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 Nov 19 at 10:15
Tesla
916426
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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 Nov 19 at 10:15
Tesla
916426
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 Nov 19 at 10:15
Tesla
916426
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 Nov 19 at 10:15
Tesla
916426
asked Nov 19 at 10:15
Tesla
916426
asked Nov 19 at 10:15
Tesla
916426
asked Nov 19 at 10:15
Tesla
916426
asked Nov 19 at 10:15
Tesla
916426
916426
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 Nov 19 at 10:18
Kavi Rama Murthy
44.2k31852
Got it, thank you very much! And why am I allowed to take the functions out of the integral?
– Tesla
Nov 19 at 10:22
1
When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
– Kavi Rama Murthy
Nov 19 at 10:24
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 Nov 19 at 10:18
Kavi Rama Murthy
44.2k31852
Got it, thank you very much! And why am I allowed to take the functions out of the integral?
– Tesla
Nov 19 at 10:22
1
When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
– Kavi Rama Murthy
Nov 19 at 10:24
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 Nov 19 at 10:18
Kavi Rama Murthy
44.2k31852
Got it, thank you very much! And why am I allowed to take the functions out of the integral?
– Tesla
Nov 19 at 10:22
1
When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
– Kavi Rama Murthy
Nov 19 at 10:24
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 Nov 19 at 10:18
Kavi Rama Murthy
44.2k31852
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 Nov 19 at 10:18
Kavi Rama Murthy
44.2k31852
answered Nov 19 at 10:18
Kavi Rama Murthy
44.2k31852
answered Nov 19 at 10:18
Kavi Rama Murthy
44.2k31852
answered Nov 19 at 10:18
Kavi Rama Murthy
44.2k31852
44.2k31852
Got it, thank you very much! And why am I allowed to take the functions out of the integral?
– Tesla
Nov 19 at 10:22
1
When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
– Kavi Rama Murthy
Nov 19 at 10:24
add a comment |
Got it, thank you very much! And why am I allowed to take the functions out of the integral?
– Tesla
Nov 19 at 10:22
1
When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
– Kavi Rama Murthy
Nov 19 at 10:24
Got it, thank you very much! And why am I allowed to take the functions out of the integral?
– Tesla
Nov 19 at 10:22
Got it, thank you very much! And why am I allowed to take the functions out of the integral?
– Tesla
Nov 19 at 10:22
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
Nov 19 at 10:24
When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out.
– Kavi Rama Murthy
Nov 19 at 10:24
add a comment |
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