Mixing
Why is it valid to break a double integral into two single integrals?
$begingroup$
The following double integral comes from a probability exercise in which I have to calculate the probability of a region by integrating the bivariate PDF over that region.
begin{align*}
int_1^inftyint_1^inftyfrac{1}{8}y_1e^{frac{-(y_1+y_2)}{2}}dy_1dy_2 &= int_1^inftyint_1^infty(frac{1}{4}y_1e^{frac{-y_1}{2}})(frac{1}{2}e^{frac{-y_2}{2}})dy_1dy_2 \ \
&= int_1^infty(frac{1}{4}y_1e^{frac{-y_1}{2}})dy_1 times int_1^infty(frac{1}{2}e^{frac{-y_2}{2}})dy_2
end{align*}
Why is it allowed to treat the double integral as the product of two single integrals? I kind of get the general idea that we can "pull" the independent terms out of an integral, but I would like a more rigorous explanation to this problem.
integration multivariable-calculus
edited Jan 21 at 23:31
Winther
20.7k33156
asked Jan 21 at 23:24
ZassZass
1032
$endgroup$
add a comment |
$begingroup$
The following double integral comes from a probability exercise in which I have to calculate the probability of a region by integrating the bivariate PDF over that region.
begin{align*}
int_1^inftyint_1^inftyfrac{1}{8}y_1e^{frac{-(y_1+y_2)}{2}}dy_1dy_2 &= int_1^inftyint_1^infty(frac{1}{4}y_1e^{frac{-y_1}{2}})(frac{1}{2}e^{frac{-y_2}{2}})dy_1dy_2 \ \
&= int_1^infty(frac{1}{4}y_1e^{frac{-y_1}{2}})dy_1 times int_1^infty(frac{1}{2}e^{frac{-y_2}{2}})dy_2
end{align*}
Why is it allowed to treat the double integral as the product of two single integrals? I kind of get the general idea that we can "pull" the independent terms out of an integral, but I would like a more rigorous explanation to this problem.
integration multivariable-calculus
edited Jan 21 at 23:31
Winther
20.7k33156
asked Jan 21 at 23:24
ZassZass
1032
$endgroup$
1
$begingroup$
A double integral can be defined as the limit of a double Riemann sum: $iint_{[0,1]^2} f(x,y){rm d}x{rm d}y = lim_{N,Mtoinfty} frac{1}{N}sum_{i=1}^Nfrac{1}{M}sum_{j=1}^M f(i/N,j/M)$ (here for simplicity over a simple integration region). If $f(i/N,j/M)$ factors as $f(i/N,j/M) = g(i/N)h(j/M)$ (i.e. $f(x,y) = g(x)h(y)$) show that the sums on the right hand side factors into the product of two Riemann sums giving you the result $iint_{[0,1]^2} f(x,y){rm d}x{rm d}y = int_0^1 g(x){rm d}xcdot int_0^1 h(y){rm d}y$.
$endgroup$
– Winther
Jan 21 at 23:35
add a comment |
$begingroup$
The following double integral comes from a probability exercise in which I have to calculate the probability of a region by integrating the bivariate PDF over that region.
begin{align*}
int_1^inftyint_1^inftyfrac{1}{8}y_1e^{frac{-(y_1+y_2)}{2}}dy_1dy_2 &= int_1^inftyint_1^infty(frac{1}{4}y_1e^{frac{-y_1}{2}})(frac{1}{2}e^{frac{-y_2}{2}})dy_1dy_2 \ \
&= int_1^infty(frac{1}{4}y_1e^{frac{-y_1}{2}})dy_1 times int_1^infty(frac{1}{2}e^{frac{-y_2}{2}})dy_2
end{align*}
Why is it allowed to treat the double integral as the product of two single integrals? I kind of get the general idea that we can "pull" the independent terms out of an integral, but I would like a more rigorous explanation to this problem.
integration multivariable-calculus
edited Jan 21 at 23:31
Winther
20.7k33156
asked Jan 21 at 23:24
ZassZass
1032
$endgroup$
The following double integral comes from a probability exercise in which I have to calculate the probability of a region by integrating the bivariate PDF over that region.
begin{align*}
int_1^inftyint_1^inftyfrac{1}{8}y_1e^{frac{-(y_1+y_2)}{2}}dy_1dy_2 &= int_1^inftyint_1^infty(frac{1}{4}y_1e^{frac{-y_1}{2}})(frac{1}{2}e^{frac{-y_2}{2}})dy_1dy_2 \ \
&= int_1^infty(frac{1}{4}y_1e^{frac{-y_1}{2}})dy_1 times int_1^infty(frac{1}{2}e^{frac{-y_2}{2}})dy_2
end{align*}
Why is it allowed to treat the double integral as the product of two single integrals? I kind of get the general idea that we can "pull" the independent terms out of an integral, but I would like a more rigorous explanation to this problem.
integration multivariable-calculus
integration multivariable-calculus
edited Jan 21 at 23:31
Winther
20.7k33156
asked Jan 21 at 23:24
ZassZass
1032
edited Jan 21 at 23:31
Winther
20.7k33156
asked Jan 21 at 23:24
ZassZass
1032
edited Jan 21 at 23:31
Winther
20.7k33156
edited Jan 21 at 23:31
Winther
20.7k33156
edited Jan 21 at 23:31
Winther
20.7k33156
20.7k33156
asked Jan 21 at 23:24
ZassZass
1032
asked Jan 21 at 23:24
ZassZass
1032
asked Jan 21 at 23:24
ZassZass
1032
1032
1
$begingroup$
A double integral can be defined as the limit of a double Riemann sum: $iint_{[0,1]^2} f(x,y){rm d}x{rm d}y = lim_{N,Mtoinfty} frac{1}{N}sum_{i=1}^Nfrac{1}{M}sum_{j=1}^M f(i/N,j/M)$ (here for simplicity over a simple integration region). If $f(i/N,j/M)$ factors as $f(i/N,j/M) = g(i/N)h(j/M)$ (i.e. $f(x,y) = g(x)h(y)$) show that the sums on the right hand side factors into the product of two Riemann sums giving you the result $iint_{[0,1]^2} f(x,y){rm d}x{rm d}y = int_0^1 g(x){rm d}xcdot int_0^1 h(y){rm d}y$.
$endgroup$
– Winther
Jan 21 at 23:35
add a comment |
1
$begingroup$
A double integral can be defined as the limit of a double Riemann sum: $iint_{[0,1]^2} f(x,y){rm d}x{rm d}y = lim_{N,Mtoinfty} frac{1}{N}sum_{i=1}^Nfrac{1}{M}sum_{j=1}^M f(i/N,j/M)$ (here for simplicity over a simple integration region). If $f(i/N,j/M)$ factors as $f(i/N,j/M) = g(i/N)h(j/M)$ (i.e. $f(x,y) = g(x)h(y)$) show that the sums on the right hand side factors into the product of two Riemann sums giving you the result $iint_{[0,1]^2} f(x,y){rm d}x{rm d}y = int_0^1 g(x){rm d}xcdot int_0^1 h(y){rm d}y$.
$endgroup$
– Winther
Jan 21 at 23:35
1
1
$begingroup$
A double integral can be defined as the limit of a double Riemann sum: $iint_{[0,1]^2} f(x,y){rm d}x{rm d}y = lim_{N,Mtoinfty} frac{1}{N}sum_{i=1}^Nfrac{1}{M}sum_{j=1}^M f(i/N,j/M)$ (here for simplicity over a simple integration region). If $f(i/N,j/M)$ factors as $f(i/N,j/M) = g(i/N)h(j/M)$ (i.e. $f(x,y) = g(x)h(y)$) show that the sums on the right hand side factors into the product of two Riemann sums giving you the result $iint_{[0,1]^2} f(x,y){rm d}x{rm d}y = int_0^1 g(x){rm d}xcdot int_0^1 h(y){rm d}y$.
$endgroup$
– Winther
Jan 21 at 23:35
$begingroup$
A double integral can be defined as the limit of a double Riemann sum: $iint_{[0,1]^2} f(x,y){rm d}x{rm d}y = lim_{N,Mtoinfty} frac{1}{N}sum_{i=1}^Nfrac{1}{M}sum_{j=1}^M f(i/N,j/M)$ (here for simplicity over a simple integration region). If $f(i/N,j/M)$ factors as $f(i/N,j/M) = g(i/N)h(j/M)$ (i.e. $f(x,y) = g(x)h(y)$) show that the sums on the right hand side factors into the product of two Riemann sums giving you the result $iint_{[0,1]^2} f(x,y){rm d}x{rm d}y = int_0^1 g(x){rm d}xcdot int_0^1 h(y){rm d}y$.
$endgroup$
– Winther
Jan 21 at 23:35
add a comment |
3 Answers
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$$int_a^b int_c^d f(x) g(y) , dy , dx = int_a^b f(x) int_c^d g(y) , dy , dx = int_a^b f(x) , dx cdot int_c^d g(y) , dy.$$
In the first equality, you can pull $f(x)$ out of the inner integral because $int_c^d Cg(y) , dy = C int_c^d g(y) , dy$.
In the second equality, you can pull the entire term $int_c^d g(y) , dy$ out of the outer integral because $int_a^b f(x) cdot C , dx = left(int_a^b f(x) , dxright) C$.
Caveats: it is important that your original integrand was separable, i.e. of the form $f(x) g(y)$. (This is related to the joint PDF of two independent random variables.) It is also important that the limits of integration of the inner integral do not depend on the outer integral's dummy variables.
answered Jan 21 at 23:30
angryavianangryavian
42.1k23381
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When you integrate w.r.t. $y_1$ you get $int_1^{infty} C frac 1 2 e^{-frac {y_2} 2}dy_2$ where $C=int_1^{infty} frac 1 4 y_1e^{-frac {y_1} 2}dy_1$. $C$ is a constant, so you can pull it out of the integral sign.
answered Jan 21 at 23:30
Kavi Rama MurthyKavi Rama Murthy
65.6k42767
$endgroup$
add a comment |
$begingroup$
Maybe an extra step would be helpful:
begin{align}
int_a^b int_c^d f(x) underbrace{g(y)}_{substack{text{pull}\ text{out front}}} , dx , dy &= int_a^b g(y) underbrace{int_c^d f(x) , dx}_{substack{text{pull}\text{out front}}} , dy \
&= int_c^d f(x) , dx int_a^b g(y) , dy.
end{align}
answered Jan 21 at 23:30
littleOlittleO
29.9k647110
$endgroup$
add a comment |
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3 Answers
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3 Answers
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$begingroup$
$$int_a^b int_c^d f(x) g(y) , dy , dx = int_a^b f(x) int_c^d g(y) , dy , dx = int_a^b f(x) , dx cdot int_c^d g(y) , dy.$$
In the first equality, you can pull $f(x)$ out of the inner integral because $int_c^d Cg(y) , dy = C int_c^d g(y) , dy$.
In the second equality, you can pull the entire term $int_c^d g(y) , dy$ out of the outer integral because $int_a^b f(x) cdot C , dx = left(int_a^b f(x) , dxright) C$.
Caveats: it is important that your original integrand was separable, i.e. of the form $f(x) g(y)$. (This is related to the joint PDF of two independent random variables.) It is also important that the limits of integration of the inner integral do not depend on the outer integral's dummy variables.
answered Jan 21 at 23:30
angryavianangryavian
42.1k23381
$endgroup$
add a comment |
$begingroup$
$$int_a^b int_c^d f(x) g(y) , dy , dx = int_a^b f(x) int_c^d g(y) , dy , dx = int_a^b f(x) , dx cdot int_c^d g(y) , dy.$$
In the first equality, you can pull $f(x)$ out of the inner integral because $int_c^d Cg(y) , dy = C int_c^d g(y) , dy$.
In the second equality, you can pull the entire term $int_c^d g(y) , dy$ out of the outer integral because $int_a^b f(x) cdot C , dx = left(int_a^b f(x) , dxright) C$.
Caveats: it is important that your original integrand was separable, i.e. of the form $f(x) g(y)$. (This is related to the joint PDF of two independent random variables.) It is also important that the limits of integration of the inner integral do not depend on the outer integral's dummy variables.
answered Jan 21 at 23:30
angryavianangryavian
42.1k23381
$endgroup$
add a comment |
$begingroup$
$$int_a^b int_c^d f(x) g(y) , dy , dx = int_a^b f(x) int_c^d g(y) , dy , dx = int_a^b f(x) , dx cdot int_c^d g(y) , dy.$$
In the first equality, you can pull $f(x)$ out of the inner integral because $int_c^d Cg(y) , dy = C int_c^d g(y) , dy$.
In the second equality, you can pull the entire term $int_c^d g(y) , dy$ out of the outer integral because $int_a^b f(x) cdot C , dx = left(int_a^b f(x) , dxright) C$.
Caveats: it is important that your original integrand was separable, i.e. of the form $f(x) g(y)$. (This is related to the joint PDF of two independent random variables.) It is also important that the limits of integration of the inner integral do not depend on the outer integral's dummy variables.
answered Jan 21 at 23:30
angryavianangryavian
42.1k23381
$endgroup$
$$int_a^b int_c^d f(x) g(y) , dy , dx = int_a^b f(x) int_c^d g(y) , dy , dx = int_a^b f(x) , dx cdot int_c^d g(y) , dy.$$
In the first equality, you can pull $f(x)$ out of the inner integral because $int_c^d Cg(y) , dy = C int_c^d g(y) , dy$.
In the second equality, you can pull the entire term $int_c^d g(y) , dy$ out of the outer integral because $int_a^b f(x) cdot C , dx = left(int_a^b f(x) , dxright) C$.
Caveats: it is important that your original integrand was separable, i.e. of the form $f(x) g(y)$. (This is related to the joint PDF of two independent random variables.) It is also important that the limits of integration of the inner integral do not depend on the outer integral's dummy variables.
answered Jan 21 at 23:30
angryavianangryavian
42.1k23381
answered Jan 21 at 23:30
angryavianangryavian
42.1k23381
answered Jan 21 at 23:30
angryavianangryavian
42.1k23381
answered Jan 21 at 23:30
angryavianangryavian
42.1k23381
42.1k23381
add a comment |
add a comment |
$begingroup$
When you integrate w.r.t. $y_1$ you get $int_1^{infty} C frac 1 2 e^{-frac {y_2} 2}dy_2$ where $C=int_1^{infty} frac 1 4 y_1e^{-frac {y_1} 2}dy_1$. $C$ is a constant, so you can pull it out of the integral sign.
answered Jan 21 at 23:30
Kavi Rama MurthyKavi Rama Murthy
65.6k42767
$endgroup$
add a comment |
$begingroup$
When you integrate w.r.t. $y_1$ you get $int_1^{infty} C frac 1 2 e^{-frac {y_2} 2}dy_2$ where $C=int_1^{infty} frac 1 4 y_1e^{-frac {y_1} 2}dy_1$. $C$ is a constant, so you can pull it out of the integral sign.
answered Jan 21 at 23:30
Kavi Rama MurthyKavi Rama Murthy
65.6k42767
$endgroup$
add a comment |
$begingroup$
When you integrate w.r.t. $y_1$ you get $int_1^{infty} C frac 1 2 e^{-frac {y_2} 2}dy_2$ where $C=int_1^{infty} frac 1 4 y_1e^{-frac {y_1} 2}dy_1$. $C$ is a constant, so you can pull it out of the integral sign.
answered Jan 21 at 23:30
Kavi Rama MurthyKavi Rama Murthy
65.6k42767
$endgroup$
When you integrate w.r.t. $y_1$ you get $int_1^{infty} C frac 1 2 e^{-frac {y_2} 2}dy_2$ where $C=int_1^{infty} frac 1 4 y_1e^{-frac {y_1} 2}dy_1$. $C$ is a constant, so you can pull it out of the integral sign.
answered Jan 21 at 23:30
Kavi Rama MurthyKavi Rama Murthy
65.6k42767
answered Jan 21 at 23:30
Kavi Rama MurthyKavi Rama Murthy
65.6k42767
answered Jan 21 at 23:30
Kavi Rama MurthyKavi Rama Murthy
65.6k42767
answered Jan 21 at 23:30
Kavi Rama MurthyKavi Rama Murthy
65.6k42767
65.6k42767
add a comment |
add a comment |
$begingroup$
Maybe an extra step would be helpful:
begin{align}
int_a^b int_c^d f(x) underbrace{g(y)}_{substack{text{pull}\ text{out front}}} , dx , dy &= int_a^b g(y) underbrace{int_c^d f(x) , dx}_{substack{text{pull}\text{out front}}} , dy \
&= int_c^d f(x) , dx int_a^b g(y) , dy.
end{align}
answered Jan 21 at 23:30
littleOlittleO
29.9k647110
$endgroup$
add a comment |
$begingroup$
Maybe an extra step would be helpful:
begin{align}
int_a^b int_c^d f(x) underbrace{g(y)}_{substack{text{pull}\ text{out front}}} , dx , dy &= int_a^b g(y) underbrace{int_c^d f(x) , dx}_{substack{text{pull}\text{out front}}} , dy \
&= int_c^d f(x) , dx int_a^b g(y) , dy.
end{align}
answered Jan 21 at 23:30
littleOlittleO
29.9k647110
$endgroup$
add a comment |
$begingroup$
Maybe an extra step would be helpful:
begin{align}
int_a^b int_c^d f(x) underbrace{g(y)}_{substack{text{pull}\ text{out front}}} , dx , dy &= int_a^b g(y) underbrace{int_c^d f(x) , dx}_{substack{text{pull}\text{out front}}} , dy \
&= int_c^d f(x) , dx int_a^b g(y) , dy.
end{align}
answered Jan 21 at 23:30
littleOlittleO
29.9k647110
$endgroup$
Maybe an extra step would be helpful:
begin{align}
int_a^b int_c^d f(x) underbrace{g(y)}_{substack{text{pull}\ text{out front}}} , dx , dy &= int_a^b g(y) underbrace{int_c^d f(x) , dx}_{substack{text{pull}\text{out front}}} , dy \
&= int_c^d f(x) , dx int_a^b g(y) , dy.
end{align}
answered Jan 21 at 23:30
littleOlittleO
29.9k647110
answered Jan 21 at 23:30
littleOlittleO
29.9k647110
answered Jan 21 at 23:30
littleOlittleO
29.9k647110
answered Jan 21 at 23:30
littleOlittleO
29.9k647110
29.9k647110
add a comment |
add a comment |
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$begingroup$
A double integral can be defined as the limit of a double Riemann sum: $iint_{[0,1]^2} f(x,y){rm d}x{rm d}y = lim_{N,Mtoinfty} frac{1}{N}sum_{i=1}^Nfrac{1}{M}sum_{j=1}^M f(i/N,j/M)$ (here for simplicity over a simple integration region). If $f(i/N,j/M)$ factors as $f(i/N,j/M) = g(i/N)h(j/M)$ (i.e. $f(x,y) = g(x)h(y)$) show that the sums on the right hand side factors into the product of two Riemann sums giving you the result $iint_{[0,1]^2} f(x,y){rm d}x{rm d}y = int_0^1 g(x){rm d}xcdot int_0^1 h(y){rm d}y$.
$endgroup$
– Winther
Jan 21 at 23:35