Simplification of conditional probability expression
$begingroup$
I have the following conditional probability expression:
$$
frac{frac{P(E|D)P(D)}{P(E)-P(E|D)P(D)}}{frac{P(bar{E}|D)P(D)}{P(bar{E})-P(bar{E}|D)P(D)}}
$$
I want to simplify it to:
$$
frac{frac{P(E|D)}{1-P(E|D)}}{frac{P(E|bar{D})}{1-P(E|bar{D})}}
$$
I already did this on my own but in an overly lengthy way and I was told this should be much more simple. We could remove $P(D)$ from both numerators but other than that I'm pretty lost. I also don't know how to "move" the negation sign from E to D in the denominator part.
Any help will be welcome. Thanks!
probability conditional-probability
asked Dec 3 '18 at 12:57
zest16zest16
33
$endgroup$
add a comment |
$begingroup$
I have the following conditional probability expression:
$$
frac{frac{P(E|D)P(D)}{P(E)-P(E|D)P(D)}}{frac{P(bar{E}|D)P(D)}{P(bar{E})-P(bar{E}|D)P(D)}}
$$
I want to simplify it to:
$$
frac{frac{P(E|D)}{1-P(E|D)}}{frac{P(E|bar{D})}{1-P(E|bar{D})}}
$$
I already did this on my own but in an overly lengthy way and I was told this should be much more simple. We could remove $P(D)$ from both numerators but other than that I'm pretty lost. I also don't know how to "move" the negation sign from E to D in the denominator part.
Any help will be welcome. Thanks!
probability conditional-probability
asked Dec 3 '18 at 12:57
zest16zest16
33
$endgroup$
add a comment |
$begingroup$
I have the following conditional probability expression:
$$
frac{frac{P(E|D)P(D)}{P(E)-P(E|D)P(D)}}{frac{P(bar{E}|D)P(D)}{P(bar{E})-P(bar{E}|D)P(D)}}
$$
I want to simplify it to:
$$
frac{frac{P(E|D)}{1-P(E|D)}}{frac{P(E|bar{D})}{1-P(E|bar{D})}}
$$
I already did this on my own but in an overly lengthy way and I was told this should be much more simple. We could remove $P(D)$ from both numerators but other than that I'm pretty lost. I also don't know how to "move" the negation sign from E to D in the denominator part.
Any help will be welcome. Thanks!
probability conditional-probability
asked Dec 3 '18 at 12:57
zest16zest16
33
$endgroup$
I have the following conditional probability expression:
$$
frac{frac{P(E|D)P(D)}{P(E)-P(E|D)P(D)}}{frac{P(bar{E}|D)P(D)}{P(bar{E})-P(bar{E}|D)P(D)}}
$$
I want to simplify it to:
$$
frac{frac{P(E|D)}{1-P(E|D)}}{frac{P(E|bar{D})}{1-P(E|bar{D})}}
$$
I already did this on my own but in an overly lengthy way and I was told this should be much more simple. We could remove $P(D)$ from both numerators but other than that I'm pretty lost. I also don't know how to "move" the negation sign from E to D in the denominator part.
Any help will be welcome. Thanks!
probability conditional-probability
probability conditional-probability
asked Dec 3 '18 at 12:57
zest16zest16
33
asked Dec 3 '18 at 12:57
zest16zest16
33
asked Dec 3 '18 at 12:57
zest16zest16
33
asked Dec 3 '18 at 12:57
zest16zest16
33
asked Dec 3 '18 at 12:57
zest16zest16
33
33
add a comment |
add a comment |
1 Answer
1
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votes
$begingroup$
Facts that might help:
begin{align}
P(bar Emid D)P(D)
&= (1 - P(Emid D))P(D) \[1ex]
P(E) - P(Emid D)P(D) &= P(E) - P(E cap D) \
&= P(E capbar D) \
&= P(Emidbar D)P(bar D)
end{align}
In case you need a derivation of the first fact:
begin{align}
P(bar Emid D)P(D) &= P(bar E cap D) \
&= P(D) - P(E cap D) \
&= P(D) - P(Emid D)P(D)\
&= (1 - P(Emid D))P(D)
end{align}
answered Dec 3 '18 at 14:22
David KDavid K
53.6k342116
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Facts that might help:
begin{align}
P(bar Emid D)P(D)
&= (1 - P(Emid D))P(D) \[1ex]
P(E) - P(Emid D)P(D) &= P(E) - P(E cap D) \
&= P(E capbar D) \
&= P(Emidbar D)P(bar D)
end{align}
In case you need a derivation of the first fact:
begin{align}
P(bar Emid D)P(D) &= P(bar E cap D) \
&= P(D) - P(E cap D) \
&= P(D) - P(Emid D)P(D)\
&= (1 - P(Emid D))P(D)
end{align}
answered Dec 3 '18 at 14:22
David KDavid K
53.6k342116
$endgroup$
add a comment |
$begingroup$
Facts that might help:
begin{align}
P(bar Emid D)P(D)
&= (1 - P(Emid D))P(D) \[1ex]
P(E) - P(Emid D)P(D) &= P(E) - P(E cap D) \
&= P(E capbar D) \
&= P(Emidbar D)P(bar D)
end{align}
In case you need a derivation of the first fact:
begin{align}
P(bar Emid D)P(D) &= P(bar E cap D) \
&= P(D) - P(E cap D) \
&= P(D) - P(Emid D)P(D)\
&= (1 - P(Emid D))P(D)
end{align}
answered Dec 3 '18 at 14:22
David KDavid K
53.6k342116
$endgroup$
add a comment |
$begingroup$
Facts that might help:
begin{align}
P(bar Emid D)P(D)
&= (1 - P(Emid D))P(D) \[1ex]
P(E) - P(Emid D)P(D) &= P(E) - P(E cap D) \
&= P(E capbar D) \
&= P(Emidbar D)P(bar D)
end{align}
In case you need a derivation of the first fact:
begin{align}
P(bar Emid D)P(D) &= P(bar E cap D) \
&= P(D) - P(E cap D) \
&= P(D) - P(Emid D)P(D)\
&= (1 - P(Emid D))P(D)
end{align}
answered Dec 3 '18 at 14:22
David KDavid K
53.6k342116
$endgroup$
Facts that might help:
begin{align}
P(bar Emid D)P(D)
&= (1 - P(Emid D))P(D) \[1ex]
P(E) - P(Emid D)P(D) &= P(E) - P(E cap D) \
&= P(E capbar D) \
&= P(Emidbar D)P(bar D)
end{align}
In case you need a derivation of the first fact:
begin{align}
P(bar Emid D)P(D) &= P(bar E cap D) \
&= P(D) - P(E cap D) \
&= P(D) - P(Emid D)P(D)\
&= (1 - P(Emid D))P(D)
end{align}
answered Dec 3 '18 at 14:22
David KDavid K
53.6k342116
answered Dec 3 '18 at 14:22
David KDavid K
53.6k342116
answered Dec 3 '18 at 14:22
David KDavid K
53.6k342116
answered Dec 3 '18 at 14:22
David KDavid K
53.6k342116
53.6k342116
add a comment |
add a comment |
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