Mixing
Find marginal density at a point [closed]
$begingroup$
Given the following joint density function:
begin{equation}
f (x,y) =
begin{cases}
2& text{} 0 le x le 1-y, 0 le y le 1\
0 &text{otherwise}
end{cases}
end{equation}
Find $f_{y}(1/2)$.
Since this is a continuous distribution, is the answer $0$. Or do I have to integrate the marginal density of $y$ from $0$ to $1/2$ or from $1/2$ to $1$.
self-learning density-function
edited Jan 31 at 13:58
Ernie060
2,940719
asked Jan 31 at 12:40
user46697user46697
228211
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closed as unclear what you're asking by Did, Cesareo, metamorphy, José Carlos Santos, Aweygan Feb 1 at 15:58
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Given the following joint density function:
begin{equation}
f (x,y) =
begin{cases}
2& text{} 0 le x le 1-y, 0 le y le 1\
0 &text{otherwise}
end{cases}
end{equation}
Find $f_{y}(1/2)$.
Since this is a continuous distribution, is the answer $0$. Or do I have to integrate the marginal density of $y$ from $0$ to $1/2$ or from $1/2$ to $1$.
self-learning density-function
edited Jan 31 at 13:58
Ernie060
2,940719
asked Jan 31 at 12:40
user46697user46697
228211
$endgroup$
closed as unclear what you're asking by Did, Cesareo, metamorphy, José Carlos Santos, Aweygan Feb 1 at 15:58
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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You have to find the marginal density $f_Y(y)=int f(x,y),dx$ first and then set $y=1/2$.
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– StubbornAtom
Jan 31 at 12:45
2
$begingroup$
$f_y(1/2)$ isn't the probability of getting $1/2$
$endgroup$
– pointguard0
Jan 31 at 13:05
add a comment |
$begingroup$
Given the following joint density function:
begin{equation}
f (x,y) =
begin{cases}
2& text{} 0 le x le 1-y, 0 le y le 1\
0 &text{otherwise}
end{cases}
end{equation}
Find $f_{y}(1/2)$.
Since this is a continuous distribution, is the answer $0$. Or do I have to integrate the marginal density of $y$ from $0$ to $1/2$ or from $1/2$ to $1$.
self-learning density-function
edited Jan 31 at 13:58
Ernie060
2,940719
asked Jan 31 at 12:40
user46697user46697
228211
$endgroup$
Given the following joint density function:
begin{equation}
f (x,y) =
begin{cases}
2& text{} 0 le x le 1-y, 0 le y le 1\
0 &text{otherwise}
end{cases}
end{equation}
Find $f_{y}(1/2)$.
Since this is a continuous distribution, is the answer $0$. Or do I have to integrate the marginal density of $y$ from $0$ to $1/2$ or from $1/2$ to $1$.
self-learning density-function
self-learning density-function
edited Jan 31 at 13:58
Ernie060
2,940719
asked Jan 31 at 12:40
user46697user46697
228211
edited Jan 31 at 13:58
Ernie060
2,940719
asked Jan 31 at 12:40
user46697user46697
228211
edited Jan 31 at 13:58
Ernie060
2,940719
edited Jan 31 at 13:58
Ernie060
2,940719
edited Jan 31 at 13:58
Ernie060
2,940719
2,940719
asked Jan 31 at 12:40
user46697user46697
228211
asked Jan 31 at 12:40
user46697user46697
228211
asked Jan 31 at 12:40
user46697user46697
228211
228211
closed as unclear what you're asking by Did, Cesareo, metamorphy, José Carlos Santos, Aweygan Feb 1 at 15:58
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Did, Cesareo, metamorphy, José Carlos Santos, Aweygan Feb 1 at 15:58
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
You have to find the marginal density $f_Y(y)=int f(x,y),dx$ first and then set $y=1/2$.
$endgroup$
– StubbornAtom
Jan 31 at 12:45
2
$begingroup$
$f_y(1/2)$ isn't the probability of getting $1/2$
$endgroup$
– pointguard0
Jan 31 at 13:05
add a comment |
$begingroup$
You have to find the marginal density $f_Y(y)=int f(x,y),dx$ first and then set $y=1/2$.
$endgroup$
– StubbornAtom
Jan 31 at 12:45
2
$begingroup$
$f_y(1/2)$ isn't the probability of getting $1/2$
$endgroup$
– pointguard0
Jan 31 at 13:05
$begingroup$
You have to find the marginal density $f_Y(y)=int f(x,y),dx$ first and then set $y=1/2$.
$endgroup$
– StubbornAtom
Jan 31 at 12:45
$begingroup$
You have to find the marginal density $f_Y(y)=int f(x,y),dx$ first and then set $y=1/2$.
$endgroup$
– StubbornAtom
Jan 31 at 12:45
2
2
$begingroup$
$f_y(1/2)$ isn't the probability of getting $1/2$
$endgroup$
– pointguard0
Jan 31 at 13:05
$begingroup$
$f_y(1/2)$ isn't the probability of getting $1/2$
$endgroup$
– pointguard0
Jan 31 at 13:05
add a comment |
1 Answer
1
active
oldest
votes
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You may first calculate the marginal density or you calculate directly:
begin{eqnarray*} f_Yleft(frac{1}{2}right)
& = & int_{-infty}^{+infty}fleft(x,frac{1}{2}right); dx \
& = & int_{0}^{1-frac{1}{2}}2; dx \
& = & 2cdot frac{1}{2} = 1 \
end{eqnarray*}
answered Jan 31 at 13:14
trancelocationtrancelocation
13.6k1829
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You may first calculate the marginal density or you calculate directly:
begin{eqnarray*} f_Yleft(frac{1}{2}right)
& = & int_{-infty}^{+infty}fleft(x,frac{1}{2}right); dx \
& = & int_{0}^{1-frac{1}{2}}2; dx \
& = & 2cdot frac{1}{2} = 1 \
end{eqnarray*}
answered Jan 31 at 13:14
trancelocationtrancelocation
13.6k1829
$endgroup$
add a comment |
$begingroup$
You may first calculate the marginal density or you calculate directly:
begin{eqnarray*} f_Yleft(frac{1}{2}right)
& = & int_{-infty}^{+infty}fleft(x,frac{1}{2}right); dx \
& = & int_{0}^{1-frac{1}{2}}2; dx \
& = & 2cdot frac{1}{2} = 1 \
end{eqnarray*}
answered Jan 31 at 13:14
trancelocationtrancelocation
13.6k1829
$endgroup$
add a comment |
$begingroup$
You may first calculate the marginal density or you calculate directly:
begin{eqnarray*} f_Yleft(frac{1}{2}right)
& = & int_{-infty}^{+infty}fleft(x,frac{1}{2}right); dx \
& = & int_{0}^{1-frac{1}{2}}2; dx \
& = & 2cdot frac{1}{2} = 1 \
end{eqnarray*}
answered Jan 31 at 13:14
trancelocationtrancelocation
13.6k1829
$endgroup$
You may first calculate the marginal density or you calculate directly:
begin{eqnarray*} f_Yleft(frac{1}{2}right)
& = & int_{-infty}^{+infty}fleft(x,frac{1}{2}right); dx \
& = & int_{0}^{1-frac{1}{2}}2; dx \
& = & 2cdot frac{1}{2} = 1 \
end{eqnarray*}
answered Jan 31 at 13:14
trancelocationtrancelocation
13.6k1829
answered Jan 31 at 13:14
trancelocationtrancelocation
13.6k1829
answered Jan 31 at 13:14
trancelocationtrancelocation
13.6k1829
answered Jan 31 at 13:14
trancelocationtrancelocation
13.6k1829
13.6k1829
add a comment |
add a comment |
$begingroup$
You have to find the marginal density $f_Y(y)=int f(x,y),dx$ first and then set $y=1/2$.
$endgroup$
– StubbornAtom
Jan 31 at 12:45
2
$begingroup$
$f_y(1/2)$ isn't the probability of getting $1/2$
$endgroup$
– pointguard0
Jan 31 at 13:05