Mixing
Finding $Var(X)$ from conditional PDF
$begingroup$
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
asked Jan 8 at 14:39
CruZCruZ
507
$endgroup$
add a comment |
$begingroup$
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
asked Jan 8 at 14:39
CruZCruZ
507
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Jan 8 at 14:39
CruZCruZ
507
$endgroup$
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
asked Jan 8 at 14:39
CruZCruZ
507
asked Jan 8 at 14:39
CruZCruZ
507
asked Jan 8 at 14:39
CruZCruZ
507
asked Jan 8 at 14:39
CruZCruZ
507
asked Jan 8 at 14:39
CruZCruZ
507
507
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Use law of total variance,
$$mathbb{Var}(X)=mathbb{E}[mathbb{Var}(X|Y)]+mathbb{Var}(mathbb{E}[X|Y])$$
answered Jan 8 at 16:01
kludgkludg
1,193611
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Use law of total variance,
$$mathbb{Var}(X)=mathbb{E}[mathbb{Var}(X|Y)]+mathbb{Var}(mathbb{E}[X|Y])$$
answered Jan 8 at 16:01
kludgkludg
1,193611
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
Use law of total variance,
$$mathbb{Var}(X)=mathbb{E}[mathbb{Var}(X|Y)]+mathbb{Var}(mathbb{E}[X|Y])$$
answered Jan 8 at 16:01
kludgkludg
1,193611
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
Use law of total variance,
$$mathbb{Var}(X)=mathbb{E}[mathbb{Var}(X|Y)]+mathbb{Var}(mathbb{E}[X|Y])$$
answered Jan 8 at 16:01
kludgkludg
1,193611
$endgroup$
Use law of total variance,
$$mathbb{Var}(X)=mathbb{E}[mathbb{Var}(X|Y)]+mathbb{Var}(mathbb{E}[X|Y])$$
answered Jan 8 at 16:01
kludgkludg
1,193611
answered Jan 8 at 16:01
kludgkludg
1,193611
answered Jan 8 at 16:01
kludgkludg
1,193611
answered Jan 8 at 16:01
kludgkludg
1,193611
1,193611
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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