar 31 at 16:21
$begingroup$
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
Mar 31 at 16:21
$begingroup$
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
Mar 31 at 16:21
$begingroup$
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
Mar 31 at 16:21
$begingroup$
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
Mar 31 at 16:23
$begingroup$
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
Mar 31 at 16:23
$begingroup$
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
Mar 31 at 16:37
$begingroup$
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
Mar 31 at 16:37
|
show 1 more comment
2 Answers
2
active
oldest
votes
2
$begingroup$
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 $E[X]=mu$):
$$2E[Xmu]=2mu E[X]=2E[X]E[X]=2E[X]^2$$
answered Mar 31 at 16:20
gunesgunes
6,7951215
$endgroup$
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
Mar 31 at 16:32
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
Mar 31 at 16:36
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
Mar 31 at 16:38
add a comment |
2
$begingroup$
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.
answered Mar 31 at 19:48
Anony-MousseAnony-Mousse
30.7k54181
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
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 $E[X]=mu$):
$$2E[Xmu]=2mu E[X]=2E[X]E[X]=2E[X]^2$$
answered Mar 31 at 16:20
gunesgunes
6,7951215
$endgroup$
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
Mar 31 at 16:32
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
Mar 31 at 16:36
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
Mar 31 at 16:38
add a comment |
2
$begingroup$
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 $E[X]=mu$):
$$2E[Xmu]=2mu E[X]=2E[X]E[X]=2E[X]^2$$
answered Mar 31 at 16:20
gunesgunes
6,7951215
$endgroup$
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
Mar 31 at 16:32
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
Mar 31 at 16:36
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
Mar 31 at 16:38
add a comment |
2
2
2
$begingroup$
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 $E[X]=mu$):
$$2E[Xmu]=2mu E[X]=2E[X]E[X]=2E[X]^2$$
answered Mar 31 at 16:20
gunesgunes
6,7951215
$endgroup$
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 $E[X]=mu$):
$$2E[Xmu]=2mu E[X]=2E[X]E[X]=2E[X]^2$$
answered Mar 31 at 16:20
gunesgunes
6,7951215
edited Mar 31 at 16:28
answered Mar 31 at 16:20
gunesgunes
6,7951215
answered Mar 31 at 16:20
gunesgunes
6,7951215
answered Mar 31 at 16:20
gunesgunes
6,7951215
6,7951215
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
Mar 31 at 16:32
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
Mar 31 at 16:36
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
Mar 31 at 16:38
add a comment |
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
Mar 31 at 16:32
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
Mar 31 at 16:36
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
Mar 31 at 16:38
1
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
Mar 31 at 16:32
$begingroup$
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
Mar 31 at 16:32
1
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
Mar 31 at 16:36
$begingroup$
@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
Mar 31 at 16:36
1
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
Mar 31 at 16:38
$begingroup$
Ah, I didn't realize that summation symbols always implied discrete :) Thanks! That makes a lot more sense now. Thank you
$endgroup$
– snarkshark
Mar 31 at 16:38
add a comment |
2
$begingroup$
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.
answered Mar 31 at 19:48
Anony-MousseAnony-Mousse
30.7k54181
$endgroup$
add a comment |
2
$begingroup$
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.
answered Mar 31 at 19:48
Anony-MousseAnony-Mousse
30.7k54181
$endgroup$
add a comment |
2
2
2
$begingroup$
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.
answered Mar 31 at 19:48
Anony-MousseAnony-Mousse
30.7k54181
$endgroup$
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.
answered Mar 31 at 19:48
Anony-MousseAnony-Mousse
30.7k54181
answered Mar 31 at 19:48
Anony-MousseAnony-Mousse
30.7k54181
answered Mar 31 at 19:48
Anony-MousseAnony-Mousse
30.7k54181
answered Mar 31 at 19:48
Anony-MousseAnony-Mousse
30.7k54181
30.7k54181
add a comment |
add a comment |
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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!)
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– Alexis
Mar 31 at 16:18
1
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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 :'(
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– snarkshark
Mar 31 at 16:21
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As an aside, you might read about the self-study tag, and edit your question (the "edit" link at lower left) to inlcude it.
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– Alexis
Mar 31 at 16:21
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Well simplifying does enter thereafter. What is $E[(-2)]$? What is $E[(X)]$? What is $E[(E[X])]$? And so, what is $E[(-2XE[X])]$?
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– Alexis
Mar 31 at 16:23
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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 ^^;;
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– snarkshark
Mar 31 at 16:37