Change of Variable of the double integral of a multivariable function
This is from chapter 3.6 of the textbook by John Rice on probability and statistics:
The continuous case is very similar. Supposing that $X$ and $Y$ are continuous random variables, we first find the cdf of $Z$ and then differentiate to find the density. Since $Z leq z$ whenever the point $(X, Y)$ is in the shaded region $R_z$ shown in Figure $3.17$, we have
begin{align}
F_Z (z) &=iintlimits_{R_z} f(x,y) dx dy \
&= int_{-infty}^infty int_{-infty}^{z-x} f(x,y) dy dx.
end{align}
In the inner integral, we make the change of variables $y=v-x$ to obtain
begin{align}
F_Z (z) &= int_{-infty}^infty int_{-infty}^z f(x,v-x) dv dx \
&=int_{-infty}^z int_{-infty}^infty f(x,v-x) dx dv.
end{align}
Differentiating, we have, if $int_{-infty}^infty f(x,z-x) dx$ is continuous at $z$,
$$f_Z (z) =int_{-infty}^infty f(x,z-x) dx.$$
I don't understand how they were applied the change of variables in the double integral. Why does the upper bound of the inner integral change from $z-x$ to $z$?
integration definite-integrals improper-integrals
edited Nov 26 at 2:51
Rócherz
2,7362721
asked Nov 26 at 2:38
Christopher
203
add a comment |
This is from chapter 3.6 of the textbook by John Rice on probability and statistics:
The continuous case is very similar. Supposing that $X$ and $Y$ are continuous random variables, we first find the cdf of $Z$ and then differentiate to find the density. Since $Z leq z$ whenever the point $(X, Y)$ is in the shaded region $R_z$ shown in Figure $3.17$, we have
begin{align}
F_Z (z) &=iintlimits_{R_z} f(x,y) dx dy \
&= int_{-infty}^infty int_{-infty}^{z-x} f(x,y) dy dx.
end{align}
In the inner integral, we make the change of variables $y=v-x$ to obtain
begin{align}
F_Z (z) &= int_{-infty}^infty int_{-infty}^z f(x,v-x) dv dx \
&=int_{-infty}^z int_{-infty}^infty f(x,v-x) dx dv.
end{align}
Differentiating, we have, if $int_{-infty}^infty f(x,z-x) dx$ is continuous at $z$,
$$f_Z (z) =int_{-infty}^infty f(x,z-x) dx.$$
I don't understand how they were applied the change of variables in the double integral. Why does the upper bound of the inner integral change from $z-x$ to $z$?
integration definite-integrals improper-integrals
edited Nov 26 at 2:51
Rócherz
2,7362721
asked Nov 26 at 2:38
Christopher
203
add a comment |
This is from chapter 3.6 of the textbook by John Rice on probability and statistics:
The continuous case is very similar. Supposing that $X$ and $Y$ are continuous random variables, we first find the cdf of $Z$ and then differentiate to find the density. Since $Z leq z$ whenever the point $(X, Y)$ is in the shaded region $R_z$ shown in Figure $3.17$, we have
begin{align}
F_Z (z) &=iintlimits_{R_z} f(x,y) dx dy \
&= int_{-infty}^infty int_{-infty}^{z-x} f(x,y) dy dx.
end{align}
In the inner integral, we make the change of variables $y=v-x$ to obtain
begin{align}
F_Z (z) &= int_{-infty}^infty int_{-infty}^z f(x,v-x) dv dx \
&=int_{-infty}^z int_{-infty}^infty f(x,v-x) dx dv.
end{align}
Differentiating, we have, if $int_{-infty}^infty f(x,z-x) dx$ is continuous at $z$,
$$f_Z (z) =int_{-infty}^infty f(x,z-x) dx.$$
I don't understand how they were applied the change of variables in the double integral. Why does the upper bound of the inner integral change from $z-x$ to $z$?
integration definite-integrals improper-integrals
edited Nov 26 at 2:51
Rócherz
2,7362721
asked Nov 26 at 2:38
Christopher
203
This is from chapter 3.6 of the textbook by John Rice on probability and statistics:
The continuous case is very similar. Supposing that $X$ and $Y$ are continuous random variables, we first find the cdf of $Z$ and then differentiate to find the density. Since $Z leq z$ whenever the point $(X, Y)$ is in the shaded region $R_z$ shown in Figure $3.17$, we have
begin{align}
F_Z (z) &=iintlimits_{R_z} f(x,y) dx dy \
&= int_{-infty}^infty int_{-infty}^{z-x} f(x,y) dy dx.
end{align}
In the inner integral, we make the change of variables $y=v-x$ to obtain
begin{align}
F_Z (z) &= int_{-infty}^infty int_{-infty}^z f(x,v-x) dv dx \
&=int_{-infty}^z int_{-infty}^infty f(x,v-x) dx dv.
end{align}
Differentiating, we have, if $int_{-infty}^infty f(x,z-x) dx$ is continuous at $z$,
$$f_Z (z) =int_{-infty}^infty f(x,z-x) dx.$$
I don't understand how they were applied the change of variables in the double integral. Why does the upper bound of the inner integral change from $z-x$ to $z$?
integration definite-integrals improper-integrals
integration definite-integrals improper-integrals
edited Nov 26 at 2:51
Rócherz
2,7362721
asked Nov 26 at 2:38
Christopher
203
edited Nov 26 at 2:51
Rócherz
2,7362721
asked Nov 26 at 2:38
Christopher
203
edited Nov 26 at 2:51
Rócherz
2,7362721
edited Nov 26 at 2:51
Rócherz
2,7362721
edited Nov 26 at 2:51
Rócherz
2,7362721
2,7362721
asked Nov 26 at 2:38
Christopher
203
asked Nov 26 at 2:38
Christopher
203
asked Nov 26 at 2:38
Christopher
203
203
add a comment |
add a comment |
1 Answer
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Because when $v=z$ you have $y=v-x = z-x$ as like in the first integral.
answered Nov 26 at 2:43
Davide M
1
add a comment |
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Because when $v=z$ you have $y=v-x = z-x$ as like in the first integral.
answered Nov 26 at 2:43
Davide M
1
add a comment |
Because when $v=z$ you have $y=v-x = z-x$ as like in the first integral.
answered Nov 26 at 2:43
Davide M
1
add a comment |
Because when $v=z$ you have $y=v-x = z-x$ as like in the first integral.
answered Nov 26 at 2:43
Davide M
1
Because when $v=z$ you have $y=v-x = z-x$ as like in the first integral.
answered Nov 26 at 2:43
Davide M
1
answered Nov 26 at 2:43
Davide M
1
answered Nov 26 at 2:43
Davide M
1
answered Nov 26 at 2:43
Davide M
1
1
add a comment |
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