How can I get the probability given the Bayesian table?
$begingroup$
Consider that $A rightarrow B$. And A has two states 0 and 1 respectively with probability of $0.6$ and $0.4$.
$$
mbox{And for} B:
left{begin{array}{rcl}
{displaystyle Prleft(B = 1 mid A = 1right)} & {displaystyle =} & {displaystyle 0.3}
\[1mm]
{displaystyle Prleft(B = 1 mid A = 0right)} & {displaystyle =} & {displaystyle 0.2}
\[1mm]
{displaystyle Prleft(B = 0 mid A = 1right)} & {displaystyle =} & {displaystyle 0.7}
\[1mm]
{displaystyle Prleft(B = 0 mid A = 0right)} & {displaystyle =} & {displaystyle 0.8}
end{array}right.
$$
What I want to get is $Pr(B=1)$, and I know the result is $frac{0.3+0.2}{0.3+0.2+0.7+0.8}$. But my question is how can I get the result provided the two tables?
My reduction:
$$
begin{align}
Pr(B=1) &= frac{Pr(B=1|A)}{Pr(B=1|A) + Pr(B=0|A)}\
& = frac{frac{Pr(B=1,A)}{Pr(A)}}{frac{Pr(B=1,A)}{Pr(A)}+frac{Pr(B=0,A)}{Pr(A)}} \
& = frac{Pr(B=1, A=0) + Pr(B=1, A=1)}{Pr(B=1, A=0) + Pr(B=1, A=1) + Pr(B=0, A=0) + Pr(B=0, A=1)}
end{align}
$$
Am I right?
Any hints or suggestions would be highly appreciated. Thank you!
bayesian bayes-theorem
edited Jan 15 at 15:55
Felix Marin
68.8k7109146
asked Jan 15 at 13:49
Lerner ZhangLerner Zhang
314218
$endgroup$
add a comment |
$begingroup$
Consider that $A rightarrow B$. And A has two states 0 and 1 respectively with probability of $0.6$ and $0.4$.
$$
mbox{And for} B:
left{begin{array}{rcl}
{displaystyle Prleft(B = 1 mid A = 1right)} & {displaystyle =} & {displaystyle 0.3}
\[1mm]
{displaystyle Prleft(B = 1 mid A = 0right)} & {displaystyle =} & {displaystyle 0.2}
\[1mm]
{displaystyle Prleft(B = 0 mid A = 1right)} & {displaystyle =} & {displaystyle 0.7}
\[1mm]
{displaystyle Prleft(B = 0 mid A = 0right)} & {displaystyle =} & {displaystyle 0.8}
end{array}right.
$$
What I want to get is $Pr(B=1)$, and I know the result is $frac{0.3+0.2}{0.3+0.2+0.7+0.8}$. But my question is how can I get the result provided the two tables?
My reduction:
$$
begin{align}
Pr(B=1) &= frac{Pr(B=1|A)}{Pr(B=1|A) + Pr(B=0|A)}\
& = frac{frac{Pr(B=1,A)}{Pr(A)}}{frac{Pr(B=1,A)}{Pr(A)}+frac{Pr(B=0,A)}{Pr(A)}} \
& = frac{Pr(B=1, A=0) + Pr(B=1, A=1)}{Pr(B=1, A=0) + Pr(B=1, A=1) + Pr(B=0, A=0) + Pr(B=0, A=1)}
end{align}
$$
Am I right?
Any hints or suggestions would be highly appreciated. Thank you!
bayesian bayes-theorem
edited Jan 15 at 15:55
Felix Marin
68.8k7109146
asked Jan 15 at 13:49
Lerner ZhangLerner Zhang
314218
$endgroup$
$begingroup$
$Pr(B = 1) = frac{Pr(B=1}{sumlimits_{i=0}^{1}Pr(B = i)}$. Its essentially the "area" of $B = 1$ divided by the entire "area". Note that you need to take into account both cases $A = 0$ and $A = 1$ when computing $Pr(B = i)$
$endgroup$
– David
Jan 15 at 13:50
add a comment |
$begingroup$
Consider that $A rightarrow B$. And A has two states 0 and 1 respectively with probability of $0.6$ and $0.4$.
$$
mbox{And for} B:
left{begin{array}{rcl}
{displaystyle Prleft(B = 1 mid A = 1right)} & {displaystyle =} & {displaystyle 0.3}
\[1mm]
{displaystyle Prleft(B = 1 mid A = 0right)} & {displaystyle =} & {displaystyle 0.2}
\[1mm]
{displaystyle Prleft(B = 0 mid A = 1right)} & {displaystyle =} & {displaystyle 0.7}
\[1mm]
{displaystyle Prleft(B = 0 mid A = 0right)} & {displaystyle =} & {displaystyle 0.8}
end{array}right.
$$
What I want to get is $Pr(B=1)$, and I know the result is $frac{0.3+0.2}{0.3+0.2+0.7+0.8}$. But my question is how can I get the result provided the two tables?
My reduction:
$$
begin{align}
Pr(B=1) &= frac{Pr(B=1|A)}{Pr(B=1|A) + Pr(B=0|A)}\
& = frac{frac{Pr(B=1,A)}{Pr(A)}}{frac{Pr(B=1,A)}{Pr(A)}+frac{Pr(B=0,A)}{Pr(A)}} \
& = frac{Pr(B=1, A=0) + Pr(B=1, A=1)}{Pr(B=1, A=0) + Pr(B=1, A=1) + Pr(B=0, A=0) + Pr(B=0, A=1)}
end{align}
$$
Am I right?
Any hints or suggestions would be highly appreciated. Thank you!
bayesian bayes-theorem
edited Jan 15 at 15:55
Felix Marin
68.8k7109146
asked Jan 15 at 13:49
Lerner ZhangLerner Zhang
314218
$endgroup$
Consider that $A rightarrow B$. And A has two states 0 and 1 respectively with probability of $0.6$ and $0.4$.
$$
mbox{And for} B:
left{begin{array}{rcl}
{displaystyle Prleft(B = 1 mid A = 1right)} & {displaystyle =} & {displaystyle 0.3}
\[1mm]
{displaystyle Prleft(B = 1 mid A = 0right)} & {displaystyle =} & {displaystyle 0.2}
\[1mm]
{displaystyle Prleft(B = 0 mid A = 1right)} & {displaystyle =} & {displaystyle 0.7}
\[1mm]
{displaystyle Prleft(B = 0 mid A = 0right)} & {displaystyle =} & {displaystyle 0.8}
end{array}right.
$$
What I want to get is $Pr(B=1)$, and I know the result is $frac{0.3+0.2}{0.3+0.2+0.7+0.8}$. But my question is how can I get the result provided the two tables?
My reduction:
$$
begin{align}
Pr(B=1) &= frac{Pr(B=1|A)}{Pr(B=1|A) + Pr(B=0|A)}\
& = frac{frac{Pr(B=1,A)}{Pr(A)}}{frac{Pr(B=1,A)}{Pr(A)}+frac{Pr(B=0,A)}{Pr(A)}} \
& = frac{Pr(B=1, A=0) + Pr(B=1, A=1)}{Pr(B=1, A=0) + Pr(B=1, A=1) + Pr(B=0, A=0) + Pr(B=0, A=1)}
end{align}
$$
Am I right?
Any hints or suggestions would be highly appreciated. Thank you!
bayesian bayes-theorem
bayesian bayes-theorem
edited Jan 15 at 15:55
Felix Marin
68.8k7109146
asked Jan 15 at 13:49
Lerner ZhangLerner Zhang
314218
edited Jan 15 at 15:55
Felix Marin
68.8k7109146
asked Jan 15 at 13:49
Lerner ZhangLerner Zhang
314218
edited Jan 15 at 15:55
Felix Marin
68.8k7109146
edited Jan 15 at 15:55
Felix Marin
68.8k7109146
edited Jan 15 at 15:55
Felix Marin
68.8k7109146
68.8k7109146
asked Jan 15 at 13:49
Lerner ZhangLerner Zhang
314218
asked Jan 15 at 13:49
Lerner ZhangLerner Zhang
314218
asked Jan 15 at 13:49
Lerner ZhangLerner Zhang
314218
314218
$begingroup$
$Pr(B = 1) = frac{Pr(B=1}{sumlimits_{i=0}^{1}Pr(B = i)}$. Its essentially the "area" of $B = 1$ divided by the entire "area". Note that you need to take into account both cases $A = 0$ and $A = 1$ when computing $Pr(B = i)$
$endgroup$
– David
Jan 15 at 13:50
add a comment |
$begingroup$
$Pr(B = 1) = frac{Pr(B=1}{sumlimits_{i=0}^{1}Pr(B = i)}$. Its essentially the "area" of $B = 1$ divided by the entire "area". Note that you need to take into account both cases $A = 0$ and $A = 1$ when computing $Pr(B = i)$
$endgroup$
– David
Jan 15 at 13:50
$begingroup$
$Pr(B = 1) = frac{Pr(B=1}{sumlimits_{i=0}^{1}Pr(B = i)}$. Its essentially the "area" of $B = 1$ divided by the entire "area". Note that you need to take into account both cases $A = 0$ and $A = 1$ when computing $Pr(B = i)$
$endgroup$
– David
Jan 15 at 13:50
$begingroup$
$Pr(B = 1) = frac{Pr(B=1}{sumlimits_{i=0}^{1}Pr(B = i)}$. Its essentially the "area" of $B = 1$ divided by the entire "area". Note that you need to take into account both cases $A = 0$ and $A = 1$ when computing $Pr(B = i)$
$endgroup$
– David
Jan 15 at 13:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I know the result is $frac{0.3+0.2}{0.3+0.2+0.7+0.8}$
It is not right and it must be:
$$begin{align}P(B=1)=&P(B=1,A=0)+P(B=1,A=1)=\
=&P(A=0)cdot P(B=1|A=0)+P(A=1)cdot P(B=1|A=1)=\
=&0.6cdot 0.2+0.4cdot 0.3=\
=&0.24.end{align}$$
answered Jan 15 at 15:46
farruhotafarruhota
21.7k2842
$endgroup$
$begingroup$
Amazing! Great!
$endgroup$
– Lerner Zhang
Jan 15 at 22:53
$begingroup$
You are right. I have read the textbook and find this: The joint distribution of two random variables has to be consistent with the marginal distribution, in that $P(x)=sum_y P(x, y)$. It seems very intuitive. What I was thinking was totally of non-sense. I was just applying the equations mechanically and unfortunately wrongly.
$endgroup$
– Lerner Zhang
Jan 17 at 6:43
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
I know the result is $frac{0.3+0.2}{0.3+0.2+0.7+0.8}$
It is not right and it must be:
$$begin{align}P(B=1)=&P(B=1,A=0)+P(B=1,A=1)=\
=&P(A=0)cdot P(B=1|A=0)+P(A=1)cdot P(B=1|A=1)=\
=&0.6cdot 0.2+0.4cdot 0.3=\
=&0.24.end{align}$$
answered Jan 15 at 15:46
farruhotafarruhota
21.7k2842
$endgroup$
$begingroup$
Amazing! Great!
$endgroup$
– Lerner Zhang
Jan 15 at 22:53
$begingroup$
You are right. I have read the textbook and find this: The joint distribution of two random variables has to be consistent with the marginal distribution, in that $P(x)=sum_y P(x, y)$. It seems very intuitive. What I was thinking was totally of non-sense. I was just applying the equations mechanically and unfortunately wrongly.
$endgroup$
– Lerner Zhang
Jan 17 at 6:43
add a comment |
$begingroup$
I know the result is $frac{0.3+0.2}{0.3+0.2+0.7+0.8}$
It is not right and it must be:
$$begin{align}P(B=1)=&P(B=1,A=0)+P(B=1,A=1)=\
=&P(A=0)cdot P(B=1|A=0)+P(A=1)cdot P(B=1|A=1)=\
=&0.6cdot 0.2+0.4cdot 0.3=\
=&0.24.end{align}$$
answered Jan 15 at 15:46
farruhotafarruhota
21.7k2842
$endgroup$
$begingroup$
Amazing! Great!
$endgroup$
– Lerner Zhang
Jan 15 at 22:53
$begingroup$
You are right. I have read the textbook and find this: The joint distribution of two random variables has to be consistent with the marginal distribution, in that $P(x)=sum_y P(x, y)$. It seems very intuitive. What I was thinking was totally of non-sense. I was just applying the equations mechanically and unfortunately wrongly.
$endgroup$
– Lerner Zhang
Jan 17 at 6:43
add a comment |
$begingroup$
I know the result is $frac{0.3+0.2}{0.3+0.2+0.7+0.8}$
It is not right and it must be:
$$begin{align}P(B=1)=&P(B=1,A=0)+P(B=1,A=1)=\
=&P(A=0)cdot P(B=1|A=0)+P(A=1)cdot P(B=1|A=1)=\
=&0.6cdot 0.2+0.4cdot 0.3=\
=&0.24.end{align}$$
answered Jan 15 at 15:46
farruhotafarruhota
21.7k2842
$endgroup$
I know the result is $frac{0.3+0.2}{0.3+0.2+0.7+0.8}$
It is not right and it must be:
$$begin{align}P(B=1)=&P(B=1,A=0)+P(B=1,A=1)=\
=&P(A=0)cdot P(B=1|A=0)+P(A=1)cdot P(B=1|A=1)=\
=&0.6cdot 0.2+0.4cdot 0.3=\
=&0.24.end{align}$$
answered Jan 15 at 15:46
farruhotafarruhota
21.7k2842
answered Jan 15 at 15:46
farruhotafarruhota
21.7k2842
answered Jan 15 at 15:46
farruhotafarruhota
21.7k2842
answered Jan 15 at 15:46
farruhotafarruhota
21.7k2842
21.7k2842
$begingroup$
Amazing! Great!
$endgroup$
– Lerner Zhang
Jan 15 at 22:53
$begingroup$
You are right. I have read the textbook and find this: The joint distribution of two random variables has to be consistent with the marginal distribution, in that $P(x)=sum_y P(x, y)$. It seems very intuitive. What I was thinking was totally of non-sense. I was just applying the equations mechanically and unfortunately wrongly.
$endgroup$
– Lerner Zhang
Jan 17 at 6:43
add a comment |
$begingroup$
Amazing! Great!
$endgroup$
– Lerner Zhang
Jan 15 at 22:53
$begingroup$
You are right. I have read the textbook and find this: The joint distribution of two random variables has to be consistent with the marginal distribution, in that $P(x)=sum_y P(x, y)$. It seems very intuitive. What I was thinking was totally of non-sense. I was just applying the equations mechanically and unfortunately wrongly.
$endgroup$
– Lerner Zhang
Jan 17 at 6:43
$begingroup$
Amazing! Great!
$endgroup$
– Lerner Zhang
Jan 15 at 22:53
$begingroup$
Amazing! Great!
$endgroup$
– Lerner Zhang
Jan 15 at 22:53
$begingroup$
You are right. I have read the textbook and find this: The joint distribution of two random variables has to be consistent with the marginal distribution, in that $P(x)=sum_y P(x, y)$. It seems very intuitive. What I was thinking was totally of non-sense. I was just applying the equations mechanically and unfortunately wrongly.
$endgroup$
– Lerner Zhang
Jan 17 at 6:43
$begingroup$
You are right. I have read the textbook and find this: The joint distribution of two random variables has to be consistent with the marginal distribution, in that $P(x)=sum_y P(x, y)$. It seems very intuitive. What I was thinking was totally of non-sense. I was just applying the equations mechanically and unfortunately wrongly.
$endgroup$
– Lerner Zhang
Jan 17 at 6:43
add a comment |
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$begingroup$
$Pr(B = 1) = frac{Pr(B=1}{sumlimits_{i=0}^{1}Pr(B = i)}$. Its essentially the "area" of $B = 1$ divided by the entire "area". Note that you need to take into account both cases $A = 0$ and $A = 1$ when computing $Pr(B = i)$
$endgroup$
– David
Jan 15 at 13:50