Define a probability distribution which satisfies independence conditions
$begingroup$
I have the following problem:
A certain hospital receives patients.
Each patient could be healthy or sick.
There are 2 doctors: Gabby, and Tully.
Each doctor writes a report with his opinion.
There are 3 random variables: P, G, T with values {0,1} when 1 indicates sick and 0 indicates healthy.
P is the real status of the patients (Sick or healthy).
P[P=0] = P[P=1] = 0.5.
G, T indicates the doctors (Gabby, Tully) opinion.
Describe a distribution above P, G, T so that G, T will be independent,
but G, T are not independent given P.
The solution should be a table with all possible values of P, G, T
with the correct distribution for each combination.
My question is how to approach such a question?
I've tried making a table like this:
P | G | T | P(G|P) | P(T|P) | P(G,T|P)
0 | 0 | 0 | 0.3 | 0.6 | 0.5
0 | 0 | 1 | 0.3 | 0.6 | 0.5
0 | 1 | 0 | 0.3 | 0.6 | 0.5
0 | 1 | 1 | 0.3 | 0.6 | 0.5
1 | 0 | 0 | 0.3 | 0.6 | 0.5
1 | 0 | 1 | 0.3 | 0.6 | 0.5
1 | 1 | 0 | 0.3 | 0.6 | 0.5
1 | 1 | 1 | 0.3 | 0.6 | 0.5
And now I have P(G,T|P) != P(G|P) * P(T|P) as needed,
(Because of P(G|P) = 0.3, P(T|P) = 0.6, P(G|P) * P(T|P) = 0.5)
But I'm not sure how to calculate P(G), P(T), and check their independence.
Will be glad to know to approach this problem. Thanks.
Edit: This is my current solution:
probability probability-distributions random-variables bayes-theorem
edited Jan 8 at 10:00
TZakrevskiy
20.2k12353
asked Dec 29 '18 at 16:44
JohnSnowTheDeveloperJohnSnowTheDeveloper
656
$endgroup$
add a comment |
$begingroup$
I have the following problem:
A certain hospital receives patients.
Each patient could be healthy or sick.
There are 2 doctors: Gabby, and Tully.
Each doctor writes a report with his opinion.
There are 3 random variables: P, G, T with values {0,1} when 1 indicates sick and 0 indicates healthy.
P is the real status of the patients (Sick or healthy).
P[P=0] = P[P=1] = 0.5.
G, T indicates the doctors (Gabby, Tully) opinion.
Describe a distribution above P, G, T so that G, T will be independent,
but G, T are not independent given P.
The solution should be a table with all possible values of P, G, T
with the correct distribution for each combination.
My question is how to approach such a question?
I've tried making a table like this:
P | G | T | P(G|P) | P(T|P) | P(G,T|P)
0 | 0 | 0 | 0.3 | 0.6 | 0.5
0 | 0 | 1 | 0.3 | 0.6 | 0.5
0 | 1 | 0 | 0.3 | 0.6 | 0.5
0 | 1 | 1 | 0.3 | 0.6 | 0.5
1 | 0 | 0 | 0.3 | 0.6 | 0.5
1 | 0 | 1 | 0.3 | 0.6 | 0.5
1 | 1 | 0 | 0.3 | 0.6 | 0.5
1 | 1 | 1 | 0.3 | 0.6 | 0.5
And now I have P(G,T|P) != P(G|P) * P(T|P) as needed,
(Because of P(G|P) = 0.3, P(T|P) = 0.6, P(G|P) * P(T|P) = 0.5)
But I'm not sure how to calculate P(G), P(T), and check their independence.
Will be glad to know to approach this problem. Thanks.
Edit: This is my current solution:
probability probability-distributions random-variables bayes-theorem
edited Jan 8 at 10:00
TZakrevskiy
20.2k12353
asked Dec 29 '18 at 16:44
JohnSnowTheDeveloperJohnSnowTheDeveloper
656
$endgroup$
$begingroup$
In the table you correctly specify all possible outcomes $(0,0,0), (0,0,1), ldots$ but you are missing the probability of each outcome! That is, what is the probability $mathbb{P}(P = 0, G=0, T=0)$? Also I am not sure about the meaning of $mathbb{P}(G|P)$. Maybe you meant $mathbb{P}(G=x|P=y)$ for all $x,y in {0,1}$?
$endgroup$
– Michh
Dec 29 '18 at 18:56
$begingroup$
First, for your question - it means P(G=x | P=y) for the values given in the table. For example in the first row, P(G=0 | P=0). Second, I've tried to add the probabilities such P(P=0,G=0,T=0) but I still dont understand how to satisfies all the constraints
$endgroup$
– JohnSnowTheDeveloper
Dec 29 '18 at 19:17
add a comment |
$begingroup$
I have the following problem:
A certain hospital receives patients.
Each patient could be healthy or sick.
There are 2 doctors: Gabby, and Tully.
Each doctor writes a report with his opinion.
There are 3 random variables: P, G, T with values {0,1} when 1 indicates sick and 0 indicates healthy.
P is the real status of the patients (Sick or healthy).
P[P=0] = P[P=1] = 0.5.
G, T indicates the doctors (Gabby, Tully) opinion.
Describe a distribution above P, G, T so that G, T will be independent,
but G, T are not independent given P.
The solution should be a table with all possible values of P, G, T
with the correct distribution for each combination.
My question is how to approach such a question?
I've tried making a table like this:
P | G | T | P(G|P) | P(T|P) | P(G,T|P)
0 | 0 | 0 | 0.3 | 0.6 | 0.5
0 | 0 | 1 | 0.3 | 0.6 | 0.5
0 | 1 | 0 | 0.3 | 0.6 | 0.5
0 | 1 | 1 | 0.3 | 0.6 | 0.5
1 | 0 | 0 | 0.3 | 0.6 | 0.5
1 | 0 | 1 | 0.3 | 0.6 | 0.5
1 | 1 | 0 | 0.3 | 0.6 | 0.5
1 | 1 | 1 | 0.3 | 0.6 | 0.5
And now I have P(G,T|P) != P(G|P) * P(T|P) as needed,
(Because of P(G|P) = 0.3, P(T|P) = 0.6, P(G|P) * P(T|P) = 0.5)
But I'm not sure how to calculate P(G), P(T), and check their independence.
Will be glad to know to approach this problem. Thanks.
Edit: This is my current solution:
probability probability-distributions random-variables bayes-theorem
edited Jan 8 at 10:00
TZakrevskiy
20.2k12353
asked Dec 29 '18 at 16:44
JohnSnowTheDeveloperJohnSnowTheDeveloper
656
$endgroup$
I have the following problem:
A certain hospital receives patients.
Each patient could be healthy or sick.
There are 2 doctors: Gabby, and Tully.
Each doctor writes a report with his opinion.
There are 3 random variables: P, G, T with values {0,1} when 1 indicates sick and 0 indicates healthy.
P is the real status of the patients (Sick or healthy).
P[P=0] = P[P=1] = 0.5.
G, T indicates the doctors (Gabby, Tully) opinion.
Describe a distribution above P, G, T so that G, T will be independent,
but G, T are not independent given P.
The solution should be a table with all possible values of P, G, T
with the correct distribution for each combination.
My question is how to approach such a question?
I've tried making a table like this:
P | G | T | P(G|P) | P(T|P) | P(G,T|P)
0 | 0 | 0 | 0.3 | 0.6 | 0.5
0 | 0 | 1 | 0.3 | 0.6 | 0.5
0 | 1 | 0 | 0.3 | 0.6 | 0.5
0 | 1 | 1 | 0.3 | 0.6 | 0.5
1 | 0 | 0 | 0.3 | 0.6 | 0.5
1 | 0 | 1 | 0.3 | 0.6 | 0.5
1 | 1 | 0 | 0.3 | 0.6 | 0.5
1 | 1 | 1 | 0.3 | 0.6 | 0.5
And now I have P(G,T|P) != P(G|P) * P(T|P) as needed,
(Because of P(G|P) = 0.3, P(T|P) = 0.6, P(G|P) * P(T|P) = 0.5)
But I'm not sure how to calculate P(G), P(T), and check their independence.
Will be glad to know to approach this problem. Thanks.
Edit: This is my current solution:
probability probability-distributions random-variables bayes-theorem
probability probability-distributions random-variables bayes-theorem
edited Jan 8 at 10:00
TZakrevskiy
20.2k12353
asked Dec 29 '18 at 16:44
JohnSnowTheDeveloperJohnSnowTheDeveloper
656
edited Jan 8 at 10:00
TZakrevskiy
20.2k12353
asked Dec 29 '18 at 16:44
JohnSnowTheDeveloperJohnSnowTheDeveloper
656
edited Jan 8 at 10:00
TZakrevskiy
20.2k12353
edited Jan 8 at 10:00
TZakrevskiy
20.2k12353
edited Jan 8 at 10:00
TZakrevskiy
20.2k12353
20.2k12353
asked Dec 29 '18 at 16:44
JohnSnowTheDeveloperJohnSnowTheDeveloper
656
asked Dec 29 '18 at 16:44
JohnSnowTheDeveloperJohnSnowTheDeveloper
656
asked Dec 29 '18 at 16:44
JohnSnowTheDeveloperJohnSnowTheDeveloper
656
656
$begingroup$
In the table you correctly specify all possible outcomes $(0,0,0), (0,0,1), ldots$ but you are missing the probability of each outcome! That is, what is the probability $mathbb{P}(P = 0, G=0, T=0)$? Also I am not sure about the meaning of $mathbb{P}(G|P)$. Maybe you meant $mathbb{P}(G=x|P=y)$ for all $x,y in {0,1}$?
$endgroup$
– Michh
Dec 29 '18 at 18:56
$begingroup$
First, for your question - it means P(G=x | P=y) for the values given in the table. For example in the first row, P(G=0 | P=0). Second, I've tried to add the probabilities such P(P=0,G=0,T=0) but I still dont understand how to satisfies all the constraints
$endgroup$
– JohnSnowTheDeveloper
Dec 29 '18 at 19:17
add a comment |
$begingroup$
In the table you correctly specify all possible outcomes $(0,0,0), (0,0,1), ldots$ but you are missing the probability of each outcome! That is, what is the probability $mathbb{P}(P = 0, G=0, T=0)$? Also I am not sure about the meaning of $mathbb{P}(G|P)$. Maybe you meant $mathbb{P}(G=x|P=y)$ for all $x,y in {0,1}$?
$endgroup$
– Michh
Dec 29 '18 at 18:56
$begingroup$
First, for your question - it means P(G=x | P=y) for the values given in the table. For example in the first row, P(G=0 | P=0). Second, I've tried to add the probabilities such P(P=0,G=0,T=0) but I still dont understand how to satisfies all the constraints
$endgroup$
– JohnSnowTheDeveloper
Dec 29 '18 at 19:17
$begingroup$
In the table you correctly specify all possible outcomes $(0,0,0), (0,0,1), ldots$ but you are missing the probability of each outcome! That is, what is the probability $mathbb{P}(P = 0, G=0, T=0)$? Also I am not sure about the meaning of $mathbb{P}(G|P)$. Maybe you meant $mathbb{P}(G=x|P=y)$ for all $x,y in {0,1}$?
$endgroup$
– Michh
Dec 29 '18 at 18:56
$begingroup$
In the table you correctly specify all possible outcomes $(0,0,0), (0,0,1), ldots$ but you are missing the probability of each outcome! That is, what is the probability $mathbb{P}(P = 0, G=0, T=0)$? Also I am not sure about the meaning of $mathbb{P}(G|P)$. Maybe you meant $mathbb{P}(G=x|P=y)$ for all $x,y in {0,1}$?
$endgroup$
– Michh
Dec 29 '18 at 18:56
$begingroup$
First, for your question - it means P(G=x | P=y) for the values given in the table. For example in the first row, P(G=0 | P=0). Second, I've tried to add the probabilities such P(P=0,G=0,T=0) but I still dont understand how to satisfies all the constraints
$endgroup$
– JohnSnowTheDeveloper
Dec 29 '18 at 19:17
$begingroup$
First, for your question - it means P(G=x | P=y) for the values given in the table. For example in the first row, P(G=0 | P=0). Second, I've tried to add the probabilities such P(P=0,G=0,T=0) but I still dont understand how to satisfies all the constraints
$endgroup$
– JohnSnowTheDeveloper
Dec 29 '18 at 19:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As Misch said, in each row of the table you should write $mathbb{P}( P = y, G= x, T = z )$, where $x,y,z$ equal $0$ or $1$. Then:
- All eight probabilities should sum to $1$ - they represent all possibilities.
- You calculate $mathbb{P}(P = y )$ by taking the sum over the four rows where $P = y$. This sum should equal $0.5$ for both $y=0$ and $y=1$.
- You calculate $mathbb{P}(G = x )$ by taking the sum over the four rows where $G = x$, and find $mathbb{P}(T = z )$ in the same way.
- You calculate $mathbb{P}(G = x, T = z )$ by taking the sum over the two rows where both $G=x$ and $T=z$.
To show that $G$ and $T$ are independent, you must show that $mathbb{P}(G = x ) mathbb{P}(T = z ) = mathbb{P}(G = x, T =z )$ for all four possible values of $(x,z)$. To show that $G$ and $T$ are not conditionally independent given $P$, it is enough to find one set of values for $(x,y,z)$ where $mathbb{P}( G = x | P = y) mathbb{P}( T = z | P = y) neq mathbb{P}( G = x, T = z | P = y)$.
As there are infinitely many solutions, it may help if you add your own constraints. For instance, you may set $mathbb{P}(G = x ) = 0.5$ for $x=0,1$, and $mathbb{P}(T = z ) = 0.5$ for $ z=0,1$.
answered Dec 29 '18 at 21:13
JAskgaardJAskgaard
1367
$endgroup$
$begingroup$
Thanks, I've edited my question to include the answer, I hope I understood you well (Feel free to fix me if I'm wrong)
$endgroup$
– JohnSnowTheDeveloper
Dec 30 '18 at 19:13
add a comment |
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$begingroup$
As Misch said, in each row of the table you should write $mathbb{P}( P = y, G= x, T = z )$, where $x,y,z$ equal $0$ or $1$. Then:
- All eight probabilities should sum to $1$ - they represent all possibilities.
- You calculate $mathbb{P}(P = y )$ by taking the sum over the four rows where $P = y$. This sum should equal $0.5$ for both $y=0$ and $y=1$.
- You calculate $mathbb{P}(G = x )$ by taking the sum over the four rows where $G = x$, and find $mathbb{P}(T = z )$ in the same way.
- You calculate $mathbb{P}(G = x, T = z )$ by taking the sum over the two rows where both $G=x$ and $T=z$.
To show that $G$ and $T$ are independent, you must show that $mathbb{P}(G = x ) mathbb{P}(T = z ) = mathbb{P}(G = x, T =z )$ for all four possible values of $(x,z)$. To show that $G$ and $T$ are not conditionally independent given $P$, it is enough to find one set of values for $(x,y,z)$ where $mathbb{P}( G = x | P = y) mathbb{P}( T = z | P = y) neq mathbb{P}( G = x, T = z | P = y)$.
As there are infinitely many solutions, it may help if you add your own constraints. For instance, you may set $mathbb{P}(G = x ) = 0.5$ for $x=0,1$, and $mathbb{P}(T = z ) = 0.5$ for $ z=0,1$.
answered Dec 29 '18 at 21:13
JAskgaardJAskgaard
1367
$endgroup$
$begingroup$
Thanks, I've edited my question to include the answer, I hope I understood you well (Feel free to fix me if I'm wrong)
$endgroup$
– JohnSnowTheDeveloper
Dec 30 '18 at 19:13
add a comment |
$begingroup$
As Misch said, in each row of the table you should write $mathbb{P}( P = y, G= x, T = z )$, where $x,y,z$ equal $0$ or $1$. Then:
- All eight probabilities should sum to $1$ - they represent all possibilities.
- You calculate $mathbb{P}(P = y )$ by taking the sum over the four rows where $P = y$. This sum should equal $0.5$ for both $y=0$ and $y=1$.
- You calculate $mathbb{P}(G = x )$ by taking the sum over the four rows where $G = x$, and find $mathbb{P}(T = z )$ in the same way.
- You calculate $mathbb{P}(G = x, T = z )$ by taking the sum over the two rows where both $G=x$ and $T=z$.
To show that $G$ and $T$ are independent, you must show that $mathbb{P}(G = x ) mathbb{P}(T = z ) = mathbb{P}(G = x, T =z )$ for all four possible values of $(x,z)$. To show that $G$ and $T$ are not conditionally independent given $P$, it is enough to find one set of values for $(x,y,z)$ where $mathbb{P}( G = x | P = y) mathbb{P}( T = z | P = y) neq mathbb{P}( G = x, T = z | P = y)$.
As there are infinitely many solutions, it may help if you add your own constraints. For instance, you may set $mathbb{P}(G = x ) = 0.5$ for $x=0,1$, and $mathbb{P}(T = z ) = 0.5$ for $ z=0,1$.
answered Dec 29 '18 at 21:13
JAskgaardJAskgaard
1367
$endgroup$
$begingroup$
Thanks, I've edited my question to include the answer, I hope I understood you well (Feel free to fix me if I'm wrong)
$endgroup$
– JohnSnowTheDeveloper
Dec 30 '18 at 19:13
add a comment |
$begingroup$
As Misch said, in each row of the table you should write $mathbb{P}( P = y, G= x, T = z )$, where $x,y,z$ equal $0$ or $1$. Then:
- All eight probabilities should sum to $1$ - they represent all possibilities.
- You calculate $mathbb{P}(P = y )$ by taking the sum over the four rows where $P = y$. This sum should equal $0.5$ for both $y=0$ and $y=1$.
- You calculate $mathbb{P}(G = x )$ by taking the sum over the four rows where $G = x$, and find $mathbb{P}(T = z )$ in the same way.
- You calculate $mathbb{P}(G = x, T = z )$ by taking the sum over the two rows where both $G=x$ and $T=z$.
To show that $G$ and $T$ are independent, you must show that $mathbb{P}(G = x ) mathbb{P}(T = z ) = mathbb{P}(G = x, T =z )$ for all four possible values of $(x,z)$. To show that $G$ and $T$ are not conditionally independent given $P$, it is enough to find one set of values for $(x,y,z)$ where $mathbb{P}( G = x | P = y) mathbb{P}( T = z | P = y) neq mathbb{P}( G = x, T = z | P = y)$.
As there are infinitely many solutions, it may help if you add your own constraints. For instance, you may set $mathbb{P}(G = x ) = 0.5$ for $x=0,1$, and $mathbb{P}(T = z ) = 0.5$ for $ z=0,1$.
answered Dec 29 '18 at 21:13
JAskgaardJAskgaard
1367
$endgroup$
As Misch said, in each row of the table you should write $mathbb{P}( P = y, G= x, T = z )$, where $x,y,z$ equal $0$ or $1$. Then:
- All eight probabilities should sum to $1$ - they represent all possibilities.
- You calculate $mathbb{P}(P = y )$ by taking the sum over the four rows where $P = y$. This sum should equal $0.5$ for both $y=0$ and $y=1$.
- You calculate $mathbb{P}(G = x )$ by taking the sum over the four rows where $G = x$, and find $mathbb{P}(T = z )$ in the same way.
- You calculate $mathbb{P}(G = x, T = z )$ by taking the sum over the two rows where both $G=x$ and $T=z$.
To show that $G$ and $T$ are independent, you must show that $mathbb{P}(G = x ) mathbb{P}(T = z ) = mathbb{P}(G = x, T =z )$ for all four possible values of $(x,z)$. To show that $G$ and $T$ are not conditionally independent given $P$, it is enough to find one set of values for $(x,y,z)$ where $mathbb{P}( G = x | P = y) mathbb{P}( T = z | P = y) neq mathbb{P}( G = x, T = z | P = y)$.
As there are infinitely many solutions, it may help if you add your own constraints. For instance, you may set $mathbb{P}(G = x ) = 0.5$ for $x=0,1$, and $mathbb{P}(T = z ) = 0.5$ for $ z=0,1$.
answered Dec 29 '18 at 21:13
JAskgaardJAskgaard
1367
edited Dec 30 '18 at 14:36
answered Dec 29 '18 at 21:13
JAskgaardJAskgaard
1367
answered Dec 29 '18 at 21:13
JAskgaardJAskgaard
1367
answered Dec 29 '18 at 21:13
JAskgaardJAskgaard
1367
1367
$begingroup$
Thanks, I've edited my question to include the answer, I hope I understood you well (Feel free to fix me if I'm wrong)
$endgroup$
– JohnSnowTheDeveloper
Dec 30 '18 at 19:13
add a comment |
$begingroup$
Thanks, I've edited my question to include the answer, I hope I understood you well (Feel free to fix me if I'm wrong)
$endgroup$
– JohnSnowTheDeveloper
Dec 30 '18 at 19:13
$begingroup$
Thanks, I've edited my question to include the answer, I hope I understood you well (Feel free to fix me if I'm wrong)
$endgroup$
– JohnSnowTheDeveloper
Dec 30 '18 at 19:13
$begingroup$
Thanks, I've edited my question to include the answer, I hope I understood you well (Feel free to fix me if I'm wrong)
$endgroup$
– JohnSnowTheDeveloper
Dec 30 '18 at 19:13
add a comment |
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$begingroup$
In the table you correctly specify all possible outcomes $(0,0,0), (0,0,1), ldots$ but you are missing the probability of each outcome! That is, what is the probability $mathbb{P}(P = 0, G=0, T=0)$? Also I am not sure about the meaning of $mathbb{P}(G|P)$. Maybe you meant $mathbb{P}(G=x|P=y)$ for all $x,y in {0,1}$?
$endgroup$
– Michh
Dec 29 '18 at 18:56
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First, for your question - it means P(G=x | P=y) for the values given in the table. For example in the first row, P(G=0 | P=0). Second, I've tried to add the probabilities such P(P=0,G=0,T=0) but I still dont understand how to satisfies all the constraints
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– JohnSnowTheDeveloper
Dec 29 '18 at 19:17