Mixing
Double integrals - how are the boundaries chosen?
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I was looking at the proof of the theorem stating that for two independent rv's $X,Y$ with density functions $f,g$ the density function of the rv $X+Y$ is the convolution of $f$ and $g$.
After having defined $A_z={(x,y) in mathbb R^2 : x+y leq z}$ for $z in mathbb R$ we get
$P(X+Y leq z) = P((X,Y) in A_z) = int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y), = , ...$
Can someone please explain the last equality
$int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y)$? I don't understand how the boundaries have been chosen and why we can just take the functions out of the integrals. Thank you.
integration convolution density-function
asked 18 hours ago
Tesla
904426
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I was looking at the proof of the theorem stating that for two independent rv's $X,Y$ with density functions $f,g$ the density function of the rv $X+Y$ is the convolution of $f$ and $g$.
After having defined $A_z={(x,y) in mathbb R^2 : x+y leq z}$ for $z in mathbb R$ we get
$P(X+Y leq z) = P((X,Y) in A_z) = int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y), = , ...$
Can someone please explain the last equality
$int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y)$? I don't understand how the boundaries have been chosen and why we can just take the functions out of the integrals. Thank you.
integration convolution density-function
asked 18 hours ago
Tesla
904426
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was looking at the proof of the theorem stating that for two independent rv's $X,Y$ with density functions $f,g$ the density function of the rv $X+Y$ is the convolution of $f$ and $g$.
After having defined $A_z={(x,y) in mathbb R^2 : x+y leq z}$ for $z in mathbb R$ we get
$P(X+Y leq z) = P((X,Y) in A_z) = int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y), = , ...$
Can someone please explain the last equality
$int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y)$? I don't understand how the boundaries have been chosen and why we can just take the functions out of the integrals. Thank you.
integration convolution density-function
asked 18 hours ago
Tesla
904426
I was looking at the proof of the theorem stating that for two independent rv's $X,Y$ with density functions $f,g$ the density function of the rv $X+Y$ is the convolution of $f$ and $g$.
After having defined $A_z={(x,y) in mathbb R^2 : x+y leq z}$ for $z in mathbb R$ we get
$P(X+Y leq z) = P((X,Y) in A_z) = int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y), = , ...$
Can someone please explain the last equality
$int_{A_z}f(x)g(y),dxdy=int_{-infty}^{infty}, dx ,f(x) int_{-infty}^{z-x}, dy , g(y)$? I don't understand how the boundaries have been chosen and why we can just take the functions out of the integrals. Thank you.
integration convolution density-function
integration convolution density-function
asked 18 hours ago
Tesla
904426
asked 18 hours ago
Tesla
904426
asked 18 hours ago
Tesla
904426
asked 18 hours ago
Tesla
904426
asked 18 hours ago
Tesla
904426
904426
add a comment |
add a comment |
1 Answer
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For fixed $x$ the inequality $x+y leq z$ is same as $y leq z-x$ so $y$ ranges from $-infty$ to $z-x$. Once the inequality $x+y leq z$ has been taken care of, there is no further restriction on $x$ so $x$ ranges from $-infty$ to $infty$. (You can also do it the other way around: keep $y$ fixed, integrate w.r.t. $x$ from $-infty$ to $z-y$ and then integrate w.r.t $y$).
answered 18 hours ago
Kavi Rama Murthy
40k31750
Got it, thank you very much! And why am I allowed to take the functions out of the integral?
– Tesla
17 hours ago
1
When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
– Kavi Rama Murthy
17 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
For fixed $x$ the inequality $x+y leq z$ is same as $y leq z-x$ so $y$ ranges from $-infty$ to $z-x$. Once the inequality $x+y leq z$ has been taken care of, there is no further restriction on $x$ so $x$ ranges from $-infty$ to $infty$. (You can also do it the other way around: keep $y$ fixed, integrate w.r.t. $x$ from $-infty$ to $z-y$ and then integrate w.r.t $y$).
answered 18 hours ago
Kavi Rama Murthy
40k31750
Got it, thank you very much! And why am I allowed to take the functions out of the integral?
– Tesla
17 hours ago
1
When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
– Kavi Rama Murthy
17 hours ago
add a comment |
up vote
1
down vote
accepted
For fixed $x$ the inequality $x+y leq z$ is same as $y leq z-x$ so $y$ ranges from $-infty$ to $z-x$. Once the inequality $x+y leq z$ has been taken care of, there is no further restriction on $x$ so $x$ ranges from $-infty$ to $infty$. (You can also do it the other way around: keep $y$ fixed, integrate w.r.t. $x$ from $-infty$ to $z-y$ and then integrate w.r.t $y$).
answered 18 hours ago
Kavi Rama Murthy
40k31750
Got it, thank you very much! And why am I allowed to take the functions out of the integral?
– Tesla
17 hours ago
1
When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
– Kavi Rama Murthy
17 hours ago
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
For fixed $x$ the inequality $x+y leq z$ is same as $y leq z-x$ so $y$ ranges from $-infty$ to $z-x$. Once the inequality $x+y leq z$ has been taken care of, there is no further restriction on $x$ so $x$ ranges from $-infty$ to $infty$. (You can also do it the other way around: keep $y$ fixed, integrate w.r.t. $x$ from $-infty$ to $z-y$ and then integrate w.r.t $y$).
answered 18 hours ago
Kavi Rama Murthy
40k31750
For fixed $x$ the inequality $x+y leq z$ is same as $y leq z-x$ so $y$ ranges from $-infty$ to $z-x$. Once the inequality $x+y leq z$ has been taken care of, there is no further restriction on $x$ so $x$ ranges from $-infty$ to $infty$. (You can also do it the other way around: keep $y$ fixed, integrate w.r.t. $x$ from $-infty$ to $z-y$ and then integrate w.r.t $y$).
answered 18 hours ago
Kavi Rama Murthy
40k31750
answered 18 hours ago
Kavi Rama Murthy
40k31750
answered 18 hours ago
Kavi Rama Murthy
40k31750
answered 18 hours ago
Kavi Rama Murthy
40k31750
40k31750
Got it, thank you very much! And why am I allowed to take the functions out of the integral?
– Tesla
17 hours ago
1
When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
– Kavi Rama Murthy
17 hours ago
add a comment |
Got it, thank you very much! And why am I allowed to take the functions out of the integral?
– Tesla
17 hours ago
1
When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
– Kavi Rama Murthy
17 hours ago
Got it, thank you very much! And why am I allowed to take the functions out of the integral?
– Tesla
17 hours ago
Got it, thank you very much! And why am I allowed to take the functions out of the integral?
– Tesla
17 hours ago
1
1
When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
– Kavi Rama Murthy
17 hours ago
When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
– Kavi Rama Murthy
17 hours ago
add a comment |
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