Mixing
Leibniz rule for double integral
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I'm attempting to differentiate the following double integral with respect to $u$:
$$I(u) = int_a^uint_b^v [(y-u) + (v - x)] f(x,y),dx ,dy$$
where $f(x,y)$ is the joint density function of RV $X$ and $Y$.
I'm trying to find $I'(u)$, that is, the derivative of $I$ with respect to $u$. I know this involves Leibniz integral rule, however I'm getting stuck on the double integral part.
If I let $g(x,y,u) = int_b^v [(y-u) + (v - x)] f(x,y),dx$ then I think the problem to solve is:
$$fracddubigg(int_a^ug(u,x,y),dybigg) = g(u,x,u) + int_a^ufracpartialpartial ug(u,x,y),dy$$
However I'm getting stuck evaluating the two expressions on the RHS.
Any help?
calculus integration multivariable-calculus derivatives
edited Jul 31 at 18:16
Michael Hardy
204k23185460
asked Jul 31 at 17:14
measure_theory
676212
add a comment |Â
up vote
3
down vote
favorite
I'm attempting to differentiate the following double integral with respect to $u$:
$$I(u) = int_a^uint_b^v [(y-u) + (v - x)] f(x,y),dx ,dy$$
where $f(x,y)$ is the joint density function of RV $X$ and $Y$.
I'm trying to find $I'(u)$, that is, the derivative of $I$ with respect to $u$. I know this involves Leibniz integral rule, however I'm getting stuck on the double integral part.
If I let $g(x,y,u) = int_b^v [(y-u) + (v - x)] f(x,y),dx$ then I think the problem to solve is:
$$fracddubigg(int_a^ug(u,x,y),dybigg) = g(u,x,u) + int_a^ufracpartialpartial ug(u,x,y),dy$$
However I'm getting stuck evaluating the two expressions on the RHS.
Any help?
calculus integration multivariable-calculus derivatives
edited Jul 31 at 18:16
Michael Hardy
204k23185460
asked Jul 31 at 17:14
measure_theory
676212
1
this is related math.stackexchange.com/questions/304937/…
– Vincent Law
Jul 31 at 17:25
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I'm attempting to differentiate the following double integral with respect to $u$:
$$I(u) = int_a^uint_b^v [(y-u) + (v - x)] f(x,y),dx ,dy$$
where $f(x,y)$ is the joint density function of RV $X$ and $Y$.
I'm trying to find $I'(u)$, that is, the derivative of $I$ with respect to $u$. I know this involves Leibniz integral rule, however I'm getting stuck on the double integral part.
If I let $g(x,y,u) = int_b^v [(y-u) + (v - x)] f(x,y),dx$ then I think the problem to solve is:
$$fracddubigg(int_a^ug(u,x,y),dybigg) = g(u,x,u) + int_a^ufracpartialpartial ug(u,x,y),dy$$
However I'm getting stuck evaluating the two expressions on the RHS.
Any help?
calculus integration multivariable-calculus derivatives
edited Jul 31 at 18:16
Michael Hardy
204k23185460
asked Jul 31 at 17:14
measure_theory
676212
I'm attempting to differentiate the following double integral with respect to $u$:
$$I(u) = int_a^uint_b^v [(y-u) + (v - x)] f(x,y),dx ,dy$$
where $f(x,y)$ is the joint density function of RV $X$ and $Y$.
I'm trying to find $I'(u)$, that is, the derivative of $I$ with respect to $u$. I know this involves Leibniz integral rule, however I'm getting stuck on the double integral part.
If I let $g(x,y,u) = int_b^v [(y-u) + (v - x)] f(x,y),dx$ then I think the problem to solve is:
$$fracddubigg(int_a^ug(u,x,y),dybigg) = g(u,x,u) + int_a^ufracpartialpartial ug(u,x,y),dy$$
However I'm getting stuck evaluating the two expressions on the RHS.
Any help?
calculus integration multivariable-calculus derivatives
edited Jul 31 at 18:16
Michael Hardy
204k23185460
asked Jul 31 at 17:14
measure_theory
676212
edited Jul 31 at 18:16
Michael Hardy
204k23185460
edited Jul 31 at 18:16
Michael Hardy
204k23185460
edited Jul 31 at 18:16
Michael Hardy
204k23185460
204k23185460
asked Jul 31 at 17:14
measure_theory
676212
asked Jul 31 at 17:14
measure_theory
676212
asked Jul 31 at 17:14
measure_theory
676212
676212
1
this is related math.stackexchange.com/questions/304937/…
– Vincent Law
Jul 31 at 17:25
add a comment |Â
1
this is related math.stackexchange.com/questions/304937/…
– Vincent Law
Jul 31 at 17:25
1
1
this is related math.stackexchange.com/questions/304937/…
– Vincent Law
Jul 31 at 17:25
this is related math.stackexchange.com/questions/304937/…
– Vincent Law
Jul 31 at 17:25
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Your approach is correct, so I'm going to let $g(u,v,y) = int_b^v [(y-u)+(v-x)]f(x,y)dx$ (note that this is not actually a function of $x$ since it gets integrated over, I think you may have had a typo). Then by the Liebnitz rule we have:
$$
fracddu int_a^u g(u,v,y)dy = g(u,v,u) + int_a^u fracddu g(u,v,y) dy
$$
The first part simplifies nicely:
$$
g(u,v,u) = int_b^v [(y-u) + (v-x)]f(x,y)dx = int_b^v (v-x)f(x,y) dx
$$
and the partial derivative of $g$:
$$
fracddu g(u,v,y) = fracdduint_b^v [(y-u) + (v-x)]f(x,y)dx = int_b^v fracddu[(y-u) + (v-x)]f(x,y)dx = -int_b^v f(x,y) dx
$$
answered Jul 31 at 17:28
Dark Malthorp
667212
I made a typo in the OP. The $f(x,y)$ should be distributed through to $(y-u)$. I think this just changes the last line of your answer a bit.
– measure_theory
Jul 31 at 17:32
OK I shall edit accordingly.
– Dark Malthorp
Jul 31 at 17:34
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Your approach is correct, so I'm going to let $g(u,v,y) = int_b^v [(y-u)+(v-x)]f(x,y)dx$ (note that this is not actually a function of $x$ since it gets integrated over, I think you may have had a typo). Then by the Liebnitz rule we have:
$$
fracddu int_a^u g(u,v,y)dy = g(u,v,u) + int_a^u fracddu g(u,v,y) dy
$$
The first part simplifies nicely:
$$
g(u,v,u) = int_b^v [(y-u) + (v-x)]f(x,y)dx = int_b^v (v-x)f(x,y) dx
$$
and the partial derivative of $g$:
$$
fracddu g(u,v,y) = fracdduint_b^v [(y-u) + (v-x)]f(x,y)dx = int_b^v fracddu[(y-u) + (v-x)]f(x,y)dx = -int_b^v f(x,y) dx
$$
answered Jul 31 at 17:28
Dark Malthorp
667212
I made a typo in the OP. The $f(x,y)$ should be distributed through to $(y-u)$. I think this just changes the last line of your answer a bit.
– measure_theory
Jul 31 at 17:32
OK I shall edit accordingly.
– Dark Malthorp
Jul 31 at 17:34
add a comment |Â
up vote
2
down vote
accepted
Your approach is correct, so I'm going to let $g(u,v,y) = int_b^v [(y-u)+(v-x)]f(x,y)dx$ (note that this is not actually a function of $x$ since it gets integrated over, I think you may have had a typo). Then by the Liebnitz rule we have:
$$
fracddu int_a^u g(u,v,y)dy = g(u,v,u) + int_a^u fracddu g(u,v,y) dy
$$
The first part simplifies nicely:
$$
g(u,v,u) = int_b^v [(y-u) + (v-x)]f(x,y)dx = int_b^v (v-x)f(x,y) dx
$$
and the partial derivative of $g$:
$$
fracddu g(u,v,y) = fracdduint_b^v [(y-u) + (v-x)]f(x,y)dx = int_b^v fracddu[(y-u) + (v-x)]f(x,y)dx = -int_b^v f(x,y) dx
$$
answered Jul 31 at 17:28
Dark Malthorp
667212
I made a typo in the OP. The $f(x,y)$ should be distributed through to $(y-u)$. I think this just changes the last line of your answer a bit.
– measure_theory
Jul 31 at 17:32
OK I shall edit accordingly.
– Dark Malthorp
Jul 31 at 17:34
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Your approach is correct, so I'm going to let $g(u,v,y) = int_b^v [(y-u)+(v-x)]f(x,y)dx$ (note that this is not actually a function of $x$ since it gets integrated over, I think you may have had a typo). Then by the Liebnitz rule we have:
$$
fracddu int_a^u g(u,v,y)dy = g(u,v,u) + int_a^u fracddu g(u,v,y) dy
$$
The first part simplifies nicely:
$$
g(u,v,u) = int_b^v [(y-u) + (v-x)]f(x,y)dx = int_b^v (v-x)f(x,y) dx
$$
and the partial derivative of $g$:
$$
fracddu g(u,v,y) = fracdduint_b^v [(y-u) + (v-x)]f(x,y)dx = int_b^v fracddu[(y-u) + (v-x)]f(x,y)dx = -int_b^v f(x,y) dx
$$
answered Jul 31 at 17:28
Dark Malthorp
667212
Your approach is correct, so I'm going to let $g(u,v,y) = int_b^v [(y-u)+(v-x)]f(x,y)dx$ (note that this is not actually a function of $x$ since it gets integrated over, I think you may have had a typo). Then by the Liebnitz rule we have:
$$
fracddu int_a^u g(u,v,y)dy = g(u,v,u) + int_a^u fracddu g(u,v,y) dy
$$
The first part simplifies nicely:
$$
g(u,v,u) = int_b^v [(y-u) + (v-x)]f(x,y)dx = int_b^v (v-x)f(x,y) dx
$$
and the partial derivative of $g$:
$$
fracddu g(u,v,y) = fracdduint_b^v [(y-u) + (v-x)]f(x,y)dx = int_b^v fracddu[(y-u) + (v-x)]f(x,y)dx = -int_b^v f(x,y) dx
$$
answered Jul 31 at 17:28
Dark Malthorp
667212
edited Jul 31 at 18:57
answered Jul 31 at 17:28
Dark Malthorp
667212
answered Jul 31 at 17:28
Dark Malthorp
667212
answered Jul 31 at 17:28
Dark Malthorp
667212
667212
I made a typo in the OP. The $f(x,y)$ should be distributed through to $(y-u)$. I think this just changes the last line of your answer a bit.
– measure_theory
Jul 31 at 17:32
OK I shall edit accordingly.
– Dark Malthorp
Jul 31 at 17:34
add a comment |Â
I made a typo in the OP. The $f(x,y)$ should be distributed through to $(y-u)$. I think this just changes the last line of your answer a bit.
– measure_theory
Jul 31 at 17:32
OK I shall edit accordingly.
– Dark Malthorp
Jul 31 at 17:34
I made a typo in the OP. The $f(x,y)$ should be distributed through to $(y-u)$. I think this just changes the last line of your answer a bit.
– measure_theory
Jul 31 at 17:32
I made a typo in the OP. The $f(x,y)$ should be distributed through to $(y-u)$. I think this just changes the last line of your answer a bit.
– measure_theory
Jul 31 at 17:32
OK I shall edit accordingly.
– Dark Malthorp
Jul 31 at 17:34
OK I shall edit accordingly.
– Dark Malthorp
Jul 31 at 17:34
add a comment |Â
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1
this is related math.stackexchange.com/questions/304937/…
– Vincent Law
Jul 31 at 17:25