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Computing $P(x)$ from $P(x,y)$
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Given the $x in mathbbR^3$ and $y in mathbbR$ following probability density function $P(x,y)$:
$$P(x,y) = begincases
1/8, & textif x in left[ -1, 1 right]^3 text and y = x_1 + 2x_2 + 2x_3 \
0, & textotherwise
endcases $$
I would like to compute $P(y|x)$. In order to so, I was thinking about computing first $P(x)$ and then through the Bayes theorem to get to $P(y|x)$. Unfortunately, I don't really know how to compute $P(x)$ in this case.
May I ask your help?
probability integration
edited Jul 25 at 15:55
Sangchul Lee
85.6k12155253
asked Jul 25 at 15:30
Tommaso Bendinelli
285
 |Â
show 12 more comments
up vote
0
down vote
favorite
Given the $x in mathbbR^3$ and $y in mathbbR$ following probability density function $P(x,y)$:
$$P(x,y) = begincases
1/8, & textif x in left[ -1, 1 right]^3 text and y = x_1 + 2x_2 + 2x_3 \
0, & textotherwise
endcases $$
I would like to compute $P(y|x)$. In order to so, I was thinking about computing first $P(x)$ and then through the Bayes theorem to get to $P(y|x)$. Unfortunately, I don't really know how to compute $P(x)$ in this case.
May I ask your help?
probability integration
edited Jul 25 at 15:55
Sangchul Lee
85.6k12155253
asked Jul 25 at 15:30
Tommaso Bendinelli
285
This is a discrete problem. There are only $8$ outcomes so if you write the eight 4-tuples you can answer this by counting.
– John Douma
Jul 25 at 15:47
ahh damm, I wrote it wrong. I meant with x being a continuous variable. Not a discrete one
– Tommaso Bendinelli
Jul 25 at 15:48
Either way, there are only a finite number of non-zero outcomes. You should be able to write $P(x)$ in cases as you did with $P(x,y)$.
– John Douma
Jul 25 at 15:50
How do you know that are only finite number of non-zero outcomes?
– Tommaso Bendinelli
Jul 25 at 15:52
1
If you are interested in the conditional distribution of $y$ given $x$, the answer is quite simple as you expected. $y$ is completely determined by $x$ by the formula $y=x_1+2x_2+2x_3$. So $$mu(dy mid x) = mathbf1_[-1,1]^3(x) delta_x_1+2x_2+2x_3(dy),$$ or if we go back to the careless notation, $$P(ymid x)=begincases1,&textif y=x_1+2x_2+2x_3\0&textotherwiseendcases$$ If you are interested in the conditional distribution of $x$ given $y$, we need to compute some pesky triple integral (which computes volume of part of $[-1,1]^3$ divided by the plane $y=x_1+2x_2+2x_3$).
– Sangchul Lee
Jul 25 at 16:34
 |Â
show 12 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given the $x in mathbbR^3$ and $y in mathbbR$ following probability density function $P(x,y)$:
$$P(x,y) = begincases
1/8, & textif x in left[ -1, 1 right]^3 text and y = x_1 + 2x_2 + 2x_3 \
0, & textotherwise
endcases $$
I would like to compute $P(y|x)$. In order to so, I was thinking about computing first $P(x)$ and then through the Bayes theorem to get to $P(y|x)$. Unfortunately, I don't really know how to compute $P(x)$ in this case.
May I ask your help?
probability integration
edited Jul 25 at 15:55
Sangchul Lee
85.6k12155253
asked Jul 25 at 15:30
Tommaso Bendinelli
285
Given the $x in mathbbR^3$ and $y in mathbbR$ following probability density function $P(x,y)$:
$$P(x,y) = begincases
1/8, & textif x in left[ -1, 1 right]^3 text and y = x_1 + 2x_2 + 2x_3 \
0, & textotherwise
endcases $$
I would like to compute $P(y|x)$. In order to so, I was thinking about computing first $P(x)$ and then through the Bayes theorem to get to $P(y|x)$. Unfortunately, I don't really know how to compute $P(x)$ in this case.
May I ask your help?
probability integration
edited Jul 25 at 15:55
Sangchul Lee
85.6k12155253
asked Jul 25 at 15:30
Tommaso Bendinelli
285
edited Jul 25 at 15:55
Sangchul Lee
85.6k12155253
edited Jul 25 at 15:55
Sangchul Lee
85.6k12155253
edited Jul 25 at 15:55
Sangchul Lee
85.6k12155253
85.6k12155253
asked Jul 25 at 15:30
Tommaso Bendinelli
285
asked Jul 25 at 15:30
Tommaso Bendinelli
285
asked Jul 25 at 15:30
Tommaso Bendinelli
285
285
This is a discrete problem. There are only $8$ outcomes so if you write the eight 4-tuples you can answer this by counting.
– John Douma
Jul 25 at 15:47
ahh damm, I wrote it wrong. I meant with x being a continuous variable. Not a discrete one
– Tommaso Bendinelli
Jul 25 at 15:48
Either way, there are only a finite number of non-zero outcomes. You should be able to write $P(x)$ in cases as you did with $P(x,y)$.
– John Douma
Jul 25 at 15:50
How do you know that are only finite number of non-zero outcomes?
– Tommaso Bendinelli
Jul 25 at 15:52
1
If you are interested in the conditional distribution of $y$ given $x$, the answer is quite simple as you expected. $y$ is completely determined by $x$ by the formula $y=x_1+2x_2+2x_3$. So $$mu(dy mid x) = mathbf1_[-1,1]^3(x) delta_x_1+2x_2+2x_3(dy),$$ or if we go back to the careless notation, $$P(ymid x)=begincases1,&textif y=x_1+2x_2+2x_3\0&textotherwiseendcases$$ If you are interested in the conditional distribution of $x$ given $y$, we need to compute some pesky triple integral (which computes volume of part of $[-1,1]^3$ divided by the plane $y=x_1+2x_2+2x_3$).
– Sangchul Lee
Jul 25 at 16:34
 |Â
show 12 more comments
This is a discrete problem. There are only $8$ outcomes so if you write the eight 4-tuples you can answer this by counting.
– John Douma
Jul 25 at 15:47
ahh damm, I wrote it wrong. I meant with x being a continuous variable. Not a discrete one
– Tommaso Bendinelli
Jul 25 at 15:48
Either way, there are only a finite number of non-zero outcomes. You should be able to write $P(x)$ in cases as you did with $P(x,y)$.
– John Douma
Jul 25 at 15:50
How do you know that are only finite number of non-zero outcomes?
– Tommaso Bendinelli
Jul 25 at 15:52
1
If you are interested in the conditional distribution of $y$ given $x$, the answer is quite simple as you expected. $y$ is completely determined by $x$ by the formula $y=x_1+2x_2+2x_3$. So $$mu(dy mid x) = mathbf1_[-1,1]^3(x) delta_x_1+2x_2+2x_3(dy),$$ or if we go back to the careless notation, $$P(ymid x)=begincases1,&textif y=x_1+2x_2+2x_3\0&textotherwiseendcases$$ If you are interested in the conditional distribution of $x$ given $y$, we need to compute some pesky triple integral (which computes volume of part of $[-1,1]^3$ divided by the plane $y=x_1+2x_2+2x_3$).
– Sangchul Lee
Jul 25 at 16:34
This is a discrete problem. There are only $8$ outcomes so if you write the eight 4-tuples you can answer this by counting.
– John Douma
Jul 25 at 15:47
This is a discrete problem. There are only $8$ outcomes so if you write the eight 4-tuples you can answer this by counting.
– John Douma
Jul 25 at 15:47
ahh damm, I wrote it wrong. I meant with x being a continuous variable. Not a discrete one
– Tommaso Bendinelli
Jul 25 at 15:48
ahh damm, I wrote it wrong. I meant with x being a continuous variable. Not a discrete one
– Tommaso Bendinelli
Jul 25 at 15:48
Either way, there are only a finite number of non-zero outcomes. You should be able to write $P(x)$ in cases as you did with $P(x,y)$.
– John Douma
Jul 25 at 15:50
Either way, there are only a finite number of non-zero outcomes. You should be able to write $P(x)$ in cases as you did with $P(x,y)$.
– John Douma
Jul 25 at 15:50
How do you know that are only finite number of non-zero outcomes?
– Tommaso Bendinelli
Jul 25 at 15:52
How do you know that are only finite number of non-zero outcomes?
– Tommaso Bendinelli
Jul 25 at 15:52
1
1
If you are interested in the conditional distribution of $y$ given $x$, the answer is quite simple as you expected. $y$ is completely determined by $x$ by the formula $y=x_1+2x_2+2x_3$. So $$mu(dy mid x) = mathbf1_[-1,1]^3(x) delta_x_1+2x_2+2x_3(dy),$$ or if we go back to the careless notation, $$P(ymid x)=begincases1,&textif y=x_1+2x_2+2x_3\0&textotherwiseendcases$$ If you are interested in the conditional distribution of $x$ given $y$, we need to compute some pesky triple integral (which computes volume of part of $[-1,1]^3$ divided by the plane $y=x_1+2x_2+2x_3$).
– Sangchul Lee
Jul 25 at 16:34
If you are interested in the conditional distribution of $y$ given $x$, the answer is quite simple as you expected. $y$ is completely determined by $x$ by the formula $y=x_1+2x_2+2x_3$. So $$mu(dy mid x) = mathbf1_[-1,1]^3(x) delta_x_1+2x_2+2x_3(dy),$$ or if we go back to the careless notation, $$P(ymid x)=begincases1,&textif y=x_1+2x_2+2x_3\0&textotherwiseendcases$$ If you are interested in the conditional distribution of $x$ given $y$, we need to compute some pesky triple integral (which computes volume of part of $[-1,1]^3$ divided by the plane $y=x_1+2x_2+2x_3$).
– Sangchul Lee
Jul 25 at 16:34
 |Â
show 12 more comments
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This is a discrete problem. There are only $8$ outcomes so if you write the eight 4-tuples you can answer this by counting.
– John Douma
Jul 25 at 15:47
ahh damm, I wrote it wrong. I meant with x being a continuous variable. Not a discrete one
– Tommaso Bendinelli
Jul 25 at 15:48
Either way, there are only a finite number of non-zero outcomes. You should be able to write $P(x)$ in cases as you did with $P(x,y)$.
– John Douma
Jul 25 at 15:50
How do you know that are only finite number of non-zero outcomes?
– Tommaso Bendinelli
Jul 25 at 15:52
1
If you are interested in the conditional distribution of $y$ given $x$, the answer is quite simple as you expected. $y$ is completely determined by $x$ by the formula $y=x_1+2x_2+2x_3$. So $$mu(dy mid x) = mathbf1_[-1,1]^3(x) delta_x_1+2x_2+2x_3(dy),$$ or if we go back to the careless notation, $$P(ymid x)=begincases1,&textif y=x_1+2x_2+2x_3\0&textotherwiseendcases$$ If you are interested in the conditional distribution of $x$ given $y$, we need to compute some pesky triple integral (which computes volume of part of $[-1,1]^3$ divided by the plane $y=x_1+2x_2+2x_3$).
– Sangchul Lee
Jul 25 at 16:34