Differentiation of a Complicated Integration
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Let, $h$ is differentiable function. $theta_1inmathbbR$, $theta_2>0$ $$F(theta_1,theta_2) = int_theta_1-theta_2^theta_1+theta_2int_theta_1-theta_2^yn(n-1)frac(y-x)^n-2(2theta_2)^nh(x,y)dxdy$$
I need to find $dfracpartial^2 Fpartial theta_1partial theta_2$. I know Leibniz Rule for these type of cases. Is there any easy way to derive this? Thanks.
calculus integration derivatives statistical-inference
asked Sep 2 at 12:09
Stat_prob_001
283112
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up vote
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Let, $h$ is differentiable function. $theta_1inmathbbR$, $theta_2>0$ $$F(theta_1,theta_2) = int_theta_1-theta_2^theta_1+theta_2int_theta_1-theta_2^yn(n-1)frac(y-x)^n-2(2theta_2)^nh(x,y)dxdy$$
I need to find $dfracpartial^2 Fpartial theta_1partial theta_2$. I know Leibniz Rule for these type of cases. Is there any easy way to derive this? Thanks.
calculus integration derivatives statistical-inference
asked Sep 2 at 12:09
Stat_prob_001
283112
Using Fubini's theorem, we can write the double integral as iterated integrals. Then, beginalign fracpartial Fpartialtheta_2&=fracn(n-1)2^nfracpartialpartialtheta_2leftint_theta_1-theta_2^y,dxrightleftint_theta_1-theta_2^theta_1+theta_2frac(y-x)^n-2theta_2^nh(x,y),dyright \&=fracn(n-1)2^nfracpartialpartialtheta_2frac(y-theta_1+theta_2)theta_2^nleftint_theta_1-theta_2^theta_1+theta_2(y-x)^n-2h(x,y),dyright endalign Use the product rule for differentiation. Leibniz rule applies as usual.
– StubbornAtom
Sep 2 at 12:35
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up vote
1
down vote
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up vote
1
down vote
favorite
Let, $h$ is differentiable function. $theta_1inmathbbR$, $theta_2>0$ $$F(theta_1,theta_2) = int_theta_1-theta_2^theta_1+theta_2int_theta_1-theta_2^yn(n-1)frac(y-x)^n-2(2theta_2)^nh(x,y)dxdy$$
I need to find $dfracpartial^2 Fpartial theta_1partial theta_2$. I know Leibniz Rule for these type of cases. Is there any easy way to derive this? Thanks.
calculus integration derivatives statistical-inference
asked Sep 2 at 12:09
Stat_prob_001
283112
Let, $h$ is differentiable function. $theta_1inmathbbR$, $theta_2>0$ $$F(theta_1,theta_2) = int_theta_1-theta_2^theta_1+theta_2int_theta_1-theta_2^yn(n-1)frac(y-x)^n-2(2theta_2)^nh(x,y)dxdy$$
I need to find $dfracpartial^2 Fpartial theta_1partial theta_2$. I know Leibniz Rule for these type of cases. Is there any easy way to derive this? Thanks.
calculus integration derivatives statistical-inference
calculus integration derivatives statistical-inference
asked Sep 2 at 12:09
Stat_prob_001
283112
asked Sep 2 at 12:09
Stat_prob_001
283112
asked Sep 2 at 12:09
Stat_prob_001
283112
asked Sep 2 at 12:09
Stat_prob_001
283112
asked Sep 2 at 12:09
Stat_prob_001
283112
283112
Using Fubini's theorem, we can write the double integral as iterated integrals. Then, beginalign fracpartial Fpartialtheta_2&=fracn(n-1)2^nfracpartialpartialtheta_2leftint_theta_1-theta_2^y,dxrightleftint_theta_1-theta_2^theta_1+theta_2frac(y-x)^n-2theta_2^nh(x,y),dyright \&=fracn(n-1)2^nfracpartialpartialtheta_2frac(y-theta_1+theta_2)theta_2^nleftint_theta_1-theta_2^theta_1+theta_2(y-x)^n-2h(x,y),dyright endalign Use the product rule for differentiation. Leibniz rule applies as usual.
– StubbornAtom
Sep 2 at 12:35
add a comment |Â
Using Fubini's theorem, we can write the double integral as iterated integrals. Then, beginalign fracpartial Fpartialtheta_2&=fracn(n-1)2^nfracpartialpartialtheta_2leftint_theta_1-theta_2^y,dxrightleftint_theta_1-theta_2^theta_1+theta_2frac(y-x)^n-2theta_2^nh(x,y),dyright \&=fracn(n-1)2^nfracpartialpartialtheta_2frac(y-theta_1+theta_2)theta_2^nleftint_theta_1-theta_2^theta_1+theta_2(y-x)^n-2h(x,y),dyright endalign Use the product rule for differentiation. Leibniz rule applies as usual.
– StubbornAtom
Sep 2 at 12:35
Using Fubini's theorem, we can write the double integral as iterated integrals. Then, beginalign fracpartial Fpartialtheta_2&=fracn(n-1)2^nfracpartialpartialtheta_2leftint_theta_1-theta_2^y,dxrightleftint_theta_1-theta_2^theta_1+theta_2frac(y-x)^n-2theta_2^nh(x,y),dyright \&=fracn(n-1)2^nfracpartialpartialtheta_2frac(y-theta_1+theta_2)theta_2^nleftint_theta_1-theta_2^theta_1+theta_2(y-x)^n-2h(x,y),dyright endalign Use the product rule for differentiation. Leibniz rule applies as usual.
– StubbornAtom
Sep 2 at 12:35
Using Fubini's theorem, we can write the double integral as iterated integrals. Then, beginalign fracpartial Fpartialtheta_2&=fracn(n-1)2^nfracpartialpartialtheta_2leftint_theta_1-theta_2^y,dxrightleftint_theta_1-theta_2^theta_1+theta_2frac(y-x)^n-2theta_2^nh(x,y),dyright \&=fracn(n-1)2^nfracpartialpartialtheta_2frac(y-theta_1+theta_2)theta_2^nleftint_theta_1-theta_2^theta_1+theta_2(y-x)^n-2h(x,y),dyright endalign Use the product rule for differentiation. Leibniz rule applies as usual.
– StubbornAtom
Sep 2 at 12:35
add a comment |Â
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Using Fubini's theorem, we can write the double integral as iterated integrals. Then, beginalign fracpartial Fpartialtheta_2&=fracn(n-1)2^nfracpartialpartialtheta_2leftint_theta_1-theta_2^y,dxrightleftint_theta_1-theta_2^theta_1+theta_2frac(y-x)^n-2theta_2^nh(x,y),dyright \&=fracn(n-1)2^nfracpartialpartialtheta_2frac(y-theta_1+theta_2)theta_2^nleftint_theta_1-theta_2^theta_1+theta_2(y-x)^n-2h(x,y),dyright endalign Use the product rule for differentiation. Leibniz rule applies as usual.
– StubbornAtom
Sep 2 at 12:35