I am wondering whether or not there exists explicit form for inverse function that is defined below.Let [$] f ( t ) [$] positive , continuous function and [$]\int_0^{+\infty} f ( t ) \,dt \: = \: +\,\infty[$] and [$]\lambda ( x )[$] is defined by equality [$] x \, = \int_0^{\lambda ( x )} f ( t ) \,dt [$]. It is obvious that inverse function [$]\lambda ( x )[$] exists. It is monotonic, positive, and continuous. I am not sure about existence of the explicit form of inverse function [$] y \, = \, F ( x ) \, = \, \int_0^{\lambda ( x )} f ( t ) \,dt [$]. I called title inverse number and then wished to edit it to change number on function but it looks like the title can not be edited?

Last edited by list1 on December 15th, 2015, 11:00 pm, edited 1 time in total.

- Cuchulainn
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QuoteOriginally posted by: list1I am wondering whether or not there exists explicit form for inverse function that is defined below.Let [$] f ( t ) [$] positive , continuous function and [$]\int_0^{+\infty} f ( t ) \,dt \: = \: +\,\infty[$] and [$]\lambda ( x )[$] is defined by equality [$] x \, = \int_0^{\lambda ( x )} f ( t ) \,dt [$]. It is obvious that inverse function [$]\lambda ( x )[$] exists. It is monotonic, positive, and continuous. I am not sure about existence of the explicit form of inverse function [$] y \, = \, F ( x ) \, = \, \int_0^{\lambda ( x )} f ( t ) \,dt [$]. This looks like a generalized quantile problem? Interesting idea.My 2 cents1. This is a nonlinear problem, so the wish to find an explicit solution is misplaced IMO. Take the special case of the Gaussian Inverse CDF. It must be solved numerically or approximately.2. "Obvious" is verboten in mathematics. The problem may have no solution or several solutions depending on the input. You could analyse it by possibly generalsing the simpler case. You would prove existence and uniques of this nonlinear (Volterra) equation using some kind of Picard iteration. Or maybe transform the limits of the integral to get it into 'standard form"?3. Do you have a concrete case to make it more clear? Can you motivate where this problem comes from please?4. And a good numerical method.5. Not sure why the integral of f must be infinity.//(A title is "write once, read many times", They made the rule to stop conniving ppeople). You can ask Admin to change the title.

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I should correct a little bit my question. Let[$] y \, = \, F (\, \lambda ( x )\, ) \, = \, \int_0^{\lambda ( x )} f ( t ) \,dt [$]Here [$] f ( t ) \, > \, 0 [$] . I am wondering whether or not exists an explicit form of inverse function [$]F^{-\,1} ( y ) [$] . In other words my question can be reduced to a simpler one[$] y \, = \, F ( x ) \, = \, \int_0^x f ( t ) \;dt [$] Whether or not exists an explicit form of inverse function [$]F^{-\,1} ( y ) [$] . The fact of existence of this function is obvious as it is monotonic.

Last edited by list1 on December 12th, 2015, 11:00 pm, edited 1 time in total.

Probably, I have got an idea. Find derivative of inverse function and integrate it locally or globally

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QuoteOriginally posted by: list1Probably, I have got an idea. Find derivative of inverse function and integrate it locally or globallyyou are not wrong

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QuoteOriginally posted by: list1I should correct a little bit my question. Let[$] y \, = \, F (\, \lambda ( x )\, ) \, = \, \int_0^{\lambda ( x )} f ( t ) \,dt [$]Here [$] f ( t ) \, > \, 0 [$] . I am wondering whether or not exists an explicit form of inverse function [$]F^{-\,1} ( y ) [$] . In other words my question can be reduced to a simpler one[$] y \, = \, F ( x ) \, = \, \int_0^x f ( t ) \;dt [$] Whether or not exists an explicit form of inverse function [$]F^{-\,1} ( y ) [$] . The fact of existence of this function is obvious as it is monotonic.No (IMO). A simple counterexample shows otherwise. See Abramowitz and Stegun 26.2.16 to 26.2.23 etc. etc.QuoteThe fact of existence of this function is obvious as it is monotonic.If you insist. But that's the second time you have said it. But I might have missed something.

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Talking about existence of inverse function I meant that if [$]f \,(\, t\, ) > 0[$] then the function[$] y \, = \, F ( x ) \, = \, \int_0^x f ( t ) \;dt [$]is a monotonic increasing function and [$] x \, \rightarrow \, y \,= \,F ( x )[$] is one-to-one correspondence and inverse function [$] x \, = \, F ^{\,-\,1} (\,y\,) [$] exists by definition.

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QuoteOriginally posted by: list1Talking about existence of inverse function I meant that if [$]f \,(\, t\, ) > 0[$] then the function[$] y \, = \, F ( x ) \, = \, \int_0^x f ( t ) \;dt [$]is a monotonic increasing function and [$] x \, \rightarrow \, y \,= \,F ( x )[$] is one-to-one correspondence and inverse function [$] x \, = \, F ^{\,-\,1} (\,y\,) [$] exists by definition.An counterexample; f(t) = t. If I did my calculus correctly I get x = + sqrt(2 y) or x = - sqrt(2 y), so the mapping is not bijective (1:1 onto). Furthermore, let y < 0 and you get pure imaginary complex numbers. There is (must be) a close connection between f(t) and y.

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QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: list1Talking about existence of inverse function I meant that if [$]f \,(\, t\, ) > 0[$] then the function[$] y \, = \, F ( x ) \, = \, \int_0^x f ( t ) \;dt [$]is a monotonic increasing function and [$] x \, \rightarrow \, y \,= \,F ( x )[$] is one-to-one correspondence and inverse function [$] x \, = \, F ^{\,-\,1} (\,y\,) [$] exists by definition.An counterexample; f(t) = t. If I did my calculus correctly I get x = + sqrt(2 y) or x = - sqrt(2 y), so the mapping is not bijective (1:1 onto). Furthermore, let y < 0 and you get pure imaginary complex numbers. There is a close connection between f(t) and y.In example [$] y = x^2[$] which corresponds to choice [$]f(\,t\,)\,=\, t[$] one should note that it is not one-to-one map [$]x\, \rightarrow\, y[$]. One value of y corresponds to two values of x and - x. In this classical we reduce domain to x [$] \ge [$] 0. This reduction of the function is monotonic and there exists inverse function x = [$] \sqrt{y}[$]. On the other hand one can use monotonic decreasing branch defined for [$] x \,\le \,0[$]. Then inverse function for this branch is x = [$]- \, \sqrt{y}[$]. Here y is always non negative which corresponds to range of the function [$]y \, = \, x\,^2[$]. This situation well known when we define inverse trigonometric functions

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Clear.And now what if f(t) = exp(-t^2) and y in (0,1). Is there an exact x? I think it's impossible.

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QuoteOriginally posted by: CuchulainnClear.And now what if f(t) = exp(-t^2) and y in (0,1). Is there an exact x? I think it's impossible.The graph of the function is symmetrical with respect to axis OY. Thus we again should separate areas [$]x \, \ge \, \, 0[$] and [$]x \, \le \, \, 0[$]. in order to figure out monotonic behaviour of the function for [$]x \, \ge \, \, 0[$] we should analyze the function [$]y\, = \, exp \, -\, x^\,{^2}[$] . Note that [$] y\, = \, -\, 2x\, exp\,-\,x\,^{\,2}\, <\, 0[$] for [$]x\, > \, 0[$]. Therefore function [$]y\, = \, exp \, -\, x^\,{^2}[$] is monotonic decreasing for [$]x\, \ge\, 0[$] and there exist inverse function for [$]y\, = \, exp \, -\, x^\,{^2}[$] for [$]x\, \ge \, 0[$]. The explicit form of it can be found by solving equation [$]y\, = \, exp \, -\, x^\,{^2}[$] for x. Hence [$] x \, = \, \sqrt {\,-\,ln\, y}[$] . Note that here [$] y \,\le \,1[$] and therefore [$]-\, ln\, y\, \ge\,0 [$]

Last edited by list1 on December 13th, 2015, 11:00 pm, edited 1 time in total.

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QuoteOriginally posted by: list1QuoteOriginally posted by: CuchulainnClear.And now what if f(t) = exp(-t^2) and y in (0,1). Is there an exact x? I think it's impossible.The graph of the function is symmetrical with respect to axis OY. Thus we again should separate areas [$]x \, \ge \, \, 0[$] and [$]x \, \le \, \, 0[$]. in order to figure out monotonic behaviour of the function for [$]x \, \ge \, \, 0[$] we should analyze the function [$]y\, = \, exp \, -\, x^\,{^2}[$] . Note that [$] y\, = \, -\, 2x\, exp\,-\,x\,^{\,2}\, <\, 0[$] for [$]x\, > \, 0[$]. Therefore function [$]y\, = \, exp \, -\, x^\,{^2}[$] is monotonic decreasing for [$]x\, \ge\, 0[$] and there exist inverse function for [$]y\, = \, exp \, -\, x^\,{^2}[$] for [$]x\, \ge \, 0[$]. The explicit form of it can be found by solving equation [$]y\, = \, exp \, -\, x^\,{^2}[$] for x. Hence [$] x \, = \, \sqrt {\,-\,ln\, y}[$] . Note that here [$] y \,\le \,1[$] and therefore [$]-\, ln\, y\, \ge\,0 [$]I miss your integral here[$] y \, = \, F ( x ) \, = \, \int_0^x f ( t ) \;dt [$]So, we still do not know if x exists, is unique and how to construct it.

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QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: list1QuoteOriginally posted by: CuchulainnClear.And now what if f(t) = exp(-t^2) and y in (0,1). Is there an exact x? I think it's impossible.The graph of the function is symmetrical with respect to axis OY. Thus we again should separate areas [$]x \, \ge \, \, 0[$] and [$]x \, \le \, \, 0[$]. in order to figure out monotonic behaviour of the function for [$]x \, \ge \, \, 0[$] we should analyze the function [$]y\, = \, exp \, -\, x^\,{^2}[$] . Note that [$] y\, = \, -\, 2x\, exp\,-\,x\,^{\,2}\, <\, 0[$] for [$]x\, > \, 0[$]. Therefore function [$]y\, = \, exp \, -\, x^\,{^2}[$] is monotonic decreasing for [$]x\, \ge\, 0[$] and there exist inverse function for [$]y\, = \, exp \, -\, x^\,{^2}[$] for [$]x\, \ge \, 0[$]. The explicit form of it can be found by solving equation [$]y\, = \, exp \, -\, x^\,{^2}[$] for x. Hence [$] x \, = \, \sqrt {\,-\,ln\, y}[$] . Note that here [$] y \,\le \,1[$] and therefore [$]-\, ln\, y\, \ge\,0 [$]I miss your integral here[$] y \, = \, F ( x ) \, = \, \int_0^x f ( t ) \;dt [$]So, we still do not know if x exists, is unique and how to construct it.1) Existence&Uniqueness: it was assumed that f ( x ) > 0. Hence the function [$]F ( x ) \, = \, \int_0^x f ( t ) \;dt [$] is monotonic increasing and continuous (differentiable). Therefore it is one-to-one map and therefore has inverse by definition. Uniqueness follows from reading one-to-one backward, ie if [$]x\, = \,g ( x ) \,\, and\,\,x \, =\,F^\,-\,1 (y) [$] two versions of inverse function they should be equal in domain of the inverse function.2) how to construct it: From equality[$]1\,=\,\frac{d}{dy} \, F ( F\,^{\,-\,1} (y)\, =\, \frac{d F ( x )}{dx} \frac{d x }{d y} [$]we arrive at[$] \frac{d x }{d y} \, =\, [\,\frac{d F ( x )}{dx}\, ]\,^{-\,1}[$]Thus if[$]y \, =\,F ( x ) \, = \, \int_0^x f ( t ) \;dt [$]Then[$]\frac{dy}{dx} \, =\,F ^{\prime}( x ) \, = \, f ( x ) ,[$][$]\frac{dx}{dy} \, =\,[\, F ^{\prime}( x )\, ] ^{\,-\,1}\, = \, f ^{\,-\,1}( x ) [$]Bearing in mind that y ( 0 ) = 0 by integrating we arrive at the formula[$]x ( y ) \, = \, \int_0^y f ^{\,-\,1}( u )\, du [$]if I did not miss something

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QuoteOriginally posted by: list1QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: list1QuoteOriginally posted by: CuchulainnClear.And now what if f(t) = exp(-t^2) and y in (0,1). Is there an exact x? I think it's impossible.The graph of the function is symmetrical with respect to axis OY. Thus we again should separate areas [$]x \, \ge \, \, 0[$] and [$]x \, \le \, \, 0[$]. in order to figure out monotonic behaviour of the function for [$]x \, \ge \, \, 0[$] we should analyze the function [$]y\, = \, exp \, -\, x^\,{^2}[$] . Note that [$] y\, = \, -\, 2x\, exp\,-\,x\,^{\,2}\, <\, 0[$] for [$]x\, > \, 0[$]. Therefore function [$]y\, = \, exp \, -\, x^\,{^2}[$] is monotonic decreasing for [$]x\, \ge\, 0[$] and there exist inverse function for [$]y\, = \, exp \, -\, x^\,{^2}[$] for [$]x\, \ge \, 0[$]. The explicit form of it can be found by solving equation [$]y\, = \, exp \, -\, x^\,{^2}[$] for x. Hence [$] x \, = \, \sqrt {\,-\,ln\, y}[$] . Note that here [$] y \,\le \,1[$] and therefore [$]-\, ln\, y\, \ge\,0 [$]I miss your integral here[$] y \, = \, F ( x ) \, = \, \int_0^x f ( t ) \;dt [$]So, we still do not know if x exists, is unique and how to construct it.1) Existence&Uniqueness: it was assumed that f ( x ) > 0. Hence the function [$]F ( x ) \, = \, \int_0^x f ( t ) \;dt [$] is monotonic increasing and continuous (differentiable). Therefore it is one-to-one map and therefore has inverse by definition. Uniqueness follows from reading one-to-one backward, ie if [$]x\, = \,g ( x ) \,\, and\,\,x \, =\,F^\,-\,1 (y) [$] two versions of inverse function they should be equal in domain of the inverse function.2) how to construct it: From equality[$]1\,=\,\frac{d}{dy} \, F ( F\,^{\,-\,1} (y)\, =\, \frac{d F ( x )}{dx} \frac{d x }{d y} [$]we arrive at[$] \frac{d x }{d y} \, =\, [\,\frac{d F ( x )}{dx}\, ]\,^{-\,1}[$]Thus if[$]y \, =\,F ( x ) \, = \, \int_0^x f ( t ) \;dt [$]Then[$]\frac{dy}{dx} \, =\,F ^{\prime}( x ) \, = \, f ( x ) ,[$][$]\frac{dx}{dy} \, =\,[\, F ^{\prime}( x )\, ] ^{\,-\,1}\, = \, f ^{\,-\,1}( x ) [$]Bearing in mind that y ( 0 ) = 0 by integrating we arrive at the formula[$]x ( y ) \, = \, \int_0^y f ^{\,-\,1}( u )\, du [$]if I did not miss somethingAnd now what if f(t) = exp(-t^2) and y in (0,1). Is there an exact x? And take f(t) = t.i.e. can you x without the following integral because the integral does not get us anywhere?[$]x ( y ) \, = \, \int_0^y f ^{\,-\,1}( u )\, du [$]BTW why is your f(t) monotonic increasing? You are excluding all the pdfs of statistics?(?)

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