Mixing
proof of product rule on conditonal probabilitiy
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From the basic product rule on conditional probability, we know the following:
p(x,y) = P(x|y)P(y).
But I cannot understand this formula:
p(x,y|z) = p(x|y,z)p(y|z).
I have tried to prove this as:
p(x,y|z) = p(x|y|z)p(y|z)
But i am confused on p(x|y|z)[don't know this notation exists or not.]. And if p(x|y|z) exists then again confused why p(x|y|z) = p(x|y,z)
probability statistics
asked Aug 7 '17 at 6:27
AnjanAnjan
133
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add a comment |
$begingroup$
From the basic product rule on conditional probability, we know the following:
p(x,y) = P(x|y)P(y).
But I cannot understand this formula:
p(x,y|z) = p(x|y,z)p(y|z).
I have tried to prove this as:
p(x,y|z) = p(x|y|z)p(y|z)
But i am confused on p(x|y|z)[don't know this notation exists or not.]. And if p(x|y|z) exists then again confused why p(x|y|z) = p(x|y,z)
probability statistics
asked Aug 7 '17 at 6:27
AnjanAnjan
133
$endgroup$
1
$begingroup$
$p(xmid y mid z)$ is not good notation, but to the extent it suggests something like "the probability (or probability density) of $X$ given $Y=y$ given $Z=z$" then this might really be "the probability (or probability density) of $X$ given $Y=y$ and $Z=z$", i.e. $p(xmid y , z)$
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– Henry
Aug 7 '17 at 6:33
1
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Related: math.stackexchange.com/questions/301207/why-is-px-yz-pyx-zpxz
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– Henry
Aug 7 '17 at 6:33
1
$begingroup$
Thanks Henry...
$endgroup$
– Anjan
Aug 7 '17 at 6:37
add a comment |
$begingroup$
From the basic product rule on conditional probability, we know the following:
p(x,y) = P(x|y)P(y).
But I cannot understand this formula:
p(x,y|z) = p(x|y,z)p(y|z).
I have tried to prove this as:
p(x,y|z) = p(x|y|z)p(y|z)
But i am confused on p(x|y|z)[don't know this notation exists or not.]. And if p(x|y|z) exists then again confused why p(x|y|z) = p(x|y,z)
probability statistics
asked Aug 7 '17 at 6:27
AnjanAnjan
133
$endgroup$
From the basic product rule on conditional probability, we know the following:
p(x,y) = P(x|y)P(y).
But I cannot understand this formula:
p(x,y|z) = p(x|y,z)p(y|z).
I have tried to prove this as:
p(x,y|z) = p(x|y|z)p(y|z)
But i am confused on p(x|y|z)[don't know this notation exists or not.]. And if p(x|y|z) exists then again confused why p(x|y|z) = p(x|y,z)
probability statistics
probability statistics
asked Aug 7 '17 at 6:27
AnjanAnjan
133
asked Aug 7 '17 at 6:27
AnjanAnjan
133
asked Aug 7 '17 at 6:27
AnjanAnjan
133
asked Aug 7 '17 at 6:27
AnjanAnjan
133
asked Aug 7 '17 at 6:27
AnjanAnjan
133
133
1
$begingroup$
$p(xmid y mid z)$ is not good notation, but to the extent it suggests something like "the probability (or probability density) of $X$ given $Y=y$ given $Z=z$" then this might really be "the probability (or probability density) of $X$ given $Y=y$ and $Z=z$", i.e. $p(xmid y , z)$
$endgroup$
– Henry
Aug 7 '17 at 6:33
1
$begingroup$
Related: math.stackexchange.com/questions/301207/why-is-px-yz-pyx-zpxz
$endgroup$
– Henry
Aug 7 '17 at 6:33
1
$begingroup$
Thanks Henry...
$endgroup$
– Anjan
Aug 7 '17 at 6:37
add a comment |
1
$begingroup$
$p(xmid y mid z)$ is not good notation, but to the extent it suggests something like "the probability (or probability density) of $X$ given $Y=y$ given $Z=z$" then this might really be "the probability (or probability density) of $X$ given $Y=y$ and $Z=z$", i.e. $p(xmid y , z)$
$endgroup$
– Henry
Aug 7 '17 at 6:33
1
$begingroup$
Related: math.stackexchange.com/questions/301207/why-is-px-yz-pyx-zpxz
$endgroup$
– Henry
Aug 7 '17 at 6:33
1
$begingroup$
Thanks Henry...
$endgroup$
– Anjan
Aug 7 '17 at 6:37
1
1
$begingroup$
$p(xmid y mid z)$ is not good notation, but to the extent it suggests something like "the probability (or probability density) of $X$ given $Y=y$ given $Z=z$" then this might really be "the probability (or probability density) of $X$ given $Y=y$ and $Z=z$", i.e. $p(xmid y , z)$
$endgroup$
– Henry
Aug 7 '17 at 6:33
$begingroup$
$p(xmid y mid z)$ is not good notation, but to the extent it suggests something like "the probability (or probability density) of $X$ given $Y=y$ given $Z=z$" then this might really be "the probability (or probability density) of $X$ given $Y=y$ and $Z=z$", i.e. $p(xmid y , z)$
$endgroup$
– Henry
Aug 7 '17 at 6:33
1
1
$begingroup$
Related: math.stackexchange.com/questions/301207/why-is-px-yz-pyx-zpxz
$endgroup$
– Henry
Aug 7 '17 at 6:33
$begingroup$
Related: math.stackexchange.com/questions/301207/why-is-px-yz-pyx-zpxz
$endgroup$
– Henry
Aug 7 '17 at 6:33
1
1
$begingroup$
Thanks Henry...
$endgroup$
– Anjan
Aug 7 '17 at 6:37
$begingroup$
Thanks Henry...
$endgroup$
– Anjan
Aug 7 '17 at 6:37
add a comment |
5 Answers
5
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Note that $p(x,ymid z) = p(x,y,z)/p(x)$, that $p(xmid y,z) = p(x,y,z)/p(y,z)$, and that $p(y,z) = p(z)p(ymid z)$ by definition. So
$$
p(xmid y,z)p(ymid z) = p(x,ymid z).
$$
answered Aug 7 '17 at 6:54
Gary MooreGary Moore
17.3k21546
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But i am confused on p(x|y|z)[don't know this notation exists or not.]. And if p(x|y|z) exists then again confused why p(x|y|z) = p(x|y,z)
The notation does not exist. The divider is not a set operator; it is a separator between the event (or random variable) and the condition over which it is being measured. The event can be joint, as can the condition, but there can only be one event-condition-divider in any measure function.
$p(xmid y,z)$ is a representation for $mathsf P(X=xmid Y=ycap Z=z)$, the probability measure for the event of $X=x$ under the condition that $Y=y$ and $Z=z$.
And such.
$$begin{array}{lll}& p(x,ymid z)&=mathsf P(X=xcap Y=ymid Z=z)\[1ex] =& dfrac{p(x,y,z)}{p(z)} &= dfrac{mathsf P(X=xcap Y=ycap Z=z)}{mathsf P(Z=z)}\[1ex]=&dfrac{p(xmid y,z),p(y,z)}{p(z)} &=dfrac{mathsf P(X=xmid Y=ycap Z=z),mathsf P(Y=ycap Z=z)}{mathsf P(Z=z)}\[1ex]=& p(xmid y,z),p(ymid z) &= mathsf P(X=xmid Y=ycap Z=z),mathsf P(Y=ymid Z=z) end{array}$$
answered Aug 7 '17 at 6:57
Graham KempGraham Kemp
86.7k43579
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$begingroup$
Sum Rule:
$$p(a) = sum_{b}p(a,b)$$
As the sum rule holds for any probability distribution it trivially holds for the distribution a|c:
$$p(a|c) = sum_{b}p(b, a|c)$$
For the product rule:
$$p(a,b)=p(a|b)p(b)$$ but does $$p(a,b|c)=p(a|b|c)p(b|c)$$?
Well presumably, they are equal if $$p(a|b|c) = p(a|b,c)$$ - but does this equality hold?
Yes: Aim to prove: $$p(a|b|c) = p(a|b,c)$$
$$p(a,b|c) = p(a|b|c)p(b|c)$$
$$p(a|b|c) = dfrac{p(a,b|c)}{p(b|c)} = dfrac{ dfrac{p(a,b,c)}{p(c)}} {dfrac{p(b,c)}{p(c)}} = dfrac{p(a,b,c)}{p(b,c)} = dfrac{p(a|b,c)p(b,c)}{p(b,c)} = p(a|b,c)$$
answered Jan 23 at 20:20
Oliver GoldsteinOliver Goldstein
111
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$begingroup$
For others' benefit, I note that this answer (and the underlying topic) is very similar to that in this post.
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– davidlowryduda♦
Jan 23 at 21:21
add a comment |
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The said notation does not exist for $x,y,z$ events.
However, the identity could be proved by
$$P(x,ymid z)=frac{P(x,y,z)}{P(z)}=frac{P(xmid y,z)P(y,z)}{P(z)}$$
$$=P(xmid y,z)frac{P(y,z)}{P(z)}=P(xmid y,z)P(ymid z)$$
answered Aug 7 '17 at 7:06
BAIBAI
1,778518
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Doubtless this derivation is included either in the body of the text or as an exercise across a great array of textbooks across numerous fields. For example, #13.16(a.) from Russell & Norvig's Artificial Intelligence, A Modern Approach, Third Edition, says,
Prove the conditionalized version of the general product rule:
$P(X,Y|e) = P(X|Y,e)P(Y|e)$.
$$P(X,Y|e) = frac{P(X,Y,e)}{P(e)}$$
$$P(X|Y,e) = frac{P(X,Y,e)}{P(Y,e)}$$
$$P(X,Y,e) = frac{P(X|Y,e)}{P(Y,e)}$$
$$P(X,Y|e) = frac{P(X|y,e)P(Y,e)}{P(e)} = P(X|Y,e)P(Y|e)$$
In the first of the 4 lines in the derivation, above, we have $P(X,Y|e) = frac{P(X,Y,e)}{P(e)}$. We can think of this, perhaps overly simplistically (but it is a nice, workable mental representation so we can see what is going on here), as $P(A|B) = P(A, B)/P(B)$, simply Bayes familiar rule, where we have replaced A with $Xcap Y$ and e with B. To map my answer back on to the original question replace X with x, Y with y, and e with z. C.f. page 9 of 10, question & answer #6, here as well. Cheers!
answered Apr 1 '18 at 23:04
E.J.E.J.
1
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5 Answers
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5 Answers
5
active
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$begingroup$
Note that $p(x,ymid z) = p(x,y,z)/p(x)$, that $p(xmid y,z) = p(x,y,z)/p(y,z)$, and that $p(y,z) = p(z)p(ymid z)$ by definition. So
$$
p(xmid y,z)p(ymid z) = p(x,ymid z).
$$
answered Aug 7 '17 at 6:54
Gary MooreGary Moore
17.3k21546
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add a comment |
$begingroup$
Note that $p(x,ymid z) = p(x,y,z)/p(x)$, that $p(xmid y,z) = p(x,y,z)/p(y,z)$, and that $p(y,z) = p(z)p(ymid z)$ by definition. So
$$
p(xmid y,z)p(ymid z) = p(x,ymid z).
$$
answered Aug 7 '17 at 6:54
Gary MooreGary Moore
17.3k21546
$endgroup$
add a comment |
$begingroup$
Note that $p(x,ymid z) = p(x,y,z)/p(x)$, that $p(xmid y,z) = p(x,y,z)/p(y,z)$, and that $p(y,z) = p(z)p(ymid z)$ by definition. So
$$
p(xmid y,z)p(ymid z) = p(x,ymid z).
$$
answered Aug 7 '17 at 6:54
Gary MooreGary Moore
17.3k21546
$endgroup$
Note that $p(x,ymid z) = p(x,y,z)/p(x)$, that $p(xmid y,z) = p(x,y,z)/p(y,z)$, and that $p(y,z) = p(z)p(ymid z)$ by definition. So
$$
p(xmid y,z)p(ymid z) = p(x,ymid z).
$$
answered Aug 7 '17 at 6:54
Gary MooreGary Moore
17.3k21546
answered Aug 7 '17 at 6:54
Gary MooreGary Moore
17.3k21546
answered Aug 7 '17 at 6:54
Gary MooreGary Moore
17.3k21546
answered Aug 7 '17 at 6:54
Gary MooreGary Moore
17.3k21546
17.3k21546
add a comment |
add a comment |
$begingroup$
But i am confused on p(x|y|z)[don't know this notation exists or not.]. And if p(x|y|z) exists then again confused why p(x|y|z) = p(x|y,z)
The notation does not exist. The divider is not a set operator; it is a separator between the event (or random variable) and the condition over which it is being measured. The event can be joint, as can the condition, but there can only be one event-condition-divider in any measure function.
$p(xmid y,z)$ is a representation for $mathsf P(X=xmid Y=ycap Z=z)$, the probability measure for the event of $X=x$ under the condition that $Y=y$ and $Z=z$.
And such.
$$begin{array}{lll}& p(x,ymid z)&=mathsf P(X=xcap Y=ymid Z=z)\[1ex] =& dfrac{p(x,y,z)}{p(z)} &= dfrac{mathsf P(X=xcap Y=ycap Z=z)}{mathsf P(Z=z)}\[1ex]=&dfrac{p(xmid y,z),p(y,z)}{p(z)} &=dfrac{mathsf P(X=xmid Y=ycap Z=z),mathsf P(Y=ycap Z=z)}{mathsf P(Z=z)}\[1ex]=& p(xmid y,z),p(ymid z) &= mathsf P(X=xmid Y=ycap Z=z),mathsf P(Y=ymid Z=z) end{array}$$
answered Aug 7 '17 at 6:57
Graham KempGraham Kemp
86.7k43579
$endgroup$
add a comment |
$begingroup$
But i am confused on p(x|y|z)[don't know this notation exists or not.]. And if p(x|y|z) exists then again confused why p(x|y|z) = p(x|y,z)
The notation does not exist. The divider is not a set operator; it is a separator between the event (or random variable) and the condition over which it is being measured. The event can be joint, as can the condition, but there can only be one event-condition-divider in any measure function.
$p(xmid y,z)$ is a representation for $mathsf P(X=xmid Y=ycap Z=z)$, the probability measure for the event of $X=x$ under the condition that $Y=y$ and $Z=z$.
And such.
$$begin{array}{lll}& p(x,ymid z)&=mathsf P(X=xcap Y=ymid Z=z)\[1ex] =& dfrac{p(x,y,z)}{p(z)} &= dfrac{mathsf P(X=xcap Y=ycap Z=z)}{mathsf P(Z=z)}\[1ex]=&dfrac{p(xmid y,z),p(y,z)}{p(z)} &=dfrac{mathsf P(X=xmid Y=ycap Z=z),mathsf P(Y=ycap Z=z)}{mathsf P(Z=z)}\[1ex]=& p(xmid y,z),p(ymid z) &= mathsf P(X=xmid Y=ycap Z=z),mathsf P(Y=ymid Z=z) end{array}$$
answered Aug 7 '17 at 6:57
Graham KempGraham Kemp
86.7k43579
$endgroup$
add a comment |
$begingroup$
But i am confused on p(x|y|z)[don't know this notation exists or not.]. And if p(x|y|z) exists then again confused why p(x|y|z) = p(x|y,z)
The notation does not exist. The divider is not a set operator; it is a separator between the event (or random variable) and the condition over which it is being measured. The event can be joint, as can the condition, but there can only be one event-condition-divider in any measure function.
$p(xmid y,z)$ is a representation for $mathsf P(X=xmid Y=ycap Z=z)$, the probability measure for the event of $X=x$ under the condition that $Y=y$ and $Z=z$.
And such.
$$begin{array}{lll}& p(x,ymid z)&=mathsf P(X=xcap Y=ymid Z=z)\[1ex] =& dfrac{p(x,y,z)}{p(z)} &= dfrac{mathsf P(X=xcap Y=ycap Z=z)}{mathsf P(Z=z)}\[1ex]=&dfrac{p(xmid y,z),p(y,z)}{p(z)} &=dfrac{mathsf P(X=xmid Y=ycap Z=z),mathsf P(Y=ycap Z=z)}{mathsf P(Z=z)}\[1ex]=& p(xmid y,z),p(ymid z) &= mathsf P(X=xmid Y=ycap Z=z),mathsf P(Y=ymid Z=z) end{array}$$
answered Aug 7 '17 at 6:57
Graham KempGraham Kemp
86.7k43579
$endgroup$
But i am confused on p(x|y|z)[don't know this notation exists or not.]. And if p(x|y|z) exists then again confused why p(x|y|z) = p(x|y,z)
The notation does not exist. The divider is not a set operator; it is a separator between the event (or random variable) and the condition over which it is being measured. The event can be joint, as can the condition, but there can only be one event-condition-divider in any measure function.
$p(xmid y,z)$ is a representation for $mathsf P(X=xmid Y=ycap Z=z)$, the probability measure for the event of $X=x$ under the condition that $Y=y$ and $Z=z$.
And such.
$$begin{array}{lll}& p(x,ymid z)&=mathsf P(X=xcap Y=ymid Z=z)\[1ex] =& dfrac{p(x,y,z)}{p(z)} &= dfrac{mathsf P(X=xcap Y=ycap Z=z)}{mathsf P(Z=z)}\[1ex]=&dfrac{p(xmid y,z),p(y,z)}{p(z)} &=dfrac{mathsf P(X=xmid Y=ycap Z=z),mathsf P(Y=ycap Z=z)}{mathsf P(Z=z)}\[1ex]=& p(xmid y,z),p(ymid z) &= mathsf P(X=xmid Y=ycap Z=z),mathsf P(Y=ymid Z=z) end{array}$$
answered Aug 7 '17 at 6:57
Graham KempGraham Kemp
86.7k43579
answered Aug 7 '17 at 6:57
Graham KempGraham Kemp
86.7k43579
answered Aug 7 '17 at 6:57
Graham KempGraham Kemp
86.7k43579
answered Aug 7 '17 at 6:57
Graham KempGraham Kemp
86.7k43579
86.7k43579
add a comment |
add a comment |
$begingroup$
Sum Rule:
$$p(a) = sum_{b}p(a,b)$$
As the sum rule holds for any probability distribution it trivially holds for the distribution a|c:
$$p(a|c) = sum_{b}p(b, a|c)$$
For the product rule:
$$p(a,b)=p(a|b)p(b)$$ but does $$p(a,b|c)=p(a|b|c)p(b|c)$$?
Well presumably, they are equal if $$p(a|b|c) = p(a|b,c)$$ - but does this equality hold?
Yes: Aim to prove: $$p(a|b|c) = p(a|b,c)$$
$$p(a,b|c) = p(a|b|c)p(b|c)$$
$$p(a|b|c) = dfrac{p(a,b|c)}{p(b|c)} = dfrac{ dfrac{p(a,b,c)}{p(c)}} {dfrac{p(b,c)}{p(c)}} = dfrac{p(a,b,c)}{p(b,c)} = dfrac{p(a|b,c)p(b,c)}{p(b,c)} = p(a|b,c)$$
answered Jan 23 at 20:20
Oliver GoldsteinOliver Goldstein
111
$endgroup$
$begingroup$
For others' benefit, I note that this answer (and the underlying topic) is very similar to that in this post.
$endgroup$
– davidlowryduda♦
Jan 23 at 21:21
add a comment |
$begingroup$
Sum Rule:
$$p(a) = sum_{b}p(a,b)$$
As the sum rule holds for any probability distribution it trivially holds for the distribution a|c:
$$p(a|c) = sum_{b}p(b, a|c)$$
For the product rule:
$$p(a,b)=p(a|b)p(b)$$ but does $$p(a,b|c)=p(a|b|c)p(b|c)$$?
Well presumably, they are equal if $$p(a|b|c) = p(a|b,c)$$ - but does this equality hold?
Yes: Aim to prove: $$p(a|b|c) = p(a|b,c)$$
$$p(a,b|c) = p(a|b|c)p(b|c)$$
$$p(a|b|c) = dfrac{p(a,b|c)}{p(b|c)} = dfrac{ dfrac{p(a,b,c)}{p(c)}} {dfrac{p(b,c)}{p(c)}} = dfrac{p(a,b,c)}{p(b,c)} = dfrac{p(a|b,c)p(b,c)}{p(b,c)} = p(a|b,c)$$
answered Jan 23 at 20:20
Oliver GoldsteinOliver Goldstein
111
$endgroup$
$begingroup$
For others' benefit, I note that this answer (and the underlying topic) is very similar to that in this post.
$endgroup$
– davidlowryduda♦
Jan 23 at 21:21
add a comment |
$begingroup$
Sum Rule:
$$p(a) = sum_{b}p(a,b)$$
As the sum rule holds for any probability distribution it trivially holds for the distribution a|c:
$$p(a|c) = sum_{b}p(b, a|c)$$
For the product rule:
$$p(a,b)=p(a|b)p(b)$$ but does $$p(a,b|c)=p(a|b|c)p(b|c)$$?
Well presumably, they are equal if $$p(a|b|c) = p(a|b,c)$$ - but does this equality hold?
Yes: Aim to prove: $$p(a|b|c) = p(a|b,c)$$
$$p(a,b|c) = p(a|b|c)p(b|c)$$
$$p(a|b|c) = dfrac{p(a,b|c)}{p(b|c)} = dfrac{ dfrac{p(a,b,c)}{p(c)}} {dfrac{p(b,c)}{p(c)}} = dfrac{p(a,b,c)}{p(b,c)} = dfrac{p(a|b,c)p(b,c)}{p(b,c)} = p(a|b,c)$$
answered Jan 23 at 20:20
Oliver GoldsteinOliver Goldstein
111
$endgroup$
Sum Rule:
$$p(a) = sum_{b}p(a,b)$$
As the sum rule holds for any probability distribution it trivially holds for the distribution a|c:
$$p(a|c) = sum_{b}p(b, a|c)$$
For the product rule:
$$p(a,b)=p(a|b)p(b)$$ but does $$p(a,b|c)=p(a|b|c)p(b|c)$$?
Well presumably, they are equal if $$p(a|b|c) = p(a|b,c)$$ - but does this equality hold?
Yes: Aim to prove: $$p(a|b|c) = p(a|b,c)$$
$$p(a,b|c) = p(a|b|c)p(b|c)$$
$$p(a|b|c) = dfrac{p(a,b|c)}{p(b|c)} = dfrac{ dfrac{p(a,b,c)}{p(c)}} {dfrac{p(b,c)}{p(c)}} = dfrac{p(a,b,c)}{p(b,c)} = dfrac{p(a|b,c)p(b,c)}{p(b,c)} = p(a|b,c)$$
answered Jan 23 at 20:20
Oliver GoldsteinOliver Goldstein
111
answered Jan 23 at 20:20
Oliver GoldsteinOliver Goldstein
111
answered Jan 23 at 20:20
Oliver GoldsteinOliver Goldstein
111
answered Jan 23 at 20:20
Oliver GoldsteinOliver Goldstein
111
111
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For others' benefit, I note that this answer (and the underlying topic) is very similar to that in this post.
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– davidlowryduda♦
Jan 23 at 21:21
add a comment |
$begingroup$
For others' benefit, I note that this answer (and the underlying topic) is very similar to that in this post.
$endgroup$
– davidlowryduda♦
Jan 23 at 21:21
$begingroup$
For others' benefit, I note that this answer (and the underlying topic) is very similar to that in this post.
$endgroup$
– davidlowryduda♦
Jan 23 at 21:21
$begingroup$
For others' benefit, I note that this answer (and the underlying topic) is very similar to that in this post.
$endgroup$
– davidlowryduda♦
Jan 23 at 21:21
add a comment |
$begingroup$
The said notation does not exist for $x,y,z$ events.
However, the identity could be proved by
$$P(x,ymid z)=frac{P(x,y,z)}{P(z)}=frac{P(xmid y,z)P(y,z)}{P(z)}$$
$$=P(xmid y,z)frac{P(y,z)}{P(z)}=P(xmid y,z)P(ymid z)$$
answered Aug 7 '17 at 7:06
BAIBAI
1,778518
$endgroup$
add a comment |
$begingroup$
The said notation does not exist for $x,y,z$ events.
However, the identity could be proved by
$$P(x,ymid z)=frac{P(x,y,z)}{P(z)}=frac{P(xmid y,z)P(y,z)}{P(z)}$$
$$=P(xmid y,z)frac{P(y,z)}{P(z)}=P(xmid y,z)P(ymid z)$$
answered Aug 7 '17 at 7:06
BAIBAI
1,778518
$endgroup$
add a comment |
$begingroup$
The said notation does not exist for $x,y,z$ events.
However, the identity could be proved by
$$P(x,ymid z)=frac{P(x,y,z)}{P(z)}=frac{P(xmid y,z)P(y,z)}{P(z)}$$
$$=P(xmid y,z)frac{P(y,z)}{P(z)}=P(xmid y,z)P(ymid z)$$
answered Aug 7 '17 at 7:06
BAIBAI
1,778518
$endgroup$
The said notation does not exist for $x,y,z$ events.
However, the identity could be proved by
$$P(x,ymid z)=frac{P(x,y,z)}{P(z)}=frac{P(xmid y,z)P(y,z)}{P(z)}$$
$$=P(xmid y,z)frac{P(y,z)}{P(z)}=P(xmid y,z)P(ymid z)$$
answered Aug 7 '17 at 7:06
BAIBAI
1,778518
answered Aug 7 '17 at 7:06
BAIBAI
1,778518
answered Aug 7 '17 at 7:06
BAIBAI
1,778518
answered Aug 7 '17 at 7:06
BAIBAI
1,778518
1,778518
add a comment |
add a comment |
$begingroup$
Doubtless this derivation is included either in the body of the text or as an exercise across a great array of textbooks across numerous fields. For example, #13.16(a.) from Russell & Norvig's Artificial Intelligence, A Modern Approach, Third Edition, says,
Prove the conditionalized version of the general product rule:
$P(X,Y|e) = P(X|Y,e)P(Y|e)$.
$$P(X,Y|e) = frac{P(X,Y,e)}{P(e)}$$
$$P(X|Y,e) = frac{P(X,Y,e)}{P(Y,e)}$$
$$P(X,Y,e) = frac{P(X|Y,e)}{P(Y,e)}$$
$$P(X,Y|e) = frac{P(X|y,e)P(Y,e)}{P(e)} = P(X|Y,e)P(Y|e)$$
In the first of the 4 lines in the derivation, above, we have $P(X,Y|e) = frac{P(X,Y,e)}{P(e)}$. We can think of this, perhaps overly simplistically (but it is a nice, workable mental representation so we can see what is going on here), as $P(A|B) = P(A, B)/P(B)$, simply Bayes familiar rule, where we have replaced A with $Xcap Y$ and e with B. To map my answer back on to the original question replace X with x, Y with y, and e with z. C.f. page 9 of 10, question & answer #6, here as well. Cheers!
answered Apr 1 '18 at 23:04
E.J.E.J.
1
$endgroup$
add a comment |
$begingroup$
Doubtless this derivation is included either in the body of the text or as an exercise across a great array of textbooks across numerous fields. For example, #13.16(a.) from Russell & Norvig's Artificial Intelligence, A Modern Approach, Third Edition, says,
Prove the conditionalized version of the general product rule:
$P(X,Y|e) = P(X|Y,e)P(Y|e)$.
$$P(X,Y|e) = frac{P(X,Y,e)}{P(e)}$$
$$P(X|Y,e) = frac{P(X,Y,e)}{P(Y,e)}$$
$$P(X,Y,e) = frac{P(X|Y,e)}{P(Y,e)}$$
$$P(X,Y|e) = frac{P(X|y,e)P(Y,e)}{P(e)} = P(X|Y,e)P(Y|e)$$
In the first of the 4 lines in the derivation, above, we have $P(X,Y|e) = frac{P(X,Y,e)}{P(e)}$. We can think of this, perhaps overly simplistically (but it is a nice, workable mental representation so we can see what is going on here), as $P(A|B) = P(A, B)/P(B)$, simply Bayes familiar rule, where we have replaced A with $Xcap Y$ and e with B. To map my answer back on to the original question replace X with x, Y with y, and e with z. C.f. page 9 of 10, question & answer #6, here as well. Cheers!
answered Apr 1 '18 at 23:04
E.J.E.J.
1
$endgroup$
add a comment |
$begingroup$
Doubtless this derivation is included either in the body of the text or as an exercise across a great array of textbooks across numerous fields. For example, #13.16(a.) from Russell & Norvig's Artificial Intelligence, A Modern Approach, Third Edition, says,
Prove the conditionalized version of the general product rule:
$P(X,Y|e) = P(X|Y,e)P(Y|e)$.
$$P(X,Y|e) = frac{P(X,Y,e)}{P(e)}$$
$$P(X|Y,e) = frac{P(X,Y,e)}{P(Y,e)}$$
$$P(X,Y,e) = frac{P(X|Y,e)}{P(Y,e)}$$
$$P(X,Y|e) = frac{P(X|y,e)P(Y,e)}{P(e)} = P(X|Y,e)P(Y|e)$$
In the first of the 4 lines in the derivation, above, we have $P(X,Y|e) = frac{P(X,Y,e)}{P(e)}$. We can think of this, perhaps overly simplistically (but it is a nice, workable mental representation so we can see what is going on here), as $P(A|B) = P(A, B)/P(B)$, simply Bayes familiar rule, where we have replaced A with $Xcap Y$ and e with B. To map my answer back on to the original question replace X with x, Y with y, and e with z. C.f. page 9 of 10, question & answer #6, here as well. Cheers!
answered Apr 1 '18 at 23:04
E.J.E.J.
1
$endgroup$
Doubtless this derivation is included either in the body of the text or as an exercise across a great array of textbooks across numerous fields. For example, #13.16(a.) from Russell & Norvig's Artificial Intelligence, A Modern Approach, Third Edition, says,
Prove the conditionalized version of the general product rule:
$P(X,Y|e) = P(X|Y,e)P(Y|e)$.
$$P(X,Y|e) = frac{P(X,Y,e)}{P(e)}$$
$$P(X|Y,e) = frac{P(X,Y,e)}{P(Y,e)}$$
$$P(X,Y,e) = frac{P(X|Y,e)}{P(Y,e)}$$
$$P(X,Y|e) = frac{P(X|y,e)P(Y,e)}{P(e)} = P(X|Y,e)P(Y|e)$$
In the first of the 4 lines in the derivation, above, we have $P(X,Y|e) = frac{P(X,Y,e)}{P(e)}$. We can think of this, perhaps overly simplistically (but it is a nice, workable mental representation so we can see what is going on here), as $P(A|B) = P(A, B)/P(B)$, simply Bayes familiar rule, where we have replaced A with $Xcap Y$ and e with B. To map my answer back on to the original question replace X with x, Y with y, and e with z. C.f. page 9 of 10, question & answer #6, here as well. Cheers!
answered Apr 1 '18 at 23:04
E.J.E.J.
1
answered Apr 1 '18 at 23:04
E.J.E.J.
1
answered Apr 1 '18 at 23:04
E.J.E.J.
1
answered Apr 1 '18 at 23:04
E.J.E.J.
1
1
add a comment |
add a comment |
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$p(xmid y mid z)$ is not good notation, but to the extent it suggests something like "the probability (or probability density) of $X$ given $Y=y$ given $Z=z$" then this might really be "the probability (or probability density) of $X$ given $Y=y$ and $Z=z$", i.e. $p(xmid y , z)$
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– Henry
Aug 7 '17 at 6:33
1
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Related: math.stackexchange.com/questions/301207/why-is-px-yz-pyx-zpxz
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– Henry
Aug 7 '17 at 6:33
1
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Thanks Henry...
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– Anjan
Aug 7 '17 at 6:37