What is $P(Amid B)(Cmid D)$? Is this the same as $P(Amid B) times P(Cmid D)$?
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I'm curious about the general case but also need to understand if $P(A|B)(C|D)$ is the same as $P(A|B) times P(C|D).$
In equation (7), it states
Is this the same as $P(u|f,s) times P(f|s)$ ?
The article is available at https://mechanicaldesign.asmedigitalcollection.asme.org/article.aspx?articleid=2297650
probability notation
edited Jan 16 at 23:08
jordan_glen
1
asked Jan 16 at 22:23
Norman MNorman M
132
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add a comment |
$begingroup$
I'm curious about the general case but also need to understand if $P(A|B)(C|D)$ is the same as $P(A|B) times P(C|D).$
In equation (7), it states
Is this the same as $P(u|f,s) times P(f|s)$ ?
The article is available at https://mechanicaldesign.asmedigitalcollection.asme.org/article.aspx?articleid=2297650
probability notation
edited Jan 16 at 23:08
jordan_glen
1
asked Jan 16 at 22:23
Norman MNorman M
132
$endgroup$
add a comment |
$begingroup$
I'm curious about the general case but also need to understand if $P(A|B)(C|D)$ is the same as $P(A|B) times P(C|D).$
In equation (7), it states
Is this the same as $P(u|f,s) times P(f|s)$ ?
The article is available at https://mechanicaldesign.asmedigitalcollection.asme.org/article.aspx?articleid=2297650
probability notation
edited Jan 16 at 23:08
jordan_glen
1
asked Jan 16 at 22:23
Norman MNorman M
132
$endgroup$
I'm curious about the general case but also need to understand if $P(A|B)(C|D)$ is the same as $P(A|B) times P(C|D).$
In equation (7), it states
Is this the same as $P(u|f,s) times P(f|s)$ ?
The article is available at https://mechanicaldesign.asmedigitalcollection.asme.org/article.aspx?articleid=2297650
probability notation
probability notation
edited Jan 16 at 23:08
jordan_glen
1
asked Jan 16 at 22:23
Norman MNorman M
132
edited Jan 16 at 23:08
jordan_glen
1
asked Jan 16 at 22:23
Norman MNorman M
132
edited Jan 16 at 23:08
jordan_glen
1
edited Jan 16 at 23:08
jordan_glen
1
edited Jan 16 at 23:08
jordan_glen
1
1
asked Jan 16 at 22:23
Norman MNorman M
132
asked Jan 16 at 22:23
Norman MNorman M
132
asked Jan 16 at 22:23
Norman MNorman M
132
132
add a comment |
add a comment |
1 Answer
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$begingroup$
Your understanding is correct. I think there is a typo in the original source. In general it holds that
begin{equation}
p(x,y|z) = p(x|y,z) p(y|z)
end{equation}
and therefore,
begin{equation}
p(x|z) = sum_y p(x,y|z) = sum_y p(x|y,z) p(y|z)
end{equation}
answered Jan 16 at 22:31
user144410user144410
1,0412719
$endgroup$
$begingroup$
Thank you for helping.
$endgroup$
– Norman M
Jan 16 at 22:33
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You are welcome. Please accept the answer if you found it helpful.
$endgroup$
– user144410
Jan 18 at 16:09
add a comment |
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1 Answer
1
active
oldest
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1 Answer
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active
oldest
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active
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$begingroup$
Your understanding is correct. I think there is a typo in the original source. In general it holds that
begin{equation}
p(x,y|z) = p(x|y,z) p(y|z)
end{equation}
and therefore,
begin{equation}
p(x|z) = sum_y p(x,y|z) = sum_y p(x|y,z) p(y|z)
end{equation}
answered Jan 16 at 22:31
user144410user144410
1,0412719
$endgroup$
$begingroup$
Thank you for helping.
$endgroup$
– Norman M
Jan 16 at 22:33
$begingroup$
You are welcome. Please accept the answer if you found it helpful.
$endgroup$
– user144410
Jan 18 at 16:09
add a comment |
$begingroup$
Your understanding is correct. I think there is a typo in the original source. In general it holds that
begin{equation}
p(x,y|z) = p(x|y,z) p(y|z)
end{equation}
and therefore,
begin{equation}
p(x|z) = sum_y p(x,y|z) = sum_y p(x|y,z) p(y|z)
end{equation}
answered Jan 16 at 22:31
user144410user144410
1,0412719
$endgroup$
$begingroup$
Thank you for helping.
$endgroup$
– Norman M
Jan 16 at 22:33
$begingroup$
You are welcome. Please accept the answer if you found it helpful.
$endgroup$
– user144410
Jan 18 at 16:09
add a comment |
$begingroup$
Your understanding is correct. I think there is a typo in the original source. In general it holds that
begin{equation}
p(x,y|z) = p(x|y,z) p(y|z)
end{equation}
and therefore,
begin{equation}
p(x|z) = sum_y p(x,y|z) = sum_y p(x|y,z) p(y|z)
end{equation}
answered Jan 16 at 22:31
user144410user144410
1,0412719
$endgroup$
Your understanding is correct. I think there is a typo in the original source. In general it holds that
begin{equation}
p(x,y|z) = p(x|y,z) p(y|z)
end{equation}
and therefore,
begin{equation}
p(x|z) = sum_y p(x,y|z) = sum_y p(x|y,z) p(y|z)
end{equation}
answered Jan 16 at 22:31
user144410user144410
1,0412719
edited Jan 16 at 22:33
answered Jan 16 at 22:31
user144410user144410
1,0412719
answered Jan 16 at 22:31
user144410user144410
1,0412719
answered Jan 16 at 22:31
user144410user144410
1,0412719
1,0412719
$begingroup$
Thank you for helping.
$endgroup$
– Norman M
Jan 16 at 22:33
$begingroup$
You are welcome. Please accept the answer if you found it helpful.
$endgroup$
– user144410
Jan 18 at 16:09
add a comment |
$begingroup$
Thank you for helping.
$endgroup$
– Norman M
Jan 16 at 22:33
$begingroup$
You are welcome. Please accept the answer if you found it helpful.
$endgroup$
– user144410
Jan 18 at 16:09
$begingroup$
Thank you for helping.
$endgroup$
– Norman M
Jan 16 at 22:33
$begingroup$
Thank you for helping.
$endgroup$
– Norman M
Jan 16 at 22:33
$begingroup$
You are welcome. Please accept the answer if you found it helpful.
$endgroup$
– user144410
Jan 18 at 16:09
$begingroup$
You are welcome. Please accept the answer if you found it helpful.
$endgroup$
– user144410
Jan 18 at 16:09
add a comment |
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