Inverse derivative
Note: $b(cdot)$ is a function, $G(cdot)$ is a cdf.
Given that $b(v) = beta$ and $b(v) = cfrac{1}{G(v)} int_{0}^{v}y_1 g(y_1)dy_1$ where $G(y_1) = Prob(Y_1 leq y_1) = F^{n-1}(y_1)$ where F is a CDF.
how do I find the derivative of the function:
$(v-beta)G(b^{-1}(beta))$ with respect to $beta$?
I know the rule is that if I have $y = f(x)$ then $x = f^{-1}(y)$ and
$$cfrac{df^{-1}(y)}{dy} = cfrac{1}{f'(x)}$$ but I am struggling to apply this my scenario.
the output should be:
$$-G(b^{-1}(beta)) + (v-b(b^{-1}(beta))g(b^{-1}(beta))cfrac{1}{b'(b^{-1}(beta))} =0$$
It is obvious for the first term, and by inspection I can even see that:
$b'(b^{-1}(beta)) = f'(x)$ in the above formula.
But as obvious as it may seem, I am having trouble seeing where exactly the $$(v-b(b^{-1}(beta))g(b^{-1}(beta))$$ is coming from, although I do note that $$G'(b^{-1}(beta)) = g(b^{-1}(beta))$$.
Any help in piecing this together is very much appreciated.
calculus probability-distributions inverse-function
asked Nov 28 '18 at 2:59
elbarto
1,538824
add a comment |
Note: $b(cdot)$ is a function, $G(cdot)$ is a cdf.
Given that $b(v) = beta$ and $b(v) = cfrac{1}{G(v)} int_{0}^{v}y_1 g(y_1)dy_1$ where $G(y_1) = Prob(Y_1 leq y_1) = F^{n-1}(y_1)$ where F is a CDF.
how do I find the derivative of the function:
$(v-beta)G(b^{-1}(beta))$ with respect to $beta$?
I know the rule is that if I have $y = f(x)$ then $x = f^{-1}(y)$ and
$$cfrac{df^{-1}(y)}{dy} = cfrac{1}{f'(x)}$$ but I am struggling to apply this my scenario.
the output should be:
$$-G(b^{-1}(beta)) + (v-b(b^{-1}(beta))g(b^{-1}(beta))cfrac{1}{b'(b^{-1}(beta))} =0$$
It is obvious for the first term, and by inspection I can even see that:
$b'(b^{-1}(beta)) = f'(x)$ in the above formula.
But as obvious as it may seem, I am having trouble seeing where exactly the $$(v-b(b^{-1}(beta))g(b^{-1}(beta))$$ is coming from, although I do note that $$G'(b^{-1}(beta)) = g(b^{-1}(beta))$$.
Any help in piecing this together is very much appreciated.
calculus probability-distributions inverse-function
asked Nov 28 '18 at 2:59
elbarto
1,538824
Chain rule + derivative of the inverse function + $b(b^{-1}(beta))=beta$ ($b$ is monotone).
– d.k.o.
Nov 28 '18 at 5:33
add a comment |
Note: $b(cdot)$ is a function, $G(cdot)$ is a cdf.
Given that $b(v) = beta$ and $b(v) = cfrac{1}{G(v)} int_{0}^{v}y_1 g(y_1)dy_1$ where $G(y_1) = Prob(Y_1 leq y_1) = F^{n-1}(y_1)$ where F is a CDF.
how do I find the derivative of the function:
$(v-beta)G(b^{-1}(beta))$ with respect to $beta$?
I know the rule is that if I have $y = f(x)$ then $x = f^{-1}(y)$ and
$$cfrac{df^{-1}(y)}{dy} = cfrac{1}{f'(x)}$$ but I am struggling to apply this my scenario.
the output should be:
$$-G(b^{-1}(beta)) + (v-b(b^{-1}(beta))g(b^{-1}(beta))cfrac{1}{b'(b^{-1}(beta))} =0$$
It is obvious for the first term, and by inspection I can even see that:
$b'(b^{-1}(beta)) = f'(x)$ in the above formula.
But as obvious as it may seem, I am having trouble seeing where exactly the $$(v-b(b^{-1}(beta))g(b^{-1}(beta))$$ is coming from, although I do note that $$G'(b^{-1}(beta)) = g(b^{-1}(beta))$$.
Any help in piecing this together is very much appreciated.
calculus probability-distributions inverse-function
asked Nov 28 '18 at 2:59
elbarto
1,538824
Note: $b(cdot)$ is a function, $G(cdot)$ is a cdf.
Given that $b(v) = beta$ and $b(v) = cfrac{1}{G(v)} int_{0}^{v}y_1 g(y_1)dy_1$ where $G(y_1) = Prob(Y_1 leq y_1) = F^{n-1}(y_1)$ where F is a CDF.
how do I find the derivative of the function:
$(v-beta)G(b^{-1}(beta))$ with respect to $beta$?
I know the rule is that if I have $y = f(x)$ then $x = f^{-1}(y)$ and
$$cfrac{df^{-1}(y)}{dy} = cfrac{1}{f'(x)}$$ but I am struggling to apply this my scenario.
the output should be:
$$-G(b^{-1}(beta)) + (v-b(b^{-1}(beta))g(b^{-1}(beta))cfrac{1}{b'(b^{-1}(beta))} =0$$
It is obvious for the first term, and by inspection I can even see that:
$b'(b^{-1}(beta)) = f'(x)$ in the above formula.
But as obvious as it may seem, I am having trouble seeing where exactly the $$(v-b(b^{-1}(beta))g(b^{-1}(beta))$$ is coming from, although I do note that $$G'(b^{-1}(beta)) = g(b^{-1}(beta))$$.
Any help in piecing this together is very much appreciated.
calculus probability-distributions inverse-function
calculus probability-distributions inverse-function
asked Nov 28 '18 at 2:59
elbarto
1,538824
asked Nov 28 '18 at 2:59
elbarto
1,538824
asked Nov 28 '18 at 2:59
elbarto
1,538824
asked Nov 28 '18 at 2:59
elbarto
1,538824
asked Nov 28 '18 at 2:59
elbarto
1,538824
1,538824
Chain rule + derivative of the inverse function + $b(b^{-1}(beta))=beta$ ($b$ is monotone).
– d.k.o.
Nov 28 '18 at 5:33
add a comment |
Chain rule + derivative of the inverse function + $b(b^{-1}(beta))=beta$ ($b$ is monotone).
– d.k.o.
Nov 28 '18 at 5:33
Chain rule + derivative of the inverse function + $b(b^{-1}(beta))=beta$ ($b$ is monotone).
– d.k.o.
Nov 28 '18 at 5:33
Chain rule + derivative of the inverse function + $b(b^{-1}(beta))=beta$ ($b$ is monotone).
– d.k.o.
Nov 28 '18 at 5:33
add a comment |
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Chain rule + derivative of the inverse function + $b(b^{-1}(beta))=beta$ ($b$ is monotone).
– d.k.o.
Nov 28 '18 at 5:33