Mixing
Calculate probability of an experiment given by a random variable
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The question: An experiment succeeds with a probability of $Z$, which is a random variable uniformly distributed over $[0,1]$. The experiment is conducted twice. Denote $X$ the random variable which equals 1 if the first experiment succeeded, and 0 otherwise. $Y$ is the same, but for the second experiment. What is $P(X=1)$ and $P(X=1, Y=1)$?
It's quite obvious that $P(X=1) = 1/2$, but I'm not sure how to prove it, and I have not the slightest clue on how to calculate $P(X=1, Y=1)$, since clearly $X,Y$ are dependent in one another.
I tried going at it with conditional probability, but it didn't help me much.
Any clues?
Thanks
probability random-variables
asked Jan 31 at 17:32
Barak BBarak B
265
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add a comment |
$begingroup$
The question: An experiment succeeds with a probability of $Z$, which is a random variable uniformly distributed over $[0,1]$. The experiment is conducted twice. Denote $X$ the random variable which equals 1 if the first experiment succeeded, and 0 otherwise. $Y$ is the same, but for the second experiment. What is $P(X=1)$ and $P(X=1, Y=1)$?
It's quite obvious that $P(X=1) = 1/2$, but I'm not sure how to prove it, and I have not the slightest clue on how to calculate $P(X=1, Y=1)$, since clearly $X,Y$ are dependent in one another.
I tried going at it with conditional probability, but it didn't help me much.
Any clues?
Thanks
probability random-variables
asked Jan 31 at 17:32
Barak BBarak B
265
$endgroup$
add a comment |
$begingroup$
The question: An experiment succeeds with a probability of $Z$, which is a random variable uniformly distributed over $[0,1]$. The experiment is conducted twice. Denote $X$ the random variable which equals 1 if the first experiment succeeded, and 0 otherwise. $Y$ is the same, but for the second experiment. What is $P(X=1)$ and $P(X=1, Y=1)$?
It's quite obvious that $P(X=1) = 1/2$, but I'm not sure how to prove it, and I have not the slightest clue on how to calculate $P(X=1, Y=1)$, since clearly $X,Y$ are dependent in one another.
I tried going at it with conditional probability, but it didn't help me much.
Any clues?
Thanks
probability random-variables
asked Jan 31 at 17:32
Barak BBarak B
265
$endgroup$
The question: An experiment succeeds with a probability of $Z$, which is a random variable uniformly distributed over $[0,1]$. The experiment is conducted twice. Denote $X$ the random variable which equals 1 if the first experiment succeeded, and 0 otherwise. $Y$ is the same, but for the second experiment. What is $P(X=1)$ and $P(X=1, Y=1)$?
It's quite obvious that $P(X=1) = 1/2$, but I'm not sure how to prove it, and I have not the slightest clue on how to calculate $P(X=1, Y=1)$, since clearly $X,Y$ are dependent in one another.
I tried going at it with conditional probability, but it didn't help me much.
Any clues?
Thanks
probability random-variables
probability random-variables
asked Jan 31 at 17:32
Barak BBarak B
265
asked Jan 31 at 17:32
Barak BBarak B
265
asked Jan 31 at 17:32
Barak BBarak B
265
asked Jan 31 at 17:32
Barak BBarak B
265
asked Jan 31 at 17:32
Barak BBarak B
265
265
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Yes, the idea is to condition on Z.
When we condition on Z , it becomes a constant and , from your description is the probability that X is 1 : $P(X=1|Z=z) = z$ . Now we use the definition of conditional probability to compute $P(X=1)$ :
$P(X=1) = int_{0}^{1}P(X=1|Z=z)P(Z=z)dz$ = 0.5
If we condition on Z , X and Y become independent (informal,if we know Z, observing X don't give any aditional information on Y ).
$P(X=1,Y=1|Z=z)=P(X=1|Z=z)P(Y=1|Z=z) = z^{2}$ .
Integrating over z , you find the final answare.
$P(X=1,Y=1) = int_{0}^{1}P(X=1|Z=z)P(Y=1|Z=z)P(Z=z)dz$ = 1/3
answered Jan 31 at 18:04
Popescu ClaudiuPopescu Claudiu
4019
$endgroup$
$begingroup$
I don't understand how from line 1 you deduced lines 2 and 3, could you clarify? The final answer shouldn't depend on Z after all, and I don't see how I'm supposed to integrate it.
$endgroup$
– Barak B
Jan 31 at 18:38
$begingroup$
I put some additional info .
$endgroup$
– Popescu Claudiu
Jan 31 at 18:53
$begingroup$
Got it, thanks alot!
$endgroup$
– Barak B
Jan 31 at 19:04
1
$begingroup$
Sorry, I'm a bit sloppy.
$endgroup$
– Popescu Claudiu
Jan 31 at 19:06
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, the idea is to condition on Z.
When we condition on Z , it becomes a constant and , from your description is the probability that X is 1 : $P(X=1|Z=z) = z$ . Now we use the definition of conditional probability to compute $P(X=1)$ :
$P(X=1) = int_{0}^{1}P(X=1|Z=z)P(Z=z)dz$ = 0.5
If we condition on Z , X and Y become independent (informal,if we know Z, observing X don't give any aditional information on Y ).
$P(X=1,Y=1|Z=z)=P(X=1|Z=z)P(Y=1|Z=z) = z^{2}$ .
Integrating over z , you find the final answare.
$P(X=1,Y=1) = int_{0}^{1}P(X=1|Z=z)P(Y=1|Z=z)P(Z=z)dz$ = 1/3
answered Jan 31 at 18:04
Popescu ClaudiuPopescu Claudiu
4019
$endgroup$
$begingroup$
I don't understand how from line 1 you deduced lines 2 and 3, could you clarify? The final answer shouldn't depend on Z after all, and I don't see how I'm supposed to integrate it.
$endgroup$
– Barak B
Jan 31 at 18:38
$begingroup$
I put some additional info .
$endgroup$
– Popescu Claudiu
Jan 31 at 18:53
$begingroup$
Got it, thanks alot!
$endgroup$
– Barak B
Jan 31 at 19:04
1
$begingroup$
Sorry, I'm a bit sloppy.
$endgroup$
– Popescu Claudiu
Jan 31 at 19:06
add a comment |
$begingroup$
Yes, the idea is to condition on Z.
When we condition on Z , it becomes a constant and , from your description is the probability that X is 1 : $P(X=1|Z=z) = z$ . Now we use the definition of conditional probability to compute $P(X=1)$ :
$P(X=1) = int_{0}^{1}P(X=1|Z=z)P(Z=z)dz$ = 0.5
If we condition on Z , X and Y become independent (informal,if we know Z, observing X don't give any aditional information on Y ).
$P(X=1,Y=1|Z=z)=P(X=1|Z=z)P(Y=1|Z=z) = z^{2}$ .
Integrating over z , you find the final answare.
$P(X=1,Y=1) = int_{0}^{1}P(X=1|Z=z)P(Y=1|Z=z)P(Z=z)dz$ = 1/3
answered Jan 31 at 18:04
Popescu ClaudiuPopescu Claudiu
4019
$endgroup$
$begingroup$
I don't understand how from line 1 you deduced lines 2 and 3, could you clarify? The final answer shouldn't depend on Z after all, and I don't see how I'm supposed to integrate it.
$endgroup$
– Barak B
Jan 31 at 18:38
$begingroup$
I put some additional info .
$endgroup$
– Popescu Claudiu
Jan 31 at 18:53
$begingroup$
Got it, thanks alot!
$endgroup$
– Barak B
Jan 31 at 19:04
1
$begingroup$
Sorry, I'm a bit sloppy.
$endgroup$
– Popescu Claudiu
Jan 31 at 19:06
add a comment |
$begingroup$
Yes, the idea is to condition on Z.
When we condition on Z , it becomes a constant and , from your description is the probability that X is 1 : $P(X=1|Z=z) = z$ . Now we use the definition of conditional probability to compute $P(X=1)$ :
$P(X=1) = int_{0}^{1}P(X=1|Z=z)P(Z=z)dz$ = 0.5
If we condition on Z , X and Y become independent (informal,if we know Z, observing X don't give any aditional information on Y ).
$P(X=1,Y=1|Z=z)=P(X=1|Z=z)P(Y=1|Z=z) = z^{2}$ .
Integrating over z , you find the final answare.
$P(X=1,Y=1) = int_{0}^{1}P(X=1|Z=z)P(Y=1|Z=z)P(Z=z)dz$ = 1/3
answered Jan 31 at 18:04
Popescu ClaudiuPopescu Claudiu
4019
$endgroup$
Yes, the idea is to condition on Z.
When we condition on Z , it becomes a constant and , from your description is the probability that X is 1 : $P(X=1|Z=z) = z$ . Now we use the definition of conditional probability to compute $P(X=1)$ :
$P(X=1) = int_{0}^{1}P(X=1|Z=z)P(Z=z)dz$ = 0.5
If we condition on Z , X and Y become independent (informal,if we know Z, observing X don't give any aditional information on Y ).
$P(X=1,Y=1|Z=z)=P(X=1|Z=z)P(Y=1|Z=z) = z^{2}$ .
Integrating over z , you find the final answare.
$P(X=1,Y=1) = int_{0}^{1}P(X=1|Z=z)P(Y=1|Z=z)P(Z=z)dz$ = 1/3
answered Jan 31 at 18:04
Popescu ClaudiuPopescu Claudiu
4019
edited Jan 31 at 19:01
answered Jan 31 at 18:04
Popescu ClaudiuPopescu Claudiu
4019
answered Jan 31 at 18:04
Popescu ClaudiuPopescu Claudiu
4019
answered Jan 31 at 18:04
Popescu ClaudiuPopescu Claudiu
4019
4019
$begingroup$
I don't understand how from line 1 you deduced lines 2 and 3, could you clarify? The final answer shouldn't depend on Z after all, and I don't see how I'm supposed to integrate it.
$endgroup$
– Barak B
Jan 31 at 18:38
$begingroup$
I put some additional info .
$endgroup$
– Popescu Claudiu
Jan 31 at 18:53
$begingroup$
Got it, thanks alot!
$endgroup$
– Barak B
Jan 31 at 19:04
1
$begingroup$
Sorry, I'm a bit sloppy.
$endgroup$
– Popescu Claudiu
Jan 31 at 19:06
add a comment |
$begingroup$
I don't understand how from line 1 you deduced lines 2 and 3, could you clarify? The final answer shouldn't depend on Z after all, and I don't see how I'm supposed to integrate it.
$endgroup$
– Barak B
Jan 31 at 18:38
$begingroup$
I put some additional info .
$endgroup$
– Popescu Claudiu
Jan 31 at 18:53
$begingroup$
Got it, thanks alot!
$endgroup$
– Barak B
Jan 31 at 19:04
1
$begingroup$
Sorry, I'm a bit sloppy.
$endgroup$
– Popescu Claudiu
Jan 31 at 19:06
$begingroup$
I don't understand how from line 1 you deduced lines 2 and 3, could you clarify? The final answer shouldn't depend on Z after all, and I don't see how I'm supposed to integrate it.
$endgroup$
– Barak B
Jan 31 at 18:38
$begingroup$
I don't understand how from line 1 you deduced lines 2 and 3, could you clarify? The final answer shouldn't depend on Z after all, and I don't see how I'm supposed to integrate it.
$endgroup$
– Barak B
Jan 31 at 18:38
$begingroup$
I put some additional info .
$endgroup$
– Popescu Claudiu
Jan 31 at 18:53
$begingroup$
I put some additional info .
$endgroup$
– Popescu Claudiu
Jan 31 at 18:53
$begingroup$
Got it, thanks alot!
$endgroup$
– Barak B
Jan 31 at 19:04
$begingroup$
Got it, thanks alot!
$endgroup$
– Barak B
Jan 31 at 19:04
1
1
$begingroup$
Sorry, I'm a bit sloppy.
$endgroup$
– Popescu Claudiu
Jan 31 at 19:06
$begingroup$
Sorry, I'm a bit sloppy.
$endgroup$
– Popescu Claudiu
Jan 31 at 19:06
add a comment |
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