Prove $mathbbV(Y) = mathbbEmathbbV(Y mid X) + mathbbVmathbbE(Y mid X)$ The 2019 Stack Overflow Developer Survey Results Are InIndependence of $min(X,Y)$ and $|X-Y|$ when $(X,Y)$ are i.i.d. exponentialA problem of regular distributionConditional expectation (mixed with an iterated expectation) $E[E(Xmid Y)mid Y]=E(Xmid Y)$Finding conditional expectation $mathbbE(Xmid (X-0.5)^2)$Prove $E[XE[YmidmathcalG]] = E[YE[XmidmathcalG]]$Breaking sides of equation to prove a probability.How to prove that a conditional pdf sums to 1?How to show that $f(x,y,mumid a,b,Sigma) = f(x,ymid mu, a,b,Sigma)cdot p(mu)$ if $(x,y)$ are MVN and $a,b,Sigma$ are hyperparameters?Gambler's ruin: $mathbbPleft X_tau=n mid X_1 = k+1right=mathbbPleft X_tau = nmid X_0 = k+1right,?$Showing $E(S^2mid bar X)=bar X$ for i.i.d Poisson random variables $X_i$
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Prove $mathbbV(Y) = mathbbEmathbbV(Y mid X) + mathbbVmathbbE(Y mid X)$
The 2019 Stack Overflow Developer Survey Results Are InIndependence of $min(X,Y)$ and $|X-Y|$ when $(X,Y)$ are i.i.d. exponentialA problem of regular distributionConditional expectation (mixed with an iterated expectation) $E[E(Xmid Y)mid Y]=E(Xmid Y)$Finding conditional expectation $mathbbE(Xmid (X-0.5)^2)$Prove $E[XE[YmidmathcalG]] = E[YE[XmidmathcalG]]$Breaking sides of equation to prove a probability.How to prove that a conditional pdf sums to 1?How to show that $f(x,y,mumid a,b,Sigma) = f(x,ymid mu, a,b,Sigma)cdot p(mu)$ if $(x,y)$ are MVN and $a,b,Sigma$ are hyperparameters?Gambler's ruin: $mathbbPleft X_tau=n mid X_1 = k+1right=mathbbPleft X_tau = nmid X_0 = k+1right,?$Showing $E(S^2mid bar X)=bar X$ for i.i.d Poisson random variables $X_i$
$begingroup$
Trying to figure out how to prove this statement
$mathbbV(Y) = mathbbEV(Ymid X) + mathbbVE(Ymid X)$
Tried expanding RHS as follows:
beginalign&= mathbbEV(Ymid X)+mathbbVE(Ymid X)\
&= mathbbE[int [y-mathbbE(Ymid X)]^2f(ymid x),dy] + int[mathbbE(Ymid X)-mathbbEmathbbE(Ymid X)]^2 f(x),dx \
&= intint (y-mathbbE(Y|X)^2f(y|x)dyf(x),dx + int[mathbbE(Y|X)-mathbbE(Y)]^2 f(x),dx;text since ;mathbbEE(Y|X) = mathbbE(Y)\
&= intint[y^2 -2ymathbbE(Y|X) + mathbbE(Y|X)^2]f(x,y),dy,dx + int[mathbbE(Y|X)^2 -2mathbbE(Y)mathbbE(Y|X)+mathbbE(Y)^2]f(x),dx\
&= cdots
endalign
probability statistics
edited Mar 23 at 0:03
Peiffap
10312
asked Mar 22 at 23:09
ZanderZander
384
$endgroup$
add a comment |
$begingroup$
Trying to figure out how to prove this statement
$mathbbV(Y) = mathbbEV(Ymid X) + mathbbVE(Ymid X)$
Tried expanding RHS as follows:
beginalign&= mathbbEV(Ymid X)+mathbbVE(Ymid X)\
&= mathbbE[int [y-mathbbE(Ymid X)]^2f(ymid x),dy] + int[mathbbE(Ymid X)-mathbbEmathbbE(Ymid X)]^2 f(x),dx \
&= intint (y-mathbbE(Y|X)^2f(y|x)dyf(x),dx + int[mathbbE(Y|X)-mathbbE(Y)]^2 f(x),dx;text since ;mathbbEE(Y|X) = mathbbE(Y)\
&= intint[y^2 -2ymathbbE(Y|X) + mathbbE(Y|X)^2]f(x,y),dy,dx + int[mathbbE(Y|X)^2 -2mathbbE(Y)mathbbE(Y|X)+mathbbE(Y)^2]f(x),dx\
&= cdots
endalign
probability statistics
edited Mar 23 at 0:03
Peiffap
10312
asked Mar 22 at 23:09
ZanderZander
384
$endgroup$
4
$begingroup$
This is the law of total variance. A proof can be found here: en.wikipedia.org/wiki/Law_of_total_variance#Proof.
$endgroup$
– Minus One-Twelfth
Mar 22 at 23:11
add a comment |
$begingroup$
Trying to figure out how to prove this statement
$mathbbV(Y) = mathbbEV(Ymid X) + mathbbVE(Ymid X)$
Tried expanding RHS as follows:
beginalign&= mathbbEV(Ymid X)+mathbbVE(Ymid X)\
&= mathbbE[int [y-mathbbE(Ymid X)]^2f(ymid x),dy] + int[mathbbE(Ymid X)-mathbbEmathbbE(Ymid X)]^2 f(x),dx \
&= intint (y-mathbbE(Y|X)^2f(y|x)dyf(x),dx + int[mathbbE(Y|X)-mathbbE(Y)]^2 f(x),dx;text since ;mathbbEE(Y|X) = mathbbE(Y)\
&= intint[y^2 -2ymathbbE(Y|X) + mathbbE(Y|X)^2]f(x,y),dy,dx + int[mathbbE(Y|X)^2 -2mathbbE(Y)mathbbE(Y|X)+mathbbE(Y)^2]f(x),dx\
&= cdots
endalign
probability statistics
edited Mar 23 at 0:03
Peiffap
10312
asked Mar 22 at 23:09
ZanderZander
384
$endgroup$
Trying to figure out how to prove this statement
$mathbbV(Y) = mathbbEV(Ymid X) + mathbbVE(Ymid X)$
Tried expanding RHS as follows:
beginalign&= mathbbEV(Ymid X)+mathbbVE(Ymid X)\
&= mathbbE[int [y-mathbbE(Ymid X)]^2f(ymid x),dy] + int[mathbbE(Ymid X)-mathbbEmathbbE(Ymid X)]^2 f(x),dx \
&= intint (y-mathbbE(Y|X)^2f(y|x)dyf(x),dx + int[mathbbE(Y|X)-mathbbE(Y)]^2 f(x),dx;text since ;mathbbEE(Y|X) = mathbbE(Y)\
&= intint[y^2 -2ymathbbE(Y|X) + mathbbE(Y|X)^2]f(x,y),dy,dx + int[mathbbE(Y|X)^2 -2mathbbE(Y)mathbbE(Y|X)+mathbbE(Y)^2]f(x),dx\
&= cdots
endalign
probability statistics
probability statistics
edited Mar 23 at 0:03
Peiffap
10312
asked Mar 22 at 23:09
ZanderZander
384
edited Mar 23 at 0:03
Peiffap
10312
asked Mar 22 at 23:09
ZanderZander
384
edited Mar 23 at 0:03
Peiffap
10312
edited Mar 23 at 0:03
Peiffap
10312
edited Mar 23 at 0:03
Peiffap
10312
10312
asked Mar 22 at 23:09
ZanderZander
384
asked Mar 22 at 23:09
ZanderZander
384
asked Mar 22 at 23:09
ZanderZander
384
384
4
$begingroup$
This is the law of total variance. A proof can be found here: en.wikipedia.org/wiki/Law_of_total_variance#Proof.
$endgroup$
– Minus One-Twelfth
Mar 22 at 23:11
add a comment |
4
$begingroup$
This is the law of total variance. A proof can be found here: en.wikipedia.org/wiki/Law_of_total_variance#Proof.
$endgroup$
– Minus One-Twelfth
Mar 22 at 23:11
4
4
$begingroup$
This is the law of total variance. A proof can be found here: en.wikipedia.org/wiki/Law_of_total_variance#Proof.
$endgroup$
– Minus One-Twelfth
Mar 22 at 23:11
$begingroup$
This is the law of total variance. A proof can be found here: en.wikipedia.org/wiki/Law_of_total_variance#Proof.
$endgroup$
– Minus One-Twelfth
Mar 22 at 23:11
add a comment |
1 Answer
1
active
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votes
$begingroup$
As @Minus One-Twelfth mentioned, this is the law of total variance.
Proof
beginalign
mathbbV[Y] &= mathbbE[Y^2] - mathbbE[Y]^2\
&= mathbbEBig[mathbbE[Y^2 mid X]Big] - mathbbEBig[mathbbE[Y mid X]Big]^2\
&= mathbbEBig[mathbbV[Y mid X] + mathbbE[Y mid X]^2Big] - mathbbEBig[mathbbE[Y mid X]Big]^2\
&= mathbbEBig[mathbbV[Y mid X]Big] + mathbbEBig[mathbbE[Y mid X]^2Big] - mathbbEBig[mathbbE[Y mid X]Big]^2\
&= mathbbEBig[mathbbV[Y mid X]Big] + mathbbVBig[mathbbE[Y mid X]Big],.
endalign
The first step is just the definition of the variance;
the second step is introducing an independent variable $X$,
the third step then applies the definition of variance, but ``the other way around''.
The fourth step uses the linearity of the expected value operator,
and the fifth and final step then reapplies the definition of the variance to group the rightmost terms into a single variance term.
answered Mar 22 at 23:33
PeiffapPeiffap
10312
$endgroup$
add a comment |
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$begingroup$
As @Minus One-Twelfth mentioned, this is the law of total variance.
Proof
beginalign
mathbbV[Y] &= mathbbE[Y^2] - mathbbE[Y]^2\
&= mathbbEBig[mathbbE[Y^2 mid X]Big] - mathbbEBig[mathbbE[Y mid X]Big]^2\
&= mathbbEBig[mathbbV[Y mid X] + mathbbE[Y mid X]^2Big] - mathbbEBig[mathbbE[Y mid X]Big]^2\
&= mathbbEBig[mathbbV[Y mid X]Big] + mathbbEBig[mathbbE[Y mid X]^2Big] - mathbbEBig[mathbbE[Y mid X]Big]^2\
&= mathbbEBig[mathbbV[Y mid X]Big] + mathbbVBig[mathbbE[Y mid X]Big],.
endalign
The first step is just the definition of the variance;
the second step is introducing an independent variable $X$,
the third step then applies the definition of variance, but ``the other way around''.
The fourth step uses the linearity of the expected value operator,
and the fifth and final step then reapplies the definition of the variance to group the rightmost terms into a single variance term.
answered Mar 22 at 23:33
PeiffapPeiffap
10312
$endgroup$
add a comment |
$begingroup$
As @Minus One-Twelfth mentioned, this is the law of total variance.
Proof
beginalign
mathbbV[Y] &= mathbbE[Y^2] - mathbbE[Y]^2\
&= mathbbEBig[mathbbE[Y^2 mid X]Big] - mathbbEBig[mathbbE[Y mid X]Big]^2\
&= mathbbEBig[mathbbV[Y mid X] + mathbbE[Y mid X]^2Big] - mathbbEBig[mathbbE[Y mid X]Big]^2\
&= mathbbEBig[mathbbV[Y mid X]Big] + mathbbEBig[mathbbE[Y mid X]^2Big] - mathbbEBig[mathbbE[Y mid X]Big]^2\
&= mathbbEBig[mathbbV[Y mid X]Big] + mathbbVBig[mathbbE[Y mid X]Big],.
endalign
The first step is just the definition of the variance;
the second step is introducing an independent variable $X$,
the third step then applies the definition of variance, but ``the other way around''.
The fourth step uses the linearity of the expected value operator,
and the fifth and final step then reapplies the definition of the variance to group the rightmost terms into a single variance term.
answered Mar 22 at 23:33
PeiffapPeiffap
10312
$endgroup$
add a comment |
$begingroup$
As @Minus One-Twelfth mentioned, this is the law of total variance.
Proof
beginalign
mathbbV[Y] &= mathbbE[Y^2] - mathbbE[Y]^2\
&= mathbbEBig[mathbbE[Y^2 mid X]Big] - mathbbEBig[mathbbE[Y mid X]Big]^2\
&= mathbbEBig[mathbbV[Y mid X] + mathbbE[Y mid X]^2Big] - mathbbEBig[mathbbE[Y mid X]Big]^2\
&= mathbbEBig[mathbbV[Y mid X]Big] + mathbbEBig[mathbbE[Y mid X]^2Big] - mathbbEBig[mathbbE[Y mid X]Big]^2\
&= mathbbEBig[mathbbV[Y mid X]Big] + mathbbVBig[mathbbE[Y mid X]Big],.
endalign
The first step is just the definition of the variance;
the second step is introducing an independent variable $X$,
the third step then applies the definition of variance, but ``the other way around''.
The fourth step uses the linearity of the expected value operator,
and the fifth and final step then reapplies the definition of the variance to group the rightmost terms into a single variance term.
answered Mar 22 at 23:33
PeiffapPeiffap
10312
$endgroup$
As @Minus One-Twelfth mentioned, this is the law of total variance.
Proof
beginalign
mathbbV[Y] &= mathbbE[Y^2] - mathbbE[Y]^2\
&= mathbbEBig[mathbbE[Y^2 mid X]Big] - mathbbEBig[mathbbE[Y mid X]Big]^2\
&= mathbbEBig[mathbbV[Y mid X] + mathbbE[Y mid X]^2Big] - mathbbEBig[mathbbE[Y mid X]Big]^2\
&= mathbbEBig[mathbbV[Y mid X]Big] + mathbbEBig[mathbbE[Y mid X]^2Big] - mathbbEBig[mathbbE[Y mid X]Big]^2\
&= mathbbEBig[mathbbV[Y mid X]Big] + mathbbVBig[mathbbE[Y mid X]Big],.
endalign
The first step is just the definition of the variance;
the second step is introducing an independent variable $X$,
the third step then applies the definition of variance, but ``the other way around''.
The fourth step uses the linearity of the expected value operator,
and the fifth and final step then reapplies the definition of the variance to group the rightmost terms into a single variance term.
answered Mar 22 at 23:33
PeiffapPeiffap
10312
edited Mar 22 at 23:39
answered Mar 22 at 23:33
PeiffapPeiffap
10312
answered Mar 22 at 23:33
PeiffapPeiffap
10312
answered Mar 22 at 23:33
PeiffapPeiffap
10312
10312
add a comment |
add a comment |
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$begingroup$
This is the law of total variance. A proof can be found here: en.wikipedia.org/wiki/Law_of_total_variance#Proof.
$endgroup$
– Minus One-Twelfth
Mar 22 at 23:11