Confusion over term “uniformly distributed” in a probability problem
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I'm trying to determine a joint PDF for random variables $X$ and $Y$ in this problem:
$(X,Y)$ is uniformly distributed on the subset of $mathbb R^2$, defined by $0<X<2$ and $0 <Y<X^3$.
I'm not sure I understand what's meant by "uniformly distributed". Are they saying all valid $(X,Y)$ combinations have equal probability? How does one even begin to define a joint PDF for this?
Would it be something like
$f_{X,Y}(x,y) = begin{cases}k,&0<X<2, 0 <Y<X^3 \0,& text{otherwise} end{cases}$
Where $k$ is some positive constant
probability probability-distributions
asked Nov 22 at 13:34
HumptyDumpty
33018
add a comment |
up vote
0
down vote
favorite
I'm trying to determine a joint PDF for random variables $X$ and $Y$ in this problem:
$(X,Y)$ is uniformly distributed on the subset of $mathbb R^2$, defined by $0<X<2$ and $0 <Y<X^3$.
I'm not sure I understand what's meant by "uniformly distributed". Are they saying all valid $(X,Y)$ combinations have equal probability? How does one even begin to define a joint PDF for this?
Would it be something like
$f_{X,Y}(x,y) = begin{cases}k,&0<X<2, 0 <Y<X^3 \0,& text{otherwise} end{cases}$
Where $k$ is some positive constant
probability probability-distributions
asked Nov 22 at 13:34
HumptyDumpty
33018
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to determine a joint PDF for random variables $X$ and $Y$ in this problem:
$(X,Y)$ is uniformly distributed on the subset of $mathbb R^2$, defined by $0<X<2$ and $0 <Y<X^3$.
I'm not sure I understand what's meant by "uniformly distributed". Are they saying all valid $(X,Y)$ combinations have equal probability? How does one even begin to define a joint PDF for this?
Would it be something like
$f_{X,Y}(x,y) = begin{cases}k,&0<X<2, 0 <Y<X^3 \0,& text{otherwise} end{cases}$
Where $k$ is some positive constant
probability probability-distributions
asked Nov 22 at 13:34
HumptyDumpty
33018
I'm trying to determine a joint PDF for random variables $X$ and $Y$ in this problem:
$(X,Y)$ is uniformly distributed on the subset of $mathbb R^2$, defined by $0<X<2$ and $0 <Y<X^3$.
I'm not sure I understand what's meant by "uniformly distributed". Are they saying all valid $(X,Y)$ combinations have equal probability? How does one even begin to define a joint PDF for this?
Would it be something like
$f_{X,Y}(x,y) = begin{cases}k,&0<X<2, 0 <Y<X^3 \0,& text{otherwise} end{cases}$
Where $k$ is some positive constant
probability probability-distributions
probability probability-distributions
asked Nov 22 at 13:34
HumptyDumpty
33018
asked Nov 22 at 13:34
HumptyDumpty
33018
asked Nov 22 at 13:34
HumptyDumpty
33018
asked Nov 22 at 13:34
HumptyDumpty
33018
asked Nov 22 at 13:34
HumptyDumpty
33018
33018
add a comment |
add a comment |
1 Answer
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up vote
2
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accepted
It means
$$f_{X,Y}(x,y) = begin{cases}k,&0<X<2, 0 <Y<X^3 \0,& text{otherwise} end{cases}$$
where $k$ is a positive constant.
To find $k$, we have to find the area of the support.
we have $$k^{-1}= int_0^2 x^3 , dx$$
answered Nov 22 at 13:37
Siong Thye Goh
97.7k1463116
What do you mean by "area of the suppose"?
– HumptyDumpty
Nov 22 at 13:40
oops, typo, i meant support
– Siong Thye Goh
Nov 22 at 13:40
I've computed $k = 1/4$. How does one interpret this result?
– HumptyDumpty
Nov 22 at 13:47
Your $k$ value is correct. hmm... if you pick two area of the size over the support and call them event $A$ and $B$, we know that the probability of $A$ is equal to the probability of $B$. It just depends on the area.
– Siong Thye Goh
Nov 22 at 13:51
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
up vote
2
down vote
accepted
It means
$$f_{X,Y}(x,y) = begin{cases}k,&0<X<2, 0 <Y<X^3 \0,& text{otherwise} end{cases}$$
where $k$ is a positive constant.
To find $k$, we have to find the area of the support.
we have $$k^{-1}= int_0^2 x^3 , dx$$
answered Nov 22 at 13:37
Siong Thye Goh
97.7k1463116
What do you mean by "area of the suppose"?
– HumptyDumpty
Nov 22 at 13:40
oops, typo, i meant support
– Siong Thye Goh
Nov 22 at 13:40
I've computed $k = 1/4$. How does one interpret this result?
– HumptyDumpty
Nov 22 at 13:47
Your $k$ value is correct. hmm... if you pick two area of the size over the support and call them event $A$ and $B$, we know that the probability of $A$ is equal to the probability of $B$. It just depends on the area.
– Siong Thye Goh
Nov 22 at 13:51
add a comment |
up vote
2
down vote
accepted
It means
$$f_{X,Y}(x,y) = begin{cases}k,&0<X<2, 0 <Y<X^3 \0,& text{otherwise} end{cases}$$
where $k$ is a positive constant.
To find $k$, we have to find the area of the support.
we have $$k^{-1}= int_0^2 x^3 , dx$$
answered Nov 22 at 13:37
Siong Thye Goh
97.7k1463116
What do you mean by "area of the suppose"?
– HumptyDumpty
Nov 22 at 13:40
oops, typo, i meant support
– Siong Thye Goh
Nov 22 at 13:40
I've computed $k = 1/4$. How does one interpret this result?
– HumptyDumpty
Nov 22 at 13:47
Your $k$ value is correct. hmm... if you pick two area of the size over the support and call them event $A$ and $B$, we know that the probability of $A$ is equal to the probability of $B$. It just depends on the area.
– Siong Thye Goh
Nov 22 at 13:51
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
It means
$$f_{X,Y}(x,y) = begin{cases}k,&0<X<2, 0 <Y<X^3 \0,& text{otherwise} end{cases}$$
where $k$ is a positive constant.
To find $k$, we have to find the area of the support.
we have $$k^{-1}= int_0^2 x^3 , dx$$
answered Nov 22 at 13:37
Siong Thye Goh
97.7k1463116
It means
$$f_{X,Y}(x,y) = begin{cases}k,&0<X<2, 0 <Y<X^3 \0,& text{otherwise} end{cases}$$
where $k$ is a positive constant.
To find $k$, we have to find the area of the support.
we have $$k^{-1}= int_0^2 x^3 , dx$$
answered Nov 22 at 13:37
Siong Thye Goh
97.7k1463116
edited Nov 22 at 13:40
answered Nov 22 at 13:37
Siong Thye Goh
97.7k1463116
answered Nov 22 at 13:37
Siong Thye Goh
97.7k1463116
answered Nov 22 at 13:37
Siong Thye Goh
97.7k1463116
97.7k1463116
What do you mean by "area of the suppose"?
– HumptyDumpty
Nov 22 at 13:40
oops, typo, i meant support
– Siong Thye Goh
Nov 22 at 13:40
I've computed $k = 1/4$. How does one interpret this result?
– HumptyDumpty
Nov 22 at 13:47
Your $k$ value is correct. hmm... if you pick two area of the size over the support and call them event $A$ and $B$, we know that the probability of $A$ is equal to the probability of $B$. It just depends on the area.
– Siong Thye Goh
Nov 22 at 13:51
add a comment |
What do you mean by "area of the suppose"?
– HumptyDumpty
Nov 22 at 13:40
oops, typo, i meant support
– Siong Thye Goh
Nov 22 at 13:40
I've computed $k = 1/4$. How does one interpret this result?
– HumptyDumpty
Nov 22 at 13:47
Your $k$ value is correct. hmm... if you pick two area of the size over the support and call them event $A$ and $B$, we know that the probability of $A$ is equal to the probability of $B$. It just depends on the area.
– Siong Thye Goh
Nov 22 at 13:51
What do you mean by "area of the suppose"?
– HumptyDumpty
Nov 22 at 13:40
What do you mean by "area of the suppose"?
– HumptyDumpty
Nov 22 at 13:40
oops, typo, i meant support
– Siong Thye Goh
Nov 22 at 13:40
oops, typo, i meant support
– Siong Thye Goh
Nov 22 at 13:40
I've computed $k = 1/4$. How does one interpret this result?
– HumptyDumpty
Nov 22 at 13:47
I've computed $k = 1/4$. How does one interpret this result?
– HumptyDumpty
Nov 22 at 13:47
Your $k$ value is correct. hmm... if you pick two area of the size over the support and call them event $A$ and $B$, we know that the probability of $A$ is equal to the probability of $B$. It just depends on the area.
– Siong Thye Goh
Nov 22 at 13:51
Your $k$ value is correct. hmm... if you pick two area of the size over the support and call them event $A$ and $B$, we know that the probability of $A$ is equal to the probability of $B$. It just depends on the area.
– Siong Thye Goh
Nov 22 at 13:51
add a comment |
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