Mixing
Let $ Xsim exp(lambda) $ and let $ (U|X=k) sim U[x-1,x+1] $ for $ xge 0 $. Why $ P(X=k) ne 0 $?
0
$begingroup$
Iv'e encountered the following definition:
$Xsim exp(lambda)$ and let $(U|X=x) sim mathcal U[x-1,x+1]$ for $xgeq 0$.
Now I know that since $X$ is continuous, $P(X=x)=0$ for any $x$. So how can we even discuss $(U|X=x)$?
probability random-variables
edited Jan 29 at 11:36
user549397
1,6541418
asked Jan 29 at 10:24
superuser123superuser123
48628
$endgroup$
add a comment |
0
$begingroup$
Iv'e encountered the following definition:
$Xsim exp(lambda)$ and let $(U|X=x) sim mathcal U[x-1,x+1]$ for $xgeq 0$.
Now I know that since $X$ is continuous, $P(X=x)=0$ for any $x$. So how can we even discuss $(U|X=x)$?
probability random-variables
edited Jan 29 at 11:36
user549397
1,6541418
asked Jan 29 at 10:24
superuser123superuser123
48628
$endgroup$
$begingroup$
I don't know where you have encountered that, but I hope you have encountered the definition of conditional distribution before that. So do you know the definition of conditional distribution?
$endgroup$
– Shashi
Jan 29 at 10:28
$begingroup$
@Shashi yes I do but I don't know how to apply that here, If $P(X=x)=0$ how does it even possible to handle?
$endgroup$
– superuser123
Jan 29 at 10:32
1
$begingroup$
Informally you may think of $U | , X = x$ as $U | , X in (x - dx, x + dx)$. This is not a rigorous way of explaining but it should help your intuition.
$endgroup$
– Harnak
Jan 29 at 10:35
1
$begingroup$
The conditional density of $(U|X=x) $ is simply $$frac{f_{U, X} (u, x) } {f_X(x)} $$ So there is nowhere where you encounter any troubles with the fact that $P(X=x) =0$. Is this clear?
$endgroup$
– Shashi
Jan 29 at 10:37
$begingroup$
@Shashi yes, thanks! :)
$endgroup$
– superuser123
Jan 29 at 10:39
add a comment |
0
0
0
$begingroup$
Iv'e encountered the following definition:
$Xsim exp(lambda)$ and let $(U|X=x) sim mathcal U[x-1,x+1]$ for $xgeq 0$.
Now I know that since $X$ is continuous, $P(X=x)=0$ for any $x$. So how can we even discuss $(U|X=x)$?
probability random-variables
edited Jan 29 at 11:36
user549397
1,6541418
asked Jan 29 at 10:24
superuser123superuser123
48628
$endgroup$
Iv'e encountered the following definition:
$Xsim exp(lambda)$ and let $(U|X=x) sim mathcal U[x-1,x+1]$ for $xgeq 0$.
Now I know that since $X$ is continuous, $P(X=x)=0$ for any $x$. So how can we even discuss $(U|X=x)$?
probability random-variables
probability random-variables
edited Jan 29 at 11:36
user549397
1,6541418
asked Jan 29 at 10:24
superuser123superuser123
48628
edited Jan 29 at 11:36
user549397
1,6541418
asked Jan 29 at 10:24
superuser123superuser123
48628
edited Jan 29 at 11:36
user549397
1,6541418
edited Jan 29 at 11:36
user549397
1,6541418
edited Jan 29 at 11:36
user549397
1,6541418
1,6541418
asked Jan 29 at 10:24
superuser123superuser123
48628
asked Jan 29 at 10:24
superuser123superuser123
48628
asked Jan 29 at 10:24
superuser123superuser123
48628
48628
$begingroup$
I don't know where you have encountered that, but I hope you have encountered the definition of conditional distribution before that. So do you know the definition of conditional distribution?
$endgroup$
– Shashi
Jan 29 at 10:28
$begingroup$
@Shashi yes I do but I don't know how to apply that here, If $P(X=x)=0$ how does it even possible to handle?
$endgroup$
– superuser123
Jan 29 at 10:32
1
$begingroup$
Informally you may think of $U | , X = x$ as $U | , X in (x - dx, x + dx)$. This is not a rigorous way of explaining but it should help your intuition.
$endgroup$
– Harnak
Jan 29 at 10:35
1
$begingroup$
The conditional density of $(U|X=x) $ is simply $$frac{f_{U, X} (u, x) } {f_X(x)} $$ So there is nowhere where you encounter any troubles with the fact that $P(X=x) =0$. Is this clear?
$endgroup$
– Shashi
Jan 29 at 10:37
$begingroup$
@Shashi yes, thanks! :)
$endgroup$
– superuser123
Jan 29 at 10:39
add a comment |
$begingroup$
I don't know where you have encountered that, but I hope you have encountered the definition of conditional distribution before that. So do you know the definition of conditional distribution?
$endgroup$
– Shashi
Jan 29 at 10:28
$begingroup$
@Shashi yes I do but I don't know how to apply that here, If $P(X=x)=0$ how does it even possible to handle?
$endgroup$
– superuser123
Jan 29 at 10:32
1
$begingroup$
Informally you may think of $U | , X = x$ as $U | , X in (x - dx, x + dx)$. This is not a rigorous way of explaining but it should help your intuition.
$endgroup$
– Harnak
Jan 29 at 10:35
1
$begingroup$
The conditional density of $(U|X=x) $ is simply $$frac{f_{U, X} (u, x) } {f_X(x)} $$ So there is nowhere where you encounter any troubles with the fact that $P(X=x) =0$. Is this clear?
$endgroup$
– Shashi
Jan 29 at 10:37
$begingroup$
@Shashi yes, thanks! :)
$endgroup$
– superuser123
Jan 29 at 10:39
$begingroup$
I don't know where you have encountered that, but I hope you have encountered the definition of conditional distribution before that. So do you know the definition of conditional distribution?
$endgroup$
– Shashi
Jan 29 at 10:28
$begingroup$
I don't know where you have encountered that, but I hope you have encountered the definition of conditional distribution before that. So do you know the definition of conditional distribution?
$endgroup$
– Shashi
Jan 29 at 10:28
$begingroup$
@Shashi yes I do but I don't know how to apply that here, If $P(X=x)=0$ how does it even possible to handle?
$endgroup$
– superuser123
Jan 29 at 10:32
$begingroup$
@Shashi yes I do but I don't know how to apply that here, If $P(X=x)=0$ how does it even possible to handle?
$endgroup$
– superuser123
Jan 29 at 10:32
1
1
$begingroup$
Informally you may think of $U | , X = x$ as $U | , X in (x - dx, x + dx)$. This is not a rigorous way of explaining but it should help your intuition.
$endgroup$
– Harnak
Jan 29 at 10:35
$begingroup$
Informally you may think of $U | , X = x$ as $U | , X in (x - dx, x + dx)$. This is not a rigorous way of explaining but it should help your intuition.
$endgroup$
– Harnak
Jan 29 at 10:35
1
1
$begingroup$
The conditional density of $(U|X=x) $ is simply $$frac{f_{U, X} (u, x) } {f_X(x)} $$ So there is nowhere where you encounter any troubles with the fact that $P(X=x) =0$. Is this clear?
$endgroup$
– Shashi
Jan 29 at 10:37
$begingroup$
The conditional density of $(U|X=x) $ is simply $$frac{f_{U, X} (u, x) } {f_X(x)} $$ So there is nowhere where you encounter any troubles with the fact that $P(X=x) =0$. Is this clear?
$endgroup$
– Shashi
Jan 29 at 10:37
$begingroup$
@Shashi yes, thanks! :)
$endgroup$
– superuser123
Jan 29 at 10:39
$begingroup$
@Shashi yes, thanks! :)
$endgroup$
– superuser123
Jan 29 at 10:39
add a comment |
1 Answer
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$E(X|Y)$ can be written as $f(Y)$ for some measurable function $f: mathbb R to mathbb R$ and $E(X|Y=x)$ is the notation used for the function $f(x)$.
answered Jan 29 at 10:33
Kavi Rama MurthyKavi Rama Murthy
71.3k53170
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1 Answer
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$begingroup$
$E(X|Y)$ can be written as $f(Y)$ for some measurable function $f: mathbb R to mathbb R$ and $E(X|Y=x)$ is the notation used for the function $f(x)$.
answered Jan 29 at 10:33
Kavi Rama MurthyKavi Rama Murthy
71.3k53170
$endgroup$
add a comment |
2
$begingroup$
$E(X|Y)$ can be written as $f(Y)$ for some measurable function $f: mathbb R to mathbb R$ and $E(X|Y=x)$ is the notation used for the function $f(x)$.
answered Jan 29 at 10:33
Kavi Rama MurthyKavi Rama Murthy
71.3k53170
$endgroup$
add a comment |
2
2
2
$begingroup$
$E(X|Y)$ can be written as $f(Y)$ for some measurable function $f: mathbb R to mathbb R$ and $E(X|Y=x)$ is the notation used for the function $f(x)$.
answered Jan 29 at 10:33
Kavi Rama MurthyKavi Rama Murthy
71.3k53170
$endgroup$
$E(X|Y)$ can be written as $f(Y)$ for some measurable function $f: mathbb R to mathbb R$ and $E(X|Y=x)$ is the notation used for the function $f(x)$.
answered Jan 29 at 10:33
Kavi Rama MurthyKavi Rama Murthy
71.3k53170
answered Jan 29 at 10:33
Kavi Rama MurthyKavi Rama Murthy
71.3k53170
answered Jan 29 at 10:33
Kavi Rama MurthyKavi Rama Murthy
71.3k53170
answered Jan 29 at 10:33
Kavi Rama MurthyKavi Rama Murthy
71.3k53170
71.3k53170
add a comment |
add a comment |
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$begingroup$
I don't know where you have encountered that, but I hope you have encountered the definition of conditional distribution before that. So do you know the definition of conditional distribution?
$endgroup$
– Shashi
Jan 29 at 10:28
$begingroup$
@Shashi yes I do but I don't know how to apply that here, If $P(X=x)=0$ how does it even possible to handle?
$endgroup$
– superuser123
Jan 29 at 10:32
1
$begingroup$
Informally you may think of $U | , X = x$ as $U | , X in (x - dx, x + dx)$. This is not a rigorous way of explaining but it should help your intuition.
$endgroup$
– Harnak
Jan 29 at 10:35
1
$begingroup$
The conditional density of $(U|X=x) $ is simply $$frac{f_{U, X} (u, x) } {f_X(x)} $$ So there is nowhere where you encounter any troubles with the fact that $P(X=x) =0$. Is this clear?
$endgroup$
– Shashi
Jan 29 at 10:37
$begingroup$
@Shashi yes, thanks! :)
$endgroup$
– superuser123
Jan 29 at 10:39