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Difference between Var(Y) and Var(Y|X)?
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What is the difference between $mathrmvar(Y)$ and $mathrmvar(Y|X)$? If $Y = c + beta X$ and $mathrmvar(X)=sigma^2$, won't both come out to be the same, i.e., $beta^2sigma^2$?
statistics variance
edited 37 mins ago
Math Lover
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up vote
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down vote
favorite
What is the difference between $mathrmvar(Y)$ and $mathrmvar(Y|X)$? If $Y = c + beta X$ and $mathrmvar(X)=sigma^2$, won't both come out to be the same, i.e., $beta^2sigma^2$?
statistics variance
edited 37 mins ago
Math Lover
12.8k21232
asked 4 hours ago
Ethan Smith
274
New contributor
Ethan Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
What is the difference between $mathrmvar(Y)$ and $mathrmvar(Y|X)$? If $Y = c + beta X$ and $mathrmvar(X)=sigma^2$, won't both come out to be the same, i.e., $beta^2sigma^2$?
statistics variance
edited 37 mins ago
Math Lover
12.8k21232
asked 4 hours ago
Ethan Smith
274
New contributor
Ethan Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What is the difference between $mathrmvar(Y)$ and $mathrmvar(Y|X)$? If $Y = c + beta X$ and $mathrmvar(X)=sigma^2$, won't both come out to be the same, i.e., $beta^2sigma^2$?
statistics variance
statistics variance
edited 37 mins ago
Math Lover
12.8k21232
asked 4 hours ago
Ethan Smith
274
New contributor
Ethan Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 37 mins ago
Math Lover
12.8k21232
asked 4 hours ago
Ethan Smith
274
New contributor
Ethan Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 37 mins ago
Math Lover
12.8k21232
edited 37 mins ago
Math Lover
12.8k21232
edited 37 mins ago
Math Lover
12.8k21232
12.8k21232
asked 4 hours ago
Ethan Smith
274
New contributor
Ethan Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Ethan Smith
274
asked 4 hours ago
Ethan Smith
274
274
New contributor
Ethan Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ethan Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ethan Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
Note that we always have $E[f(X)|X] = f(X)$. Loosely speaking, given $X$ there is no randomness in $f(X)$, so we expect the conditional variance to be zero.
Let $f(x) = c+ beta x$.
begineqnarray
operatornamevar (f(X)|X) &=& E [ (f(X)-E[f(X)|X])^2 | X] \
&=& E [ (f(X)-f(X))^2 | X] \
&=& E[0 | X] \
&=& 0
endeqnarray
On the other hand (note that in this case we have $E[f(X)] = f(EX)$).
begineqnarray
operatornamevar (f(X)) &=& E [ (f(X)-E[f(X)])^2 ] \
&=& E [ (f(X)-f(EX))^2] \
&=& E[(beta(X-EX))^2] \
&=& beta ^2 E[(X-EX)^2] \
&=& beta^2 operatornamevar X \
&=& beta^2 sigma^2
endeqnarray
answered 3 hours ago
copper.hat
124k558156
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up vote
2
down vote
$operatornameVar(Y)=mathbbE(Y-mathbbE(Y))^2$ is a number. Var(Y|X) is the conditional variance of $Y$ given $X$:
$$
operatornameVar(Ymid X):=mathbbE[(Y-mathbbE[Ymid X])^2mid X]
$$
is a function of the random variable $X$.
answered 3 hours ago
user10354138
1,5125
add a comment |Â
up vote
2
down vote
Note that given the value of $X$, $Y$ ceases to be random (presuming $beta$ and $c$ are constants). Therefore, $$mathrmvar(Y|X) = 0.$$
On the other hand, if $X$ is not known then $Y$ can take on different values. Therefore, $mathrmvar(Y) neq 0$. In fact, $mathrmvar(Y) = beta^2 mathrmvar(X)=beta^2 sigma^2$.
answered 3 hours ago
Math Lover
12.8k21232
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Note that we always have $E[f(X)|X] = f(X)$. Loosely speaking, given $X$ there is no randomness in $f(X)$, so we expect the conditional variance to be zero.
Let $f(x) = c+ beta x$.
begineqnarray
operatornamevar (f(X)|X) &=& E [ (f(X)-E[f(X)|X])^2 | X] \
&=& E [ (f(X)-f(X))^2 | X] \
&=& E[0 | X] \
&=& 0
endeqnarray
On the other hand (note that in this case we have $E[f(X)] = f(EX)$).
begineqnarray
operatornamevar (f(X)) &=& E [ (f(X)-E[f(X)])^2 ] \
&=& E [ (f(X)-f(EX))^2] \
&=& E[(beta(X-EX))^2] \
&=& beta ^2 E[(X-EX)^2] \
&=& beta^2 operatornamevar X \
&=& beta^2 sigma^2
endeqnarray
answered 3 hours ago
copper.hat
124k558156
add a comment |Â
up vote
4
down vote
accepted
Note that we always have $E[f(X)|X] = f(X)$. Loosely speaking, given $X$ there is no randomness in $f(X)$, so we expect the conditional variance to be zero.
Let $f(x) = c+ beta x$.
begineqnarray
operatornamevar (f(X)|X) &=& E [ (f(X)-E[f(X)|X])^2 | X] \
&=& E [ (f(X)-f(X))^2 | X] \
&=& E[0 | X] \
&=& 0
endeqnarray
On the other hand (note that in this case we have $E[f(X)] = f(EX)$).
begineqnarray
operatornamevar (f(X)) &=& E [ (f(X)-E[f(X)])^2 ] \
&=& E [ (f(X)-f(EX))^2] \
&=& E[(beta(X-EX))^2] \
&=& beta ^2 E[(X-EX)^2] \
&=& beta^2 operatornamevar X \
&=& beta^2 sigma^2
endeqnarray
answered 3 hours ago
copper.hat
124k558156
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Note that we always have $E[f(X)|X] = f(X)$. Loosely speaking, given $X$ there is no randomness in $f(X)$, so we expect the conditional variance to be zero.
Let $f(x) = c+ beta x$.
begineqnarray
operatornamevar (f(X)|X) &=& E [ (f(X)-E[f(X)|X])^2 | X] \
&=& E [ (f(X)-f(X))^2 | X] \
&=& E[0 | X] \
&=& 0
endeqnarray
On the other hand (note that in this case we have $E[f(X)] = f(EX)$).
begineqnarray
operatornamevar (f(X)) &=& E [ (f(X)-E[f(X)])^2 ] \
&=& E [ (f(X)-f(EX))^2] \
&=& E[(beta(X-EX))^2] \
&=& beta ^2 E[(X-EX)^2] \
&=& beta^2 operatornamevar X \
&=& beta^2 sigma^2
endeqnarray
answered 3 hours ago
copper.hat
124k558156
Note that we always have $E[f(X)|X] = f(X)$. Loosely speaking, given $X$ there is no randomness in $f(X)$, so we expect the conditional variance to be zero.
Let $f(x) = c+ beta x$.
begineqnarray
operatornamevar (f(X)|X) &=& E [ (f(X)-E[f(X)|X])^2 | X] \
&=& E [ (f(X)-f(X))^2 | X] \
&=& E[0 | X] \
&=& 0
endeqnarray
On the other hand (note that in this case we have $E[f(X)] = f(EX)$).
begineqnarray
operatornamevar (f(X)) &=& E [ (f(X)-E[f(X)])^2 ] \
&=& E [ (f(X)-f(EX))^2] \
&=& E[(beta(X-EX))^2] \
&=& beta ^2 E[(X-EX)^2] \
&=& beta^2 operatornamevar X \
&=& beta^2 sigma^2
endeqnarray
answered 3 hours ago
copper.hat
124k558156
edited 3 hours ago
answered 3 hours ago
copper.hat
124k558156
answered 3 hours ago
copper.hat
124k558156
answered 3 hours ago
copper.hat
124k558156
124k558156
add a comment |Â
add a comment |Â
up vote
2
down vote
$operatornameVar(Y)=mathbbE(Y-mathbbE(Y))^2$ is a number. Var(Y|X) is the conditional variance of $Y$ given $X$:
$$
operatornameVar(Ymid X):=mathbbE[(Y-mathbbE[Ymid X])^2mid X]
$$
is a function of the random variable $X$.
answered 3 hours ago
user10354138
1,5125
add a comment |Â
up vote
2
down vote
$operatornameVar(Y)=mathbbE(Y-mathbbE(Y))^2$ is a number. Var(Y|X) is the conditional variance of $Y$ given $X$:
$$
operatornameVar(Ymid X):=mathbbE[(Y-mathbbE[Ymid X])^2mid X]
$$
is a function of the random variable $X$.
answered 3 hours ago
user10354138
1,5125
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$operatornameVar(Y)=mathbbE(Y-mathbbE(Y))^2$ is a number. Var(Y|X) is the conditional variance of $Y$ given $X$:
$$
operatornameVar(Ymid X):=mathbbE[(Y-mathbbE[Ymid X])^2mid X]
$$
is a function of the random variable $X$.
answered 3 hours ago
user10354138
1,5125
$operatornameVar(Y)=mathbbE(Y-mathbbE(Y))^2$ is a number. Var(Y|X) is the conditional variance of $Y$ given $X$:
$$
operatornameVar(Ymid X):=mathbbE[(Y-mathbbE[Ymid X])^2mid X]
$$
is a function of the random variable $X$.
answered 3 hours ago
user10354138
1,5125
answered 3 hours ago
user10354138
1,5125
answered 3 hours ago
user10354138
1,5125
answered 3 hours ago
user10354138
1,5125
1,5125
add a comment |Â
add a comment |Â
up vote
2
down vote
Note that given the value of $X$, $Y$ ceases to be random (presuming $beta$ and $c$ are constants). Therefore, $$mathrmvar(Y|X) = 0.$$
On the other hand, if $X$ is not known then $Y$ can take on different values. Therefore, $mathrmvar(Y) neq 0$. In fact, $mathrmvar(Y) = beta^2 mathrmvar(X)=beta^2 sigma^2$.
answered 3 hours ago
Math Lover
12.8k21232
add a comment |Â
up vote
2
down vote
Note that given the value of $X$, $Y$ ceases to be random (presuming $beta$ and $c$ are constants). Therefore, $$mathrmvar(Y|X) = 0.$$
On the other hand, if $X$ is not known then $Y$ can take on different values. Therefore, $mathrmvar(Y) neq 0$. In fact, $mathrmvar(Y) = beta^2 mathrmvar(X)=beta^2 sigma^2$.
answered 3 hours ago
Math Lover
12.8k21232
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Note that given the value of $X$, $Y$ ceases to be random (presuming $beta$ and $c$ are constants). Therefore, $$mathrmvar(Y|X) = 0.$$
On the other hand, if $X$ is not known then $Y$ can take on different values. Therefore, $mathrmvar(Y) neq 0$. In fact, $mathrmvar(Y) = beta^2 mathrmvar(X)=beta^2 sigma^2$.
answered 3 hours ago
Math Lover
12.8k21232
Note that given the value of $X$, $Y$ ceases to be random (presuming $beta$ and $c$ are constants). Therefore, $$mathrmvar(Y|X) = 0.$$
On the other hand, if $X$ is not known then $Y$ can take on different values. Therefore, $mathrmvar(Y) neq 0$. In fact, $mathrmvar(Y) = beta^2 mathrmvar(X)=beta^2 sigma^2$.
answered 3 hours ago
Math Lover
12.8k21232
answered 3 hours ago
Math Lover
12.8k21232
answered 3 hours ago
Math Lover
12.8k21232
answered 3 hours ago
Math Lover
12.8k21232
12.8k21232
add a comment |Â
add a comment |Â
Ethan Smith is a new contributor. Be nice, and check out our Code of Conduct.
Ethan Smith is a new contributor. Be nice, and check out our Code of Conduct.
Ethan Smith is a new contributor. Be nice, and check out our Code of Conduct.
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