Mixing
Integration of PDF's to find valid constant
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I'm studying probability theory and came across and exercise problem that I would like to ask help with. Here's the specific problem that is in question:
Let $X$ and $Y$ have joint PDF
$$f_{X, Y}(x, y) = cxy$$
for $(0 lt x lt y lt 1)$. Find $c$ to make this a valid PDF.
It's honestly a simple problem, but I am struggling to grasp some techniques regarding the integration of the joint PDF, which is leading me to get an incorrect answer.
My approach
Since a PDF must integrate to $1$ and must be nonnegative in order to be valid,
begin{align}
int int f_{X, Y}(x, y) & = int_{y=x}^{y=1} int_{x=0}^{x=y} cxy dxdy \
& = int_{y=x}^{y=1}frac{c}{2}y^3dy \
& = frac{c(1-x^4)}{8}
end{align}
Therefore, I concluded that $c = frac{8}{1-x^4}$. However, the solution that I was given has taken a different approach of switching the order of integration and also using different integration limits, leading to a completely different result.
Correct solution
begin{align}
int int f_{X, Y}(x, y) & = int_{x=0}^{x=1} left( int_{y=x}^{y=1} cxy dy right) dx \
& = int_{x=0}^{x=1} frac{cx-cx^3}{2} dx \
& = frac{c}{4} \
end{align}
This gives us $c = 4$.
I have two specific questions:
Why did the correct solution switch the order of integration?
The limits of integration that I set were based on the initial condition of $(0 lt x lt y lt 1)$. That's why I had the limits for $x$ set to $x = 0$ and $x = y$. However, the correct solution set it to $x = 0$ and $x = 1$. Why is this so? Is there something that I missed?
Any feedback is appreciated. Thank you.
EDIT
The "correct" solution is actually incorrect. The last integral calculation should give us $frac{c}{8}$, leading us to $c = 8$, not $4$.
probability integration density-function
asked 2 days ago
Sean
21510
add a comment |
up vote
0
down vote
favorite
I'm studying probability theory and came across and exercise problem that I would like to ask help with. Here's the specific problem that is in question:
Let $X$ and $Y$ have joint PDF
$$f_{X, Y}(x, y) = cxy$$
for $(0 lt x lt y lt 1)$. Find $c$ to make this a valid PDF.
It's honestly a simple problem, but I am struggling to grasp some techniques regarding the integration of the joint PDF, which is leading me to get an incorrect answer.
My approach
Since a PDF must integrate to $1$ and must be nonnegative in order to be valid,
begin{align}
int int f_{X, Y}(x, y) & = int_{y=x}^{y=1} int_{x=0}^{x=y} cxy dxdy \
& = int_{y=x}^{y=1}frac{c}{2}y^3dy \
& = frac{c(1-x^4)}{8}
end{align}
Therefore, I concluded that $c = frac{8}{1-x^4}$. However, the solution that I was given has taken a different approach of switching the order of integration and also using different integration limits, leading to a completely different result.
Correct solution
begin{align}
int int f_{X, Y}(x, y) & = int_{x=0}^{x=1} left( int_{y=x}^{y=1} cxy dy right) dx \
& = int_{x=0}^{x=1} frac{cx-cx^3}{2} dx \
& = frac{c}{4} \
end{align}
This gives us $c = 4$.
I have two specific questions:
Why did the correct solution switch the order of integration?
The limits of integration that I set were based on the initial condition of $(0 lt x lt y lt 1)$. That's why I had the limits for $x$ set to $x = 0$ and $x = y$. However, the correct solution set it to $x = 0$ and $x = 1$. Why is this so? Is there something that I missed?
Any feedback is appreciated. Thank you.
EDIT
The "correct" solution is actually incorrect. The last integral calculation should give us $frac{c}{8}$, leading us to $c = 8$, not $4$.
probability integration density-function
asked 2 days ago
Sean
21510
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm studying probability theory and came across and exercise problem that I would like to ask help with. Here's the specific problem that is in question:
Let $X$ and $Y$ have joint PDF
$$f_{X, Y}(x, y) = cxy$$
for $(0 lt x lt y lt 1)$. Find $c$ to make this a valid PDF.
It's honestly a simple problem, but I am struggling to grasp some techniques regarding the integration of the joint PDF, which is leading me to get an incorrect answer.
My approach
Since a PDF must integrate to $1$ and must be nonnegative in order to be valid,
begin{align}
int int f_{X, Y}(x, y) & = int_{y=x}^{y=1} int_{x=0}^{x=y} cxy dxdy \
& = int_{y=x}^{y=1}frac{c}{2}y^3dy \
& = frac{c(1-x^4)}{8}
end{align}
Therefore, I concluded that $c = frac{8}{1-x^4}$. However, the solution that I was given has taken a different approach of switching the order of integration and also using different integration limits, leading to a completely different result.
Correct solution
begin{align}
int int f_{X, Y}(x, y) & = int_{x=0}^{x=1} left( int_{y=x}^{y=1} cxy dy right) dx \
& = int_{x=0}^{x=1} frac{cx-cx^3}{2} dx \
& = frac{c}{4} \
end{align}
This gives us $c = 4$.
I have two specific questions:
Why did the correct solution switch the order of integration?
The limits of integration that I set were based on the initial condition of $(0 lt x lt y lt 1)$. That's why I had the limits for $x$ set to $x = 0$ and $x = y$. However, the correct solution set it to $x = 0$ and $x = 1$. Why is this so? Is there something that I missed?
Any feedback is appreciated. Thank you.
EDIT
The "correct" solution is actually incorrect. The last integral calculation should give us $frac{c}{8}$, leading us to $c = 8$, not $4$.
probability integration density-function
asked 2 days ago
Sean
21510
I'm studying probability theory and came across and exercise problem that I would like to ask help with. Here's the specific problem that is in question:
Let $X$ and $Y$ have joint PDF
$$f_{X, Y}(x, y) = cxy$$
for $(0 lt x lt y lt 1)$. Find $c$ to make this a valid PDF.
It's honestly a simple problem, but I am struggling to grasp some techniques regarding the integration of the joint PDF, which is leading me to get an incorrect answer.
My approach
Since a PDF must integrate to $1$ and must be nonnegative in order to be valid,
begin{align}
int int f_{X, Y}(x, y) & = int_{y=x}^{y=1} int_{x=0}^{x=y} cxy dxdy \
& = int_{y=x}^{y=1}frac{c}{2}y^3dy \
& = frac{c(1-x^4)}{8}
end{align}
Therefore, I concluded that $c = frac{8}{1-x^4}$. However, the solution that I was given has taken a different approach of switching the order of integration and also using different integration limits, leading to a completely different result.
Correct solution
begin{align}
int int f_{X, Y}(x, y) & = int_{x=0}^{x=1} left( int_{y=x}^{y=1} cxy dy right) dx \
& = int_{x=0}^{x=1} frac{cx-cx^3}{2} dx \
& = frac{c}{4} \
end{align}
This gives us $c = 4$.
I have two specific questions:
Why did the correct solution switch the order of integration?
The limits of integration that I set were based on the initial condition of $(0 lt x lt y lt 1)$. That's why I had the limits for $x$ set to $x = 0$ and $x = y$. However, the correct solution set it to $x = 0$ and $x = 1$. Why is this so? Is there something that I missed?
Any feedback is appreciated. Thank you.
EDIT
The "correct" solution is actually incorrect. The last integral calculation should give us $frac{c}{8}$, leading us to $c = 8$, not $4$.
probability integration density-function
probability integration density-function
asked 2 days ago
Sean
21510
asked 2 days ago
Sean
21510
edited yesterday
asked 2 days ago
Sean
21510
asked 2 days ago
Sean
21510
asked 2 days ago
Sean
21510
21510
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Your integral needs to reflect the constraints $ 0 < x,y <1$ and $x<y$. If you think about it, this means that x could take any value in (0,1) and moreover, for any such x, y can take any value in (x,1) and that’s why your solution is not correct and their solution is.
In your solution you are fixing x, but what even is the value of x you are fixing it to?
answered 2 days ago
Sorin Tirc
5039
That's what I initially thought, but if we take into account the constraint that $x lt y$, then wouldn't it make more sense to set the integration limit as $y$? I'm having trouble understanding why $x$ would be able to take on any value in interval $(0, 1)$ when it's constrained by the value of $y$.
– Sean
2 days ago
Are you saying that because the value of $y$ is not yet determined, $x$ may take on any value in $(0, 1)$?
– Sean
2 days ago
1
You don’t have to consider the value of y first. You can consider either the value of x first, in which case y is constrained to be between x and 1 or you could consider y first and then x is constrained to be within 0 and y and you get two perfectly equivalent solutions
– Sorin Tirc
2 days ago
1
What matters is that the set {$0<x<y<1$} can be written as {$0<x<1$, $x<y<1$}
– Sorin Tirc
2 days ago
1
Yes, absolutely ;)
– Sorin Tirc
2 days ago
|
show 6 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your integral needs to reflect the constraints $ 0 < x,y <1$ and $x<y$. If you think about it, this means that x could take any value in (0,1) and moreover, for any such x, y can take any value in (x,1) and that’s why your solution is not correct and their solution is.
In your solution you are fixing x, but what even is the value of x you are fixing it to?
answered 2 days ago
Sorin Tirc
5039
That's what I initially thought, but if we take into account the constraint that $x lt y$, then wouldn't it make more sense to set the integration limit as $y$? I'm having trouble understanding why $x$ would be able to take on any value in interval $(0, 1)$ when it's constrained by the value of $y$.
– Sean
2 days ago
Are you saying that because the value of $y$ is not yet determined, $x$ may take on any value in $(0, 1)$?
– Sean
2 days ago
1
You don’t have to consider the value of y first. You can consider either the value of x first, in which case y is constrained to be between x and 1 or you could consider y first and then x is constrained to be within 0 and y and you get two perfectly equivalent solutions
– Sorin Tirc
2 days ago
1
What matters is that the set {$0<x<y<1$} can be written as {$0<x<1$, $x<y<1$}
– Sorin Tirc
2 days ago
1
Yes, absolutely ;)
– Sorin Tirc
2 days ago
|
show 6 more comments
up vote
1
down vote
accepted
Your integral needs to reflect the constraints $ 0 < x,y <1$ and $x<y$. If you think about it, this means that x could take any value in (0,1) and moreover, for any such x, y can take any value in (x,1) and that’s why your solution is not correct and their solution is.
In your solution you are fixing x, but what even is the value of x you are fixing it to?
answered 2 days ago
Sorin Tirc
5039
That's what I initially thought, but if we take into account the constraint that $x lt y$, then wouldn't it make more sense to set the integration limit as $y$? I'm having trouble understanding why $x$ would be able to take on any value in interval $(0, 1)$ when it's constrained by the value of $y$.
– Sean
2 days ago
Are you saying that because the value of $y$ is not yet determined, $x$ may take on any value in $(0, 1)$?
– Sean
2 days ago
1
You don’t have to consider the value of y first. You can consider either the value of x first, in which case y is constrained to be between x and 1 or you could consider y first and then x is constrained to be within 0 and y and you get two perfectly equivalent solutions
– Sorin Tirc
2 days ago
1
What matters is that the set {$0<x<y<1$} can be written as {$0<x<1$, $x<y<1$}
– Sorin Tirc
2 days ago
1
Yes, absolutely ;)
– Sorin Tirc
2 days ago
|
show 6 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your integral needs to reflect the constraints $ 0 < x,y <1$ and $x<y$. If you think about it, this means that x could take any value in (0,1) and moreover, for any such x, y can take any value in (x,1) and that’s why your solution is not correct and their solution is.
In your solution you are fixing x, but what even is the value of x you are fixing it to?
answered 2 days ago
Sorin Tirc
5039
Your integral needs to reflect the constraints $ 0 < x,y <1$ and $x<y$. If you think about it, this means that x could take any value in (0,1) and moreover, for any such x, y can take any value in (x,1) and that’s why your solution is not correct and their solution is.
In your solution you are fixing x, but what even is the value of x you are fixing it to?
answered 2 days ago
Sorin Tirc
5039
answered 2 days ago
Sorin Tirc
5039
answered 2 days ago
Sorin Tirc
5039
answered 2 days ago
Sorin Tirc
5039
5039
That's what I initially thought, but if we take into account the constraint that $x lt y$, then wouldn't it make more sense to set the integration limit as $y$? I'm having trouble understanding why $x$ would be able to take on any value in interval $(0, 1)$ when it's constrained by the value of $y$.
– Sean
2 days ago
Are you saying that because the value of $y$ is not yet determined, $x$ may take on any value in $(0, 1)$?
– Sean
2 days ago
1
You don’t have to consider the value of y first. You can consider either the value of x first, in which case y is constrained to be between x and 1 or you could consider y first and then x is constrained to be within 0 and y and you get two perfectly equivalent solutions
– Sorin Tirc
2 days ago
1
What matters is that the set {$0<x<y<1$} can be written as {$0<x<1$, $x<y<1$}
– Sorin Tirc
2 days ago
1
Yes, absolutely ;)
– Sorin Tirc
2 days ago
|
show 6 more comments
That's what I initially thought, but if we take into account the constraint that $x lt y$, then wouldn't it make more sense to set the integration limit as $y$? I'm having trouble understanding why $x$ would be able to take on any value in interval $(0, 1)$ when it's constrained by the value of $y$.
– Sean
2 days ago
Are you saying that because the value of $y$ is not yet determined, $x$ may take on any value in $(0, 1)$?
– Sean
2 days ago
1
You don’t have to consider the value of y first. You can consider either the value of x first, in which case y is constrained to be between x and 1 or you could consider y first and then x is constrained to be within 0 and y and you get two perfectly equivalent solutions
– Sorin Tirc
2 days ago
1
What matters is that the set {$0<x<y<1$} can be written as {$0<x<1$, $x<y<1$}
– Sorin Tirc
2 days ago
1
Yes, absolutely ;)
– Sorin Tirc
2 days ago
That's what I initially thought, but if we take into account the constraint that $x lt y$, then wouldn't it make more sense to set the integration limit as $y$? I'm having trouble understanding why $x$ would be able to take on any value in interval $(0, 1)$ when it's constrained by the value of $y$.
– Sean
2 days ago
That's what I initially thought, but if we take into account the constraint that $x lt y$, then wouldn't it make more sense to set the integration limit as $y$? I'm having trouble understanding why $x$ would be able to take on any value in interval $(0, 1)$ when it's constrained by the value of $y$.
– Sean
2 days ago
Are you saying that because the value of $y$ is not yet determined, $x$ may take on any value in $(0, 1)$?
– Sean
2 days ago
Are you saying that because the value of $y$ is not yet determined, $x$ may take on any value in $(0, 1)$?
– Sean
2 days ago
1
1
You don’t have to consider the value of y first. You can consider either the value of x first, in which case y is constrained to be between x and 1 or you could consider y first and then x is constrained to be within 0 and y and you get two perfectly equivalent solutions
– Sorin Tirc
2 days ago
You don’t have to consider the value of y first. You can consider either the value of x first, in which case y is constrained to be between x and 1 or you could consider y first and then x is constrained to be within 0 and y and you get two perfectly equivalent solutions
– Sorin Tirc
2 days ago
1
1
What matters is that the set {$0<x<y<1$} can be written as {$0<x<1$, $x<y<1$}
– Sorin Tirc
2 days ago
What matters is that the set {$0<x<y<1$} can be written as {$0<x<1$, $x<y<1$}
– Sorin Tirc
2 days ago
1
1
Yes, absolutely ;)
– Sorin Tirc
2 days ago
Yes, absolutely ;)
– Sorin Tirc
2 days ago
|
show 6 more comments
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