Finding $Var(X)$ from conditional PDF












0












$begingroup$


Let $X$ and $Y$ be random variables such that $X vert Y=y$ is normal distributed as $N(y,1)$ and Y is a continues random variable with PDF:



$f_Y(y)=3y^2$



for $0<y<1$



and $0$ otherwise.



Find $Var(X)$




My idea is to find the marginal PDF for X and from there it's easy to find the variance. Now since the conditional is normal distributed, I should be able to find the simulatanous PDF by multiplying the the conditional PDF by the marginal PDF for Y. However when I try to use integration to find the marginal PDF for X with limits 0 and infinity, I get an uncomputable integral.... Why is this happening and/or is there another easier method? Thanks in advance!






probability normal-distribution conditional-probability







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$endgroup$












  • $begingroup$
    You can find the variance from $E(X)=E, [E(Xmid Y)]$ and $E(X^2)=E, [E(X^2mid Y)]$.
    $endgroup$
    – StubbornAtom
    Jan 8 at 14:58
















0












$begingroup$


Let $X$ and $Y$ be random variables such that $X vert Y=y$ is normal distributed as $N(y,1)$ and Y is a continues random variable with PDF:



$f_Y(y)=3y^2$



for $0<y<1$



and $0$ otherwise.



Find $Var(X)$




My idea is to find the marginal PDF for X and from there it's easy to find the variance. Now since the conditional is normal distributed, I should be able to find the simulatanous PDF by multiplying the the conditional PDF by the marginal PDF for Y. However when I try to use integration to find the marginal PDF for X with limits 0 and infinity, I get an uncomputable integral.... Why is this happening and/or is there another easier method? Thanks in advance!






probability normal-distribution conditional-probability







share|cite|improve this question









$endgroup$












  • $begingroup$
    You can find the variance from $E(X)=E, [E(Xmid Y)]$ and $E(X^2)=E, [E(X^2mid Y)]$.
    $endgroup$
    – StubbornAtom
    Jan 8 at 14:58














0












0








0





$begingroup$


Let $X$ and $Y$ be random variables such that $X vert Y=y$ is normal distributed as $N(y,1)$ and Y is a continues random variable with PDF:



$f_Y(y)=3y^2$



for $0<y<1$



and $0$ otherwise.



Find $Var(X)$




My idea is to find the marginal PDF for X and from there it's easy to find the variance. Now since the conditional is normal distributed, I should be able to find the simulatanous PDF by multiplying the the conditional PDF by the marginal PDF for Y. However when I try to use integration to find the marginal PDF for X with limits 0 and infinity, I get an uncomputable integral.... Why is this happening and/or is there another easier method? Thanks in advance!






probability normal-distribution conditional-probability







share|cite|improve this question









$endgroup$




Let $X$ and $Y$ be random variables such that $X vert Y=y$ is normal distributed as $N(y,1)$ and Y is a continues random variable with PDF:



$f_Y(y)=3y^2$



for $0<y<1$



and $0$ otherwise.



Find $Var(X)$




My idea is to find the marginal PDF for X and from there it's easy to find the variance. Now since the conditional is normal distributed, I should be able to find the simulatanous PDF by multiplying the the conditional PDF by the marginal PDF for Y. However when I try to use integration to find the marginal PDF for X with limits 0 and infinity, I get an uncomputable integral.... Why is this happening and/or is there another easier method? Thanks in advance!






probability normal-distribution conditional-probability




probability normal-distribution conditional-probability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 8 at 14:39









CruZCruZ

406




406












  • $begingroup$
    You can find the variance from $E(X)=E, [E(Xmid Y)]$ and $E(X^2)=E, [E(X^2mid Y)]$.
    $endgroup$
    – StubbornAtom
    Jan 8 at 14:58


















  • $begingroup$
    You can find the variance from $E(X)=E, [E(Xmid Y)]$ and $E(X^2)=E, [E(X^2mid Y)]$.
    $endgroup$
    – StubbornAtom
    Jan 8 at 14:58
















$begingroup$
You can find the variance from $E(X)=E, [E(Xmid Y)]$ and $E(X^2)=E, [E(X^2mid Y)]$.
$endgroup$
– StubbornAtom
Jan 8 at 14:58




$begingroup$
You can find the variance from $E(X)=E, [E(Xmid Y)]$ and $E(X^2)=E, [E(X^2mid Y)]$.
$endgroup$
– StubbornAtom
Jan 8 at 14:58










1 Answer
1






active

oldest

votes


















1












$begingroup$

Use law of total variance,
$$mathbb{Var}(X)=mathbb{E}[mathbb{Var}(X|Y)]+mathbb{Var}(mathbb{E}[X|Y])$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay so E[Var(XIY)] must be E[1]=1 since we know it's normal distributed with variance = 1. But how do I calculate Var(E[XIY])...? Is it using LOTUS? And wouldn't I also need to use Y's PDF in some way?
    $endgroup$
    – CruZ
    Jan 8 at 16:54






  • 1




    $begingroup$
    @CruZ You are almost there. Think what random variable $mathbb{E}[X|Y]$ is, and find its variance.
    $endgroup$
    – kludg
    Jan 8 at 17:04












  • $begingroup$
    Okay so E[XIY]=y so I just have to calculate Var(Y)=E[Y^2]-(E[Y])^2? And I can do that using LOTUS! Is that correct?
    $endgroup$
    – CruZ
    Jan 8 at 17:21










  • $begingroup$
    Looks reasonable to me, I would solve it so.
    $endgroup$
    – kludg
    Jan 8 at 17:24











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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Use law of total variance,
$$mathbb{Var}(X)=mathbb{E}[mathbb{Var}(X|Y)]+mathbb{Var}(mathbb{E}[X|Y])$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay so E[Var(XIY)] must be E[1]=1 since we know it's normal distributed with variance = 1. But how do I calculate Var(E[XIY])...? Is it using LOTUS? And wouldn't I also need to use Y's PDF in some way?
    $endgroup$
    – CruZ
    Jan 8 at 16:54






  • 1




    $begingroup$
    @CruZ You are almost there. Think what random variable $mathbb{E}[X|Y]$ is, and find its variance.
    $endgroup$
    – kludg
    Jan 8 at 17:04












  • $begingroup$
    Okay so E[XIY]=y so I just have to calculate Var(Y)=E[Y^2]-(E[Y])^2? And I can do that using LOTUS! Is that correct?
    $endgroup$
    – CruZ
    Jan 8 at 17:21










  • $begingroup$
    Looks reasonable to me, I would solve it so.
    $endgroup$
    – kludg
    Jan 8 at 17:24
















1












$begingroup$

Use law of total variance,
$$mathbb{Var}(X)=mathbb{E}[mathbb{Var}(X|Y)]+mathbb{Var}(mathbb{E}[X|Y])$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay so E[Var(XIY)] must be E[1]=1 since we know it's normal distributed with variance = 1. But how do I calculate Var(E[XIY])...? Is it using LOTUS? And wouldn't I also need to use Y's PDF in some way?
    $endgroup$
    – CruZ
    Jan 8 at 16:54






  • 1




    $begingroup$
    @CruZ You are almost there. Think what random variable $mathbb{E}[X|Y]$ is, and find its variance.
    $endgroup$
    – kludg
    Jan 8 at 17:04












  • $begingroup$
    Okay so E[XIY]=y so I just have to calculate Var(Y)=E[Y^2]-(E[Y])^2? And I can do that using LOTUS! Is that correct?
    $endgroup$
    – CruZ
    Jan 8 at 17:21










  • $begingroup$
    Looks reasonable to me, I would solve it so.
    $endgroup$
    – kludg
    Jan 8 at 17:24














1












1








1





$begingroup$

Use law of total variance,
$$mathbb{Var}(X)=mathbb{E}[mathbb{Var}(X|Y)]+mathbb{Var}(mathbb{E}[X|Y])$$






share|cite|improve this answer









$endgroup$



Use law of total variance,
$$mathbb{Var}(X)=mathbb{E}[mathbb{Var}(X|Y)]+mathbb{Var}(mathbb{E}[X|Y])$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 8 at 16:01









kludgkludg

1,183611




1,183611












  • $begingroup$
    Okay so E[Var(XIY)] must be E[1]=1 since we know it's normal distributed with variance = 1. But how do I calculate Var(E[XIY])...? Is it using LOTUS? And wouldn't I also need to use Y's PDF in some way?
    $endgroup$
    – CruZ
    Jan 8 at 16:54






  • 1




    $begingroup$
    @CruZ You are almost there. Think what random variable $mathbb{E}[X|Y]$ is, and find its variance.
    $endgroup$
    – kludg
    Jan 8 at 17:04












  • $begingroup$
    Okay so E[XIY]=y so I just have to calculate Var(Y)=E[Y^2]-(E[Y])^2? And I can do that using LOTUS! Is that correct?
    $endgroup$
    – CruZ
    Jan 8 at 17:21










  • $begingroup$
    Looks reasonable to me, I would solve it so.
    $endgroup$
    – kludg
    Jan 8 at 17:24


















  • $begingroup$
    Okay so E[Var(XIY)] must be E[1]=1 since we know it's normal distributed with variance = 1. But how do I calculate Var(E[XIY])...? Is it using LOTUS? And wouldn't I also need to use Y's PDF in some way?
    $endgroup$
    – CruZ
    Jan 8 at 16:54






  • 1




    $begingroup$
    @CruZ You are almost there. Think what random variable $mathbb{E}[X|Y]$ is, and find its variance.
    $endgroup$
    – kludg
    Jan 8 at 17:04












  • $begingroup$
    Okay so E[XIY]=y so I just have to calculate Var(Y)=E[Y^2]-(E[Y])^2? And I can do that using LOTUS! Is that correct?
    $endgroup$
    – CruZ
    Jan 8 at 17:21










  • $begingroup$
    Looks reasonable to me, I would solve it so.
    $endgroup$
    – kludg
    Jan 8 at 17:24
















$begingroup$
Okay so E[Var(XIY)] must be E[1]=1 since we know it's normal distributed with variance = 1. But how do I calculate Var(E[XIY])...? Is it using LOTUS? And wouldn't I also need to use Y's PDF in some way?
$endgroup$
– CruZ
Jan 8 at 16:54




$begingroup$
Okay so E[Var(XIY)] must be E[1]=1 since we know it's normal distributed with variance = 1. But how do I calculate Var(E[XIY])...? Is it using LOTUS? And wouldn't I also need to use Y's PDF in some way?
$endgroup$
– CruZ
Jan 8 at 16:54




1




1




$begingroup$
@CruZ You are almost there. Think what random variable $mathbb{E}[X|Y]$ is, and find its variance.
$endgroup$
– kludg
Jan 8 at 17:04






$begingroup$
@CruZ You are almost there. Think what random variable $mathbb{E}[X|Y]$ is, and find its variance.
$endgroup$
– kludg
Jan 8 at 17:04














$begingroup$
Okay so E[XIY]=y so I just have to calculate Var(Y)=E[Y^2]-(E[Y])^2? And I can do that using LOTUS! Is that correct?
$endgroup$
– CruZ
Jan 8 at 17:21




$begingroup$
Okay so E[XIY]=y so I just have to calculate Var(Y)=E[Y^2]-(E[Y])^2? And I can do that using LOTUS! Is that correct?
$endgroup$
– CruZ
Jan 8 at 17:21












$begingroup$
Looks reasonable to me, I would solve it so.
$endgroup$
– kludg
Jan 8 at 17:24




$begingroup$
Looks reasonable to me, I would solve it so.
$endgroup$
– kludg
Jan 8 at 17:24


















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