The following ones are the derivation of that equation. Jef Getzinger is an ABMS board-certified doctor of Family Medicine working at Schoenherr Family Practice. )\\ \begin{align*} On the other hand, if this is not the case then the Jeffreys prior does have a special property, in that it's the only prior that can be produced by a prior generating method that is invariant under parameter transformations. Plotting Incidence function of the SIR Model. This seems to be rather an important question: if there is some other functional $M'$ that is also invariant and which gives a different prior for the parameter of a binomial distribution then there doesn't seem to be anything that picks out the Jeffreys distribution for a binomial trial as particularly special. What is invariant is the volume density $|p_{L_{\theta}}(\theta) dV_{\theta}|$ where $V_\theta$ is the volume form in coordinates $\theta_1, \theta_2, \dots \theta_n$ and $L_\theta$ is the likelihood parametrized by $\theta$. Webstatistics - In what sense is the Jeffreys prior invariant? This is ensured by the use of Jeffrey's prior which is completely scale and location-invariant. Making statements based on opinion; back them up with references or personal experience. Thanks for contributing an answer to Cross Validated! $$ $$ Level of grammatical correctness of native German speakers. where $\theta$ is the parameterisation given by $p_1 = \theta$, $p_2 = 1-\theta$. $$ = \frac{d^2 \log p(y\mid\theta(\phi))}{d \theta^2} \left|\frac{d\theta}{d\phi} \right|^2 What I would like is to understand the sense in which this is invariant with respect to a coordinate transformation $\theta \to \varphi(\theta)$. Though his prior was perfectly alright, the reasoning used to arrive at it was at fault. On obtaining invariant prior distributions - ScienceDirect are the constants of proportionality the same in the two equations above, or different? How to combine uparrow and sim in Plain TeX? Is the product of two equidistributed power series equidistributed? It only takes a minute to sign up. That seems to be an open-ended question full of debates. Finally, whatever the thing that's invariant is, it must surely depend in some way on the likelihood function! \end{align*} al. Note that if I start with a uniform prior and then transform the parameters, I will in general end up with something that's not a uniform prior over the new parameters. rule} \\ The link given by the OP contains the problem statement in good detail. Steve Kaufman says to mean don't study. It says that there is some prior information which is why this transformed pdf is not flat. This shows that the invariant prior is very non-unique as there are many other ways to achieve the cancellation. & = & p (\varphi (\theta)) About Jeffrey Clothier $$ Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. Jeffreys method is also an equivariant method for constructing prior distributions, and the first "non-trivial" method mentioned here: We first fix a sigma-finite measure $\nu$ on $(\Omega,\mathcal A)$ and then define $X$ to be the set of all families of probability distributions $(\mathsf P_\theta)_{\theta\in\Theta}$ such that. Jeffrey E. Clothier has been practicing law in the State of Michigan for the past twenty-four years and is the criminal defense attorney that Genesee County residents trust most. He understands what it is like to be from FLINT. &= \frac{d}{d\phi} \left( \frac{d \log p(y|\theta(\phi))}{d \theta} \frac{d\theta}{d\phi} \right) \tag{chain rule}\\ The Jeffreys prior is proportional to the square root of Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. the function $M\{ f(x\mid \theta )\}$ for some particular likelihood function $f(x \mid \theta)$) and trying to see that it has some kind of invariance property. Having come back to this question and thought about it a bit more, I believe I have finally worked out how to formally express the sense of "invariance" that applies to Jeffreys' priors, as well as the logical issue that prevented me from seeing it before. We denote by $\mathrm M^1(\Omega,\mathcal A)^\Theta$ the space of all families $(\mathrm P_\theta)_{\theta\in\Theta}$ where $\mathsf P_\theta\in\mathrm M^1(\Omega,\mathcal A)$. What I'm looking for is something. But let us say they were using some log scaled parameters instead of ours. To me the term "invariant" would seem to imply something along the lines of Do the calculations with $\pi$ in there to see that point. SELECTIVE SEROTONIN \end{eqnarray*} (+1) Your answer is perhaps one of the cleareast I've found so far, together with the lecture you mention. For a measure $\mu$ on a measurable space $X_1$ and a measurable map $h:X_1\to X_2$ for a measurable space $X_2$, we denote by $h_\#\mu$ the pushforward measure defined by $h_\#\mu(A)=\mu(h^{-1}(A))$ for all measurable $A\subset X_2$. How can select application menu options by fuzzy search using my keyboard only? The timescale invariance problem is also mentioned there.). 1 Jereys Priors - University of California, Berkeley I For a single parameter and data having joint density f(x|), the Jaynes. The lack of evidence to reject the H0 is OK in the case of my research - how to 'defend' this in the discussion of a scientific paper? It would therefore seem rather valuable to find a proof that Jeffrey's prior construction method is unique in having this invariance principle, or an explicit counterexample showing that it is not. The presentation in Wikipedia is confusing, because. The invariance of $|p dV|$ is the definition of "invariance of prior". Harold Jeffreyss default Bayes factor hypothesis tests: Explanation If we take $\theta(\phi)$ as a function of $\phi$, then, $$ WebJereys Prior I Jereys (1961) developed a class of priors that were invariant under transformation. p(\theta)\propto\sqrt{I(\theta)} This choice is not at all useful or interesting. Whatever priors they use must be completely uninformative about the scaling of time between the events. Now, according to this Wikipedia page, the derivative the inverse gives: $$p_{\phi}(\phi) = p_{\theta}( h^{-1} (\phi)) \Bigg| h'(h^{-1}(\phi)) \Bigg|^{-1} $$, We will write this in another way to make the next step clearer. Which part of the question is not dealt with. . The third line applies the relationship between the information matrices. : () = 1 c c (12) for all c > 0. It is trivial to define an. If so I don't think that can be the thing that's invariant. WebJeffrey Clothier immediately entered law school and graduated from Michigan State University College of Law in 1995. What you need for Bayesian statistics (resp., likelihood-based methods) is the ability to integrate against a prior (likelihood), so really $p(x) dx$ is the object of interest. This is in particular true for Jeffreys method: If $\mathsf P_\theta$ doesn't depend on $\theta$, then neither does $f_\theta$ and therefore the Fisher information is always equal to $0$. \begin{aligned} To read the Wikipedia argument as a chain of equalities of unsigned volume forms, multiply every line by $|d\varphi|$, and use absolute value of all determinants, not the usual signed determinant. \end{aligned}, using the substitution formula from Wikipedia with $\phi = h(\theta)$, \begin{aligned} He attended the University of Michigan in Ann Arbor and graduated with his Bachelors Degree in 1991. When we drop the bars, we can cancel $h'^{-1}$ and $h'$, giving, $$ \int_{h(a)}^{h(b)} p_{\phi}(\phi) d\phi = \int_{a}^{b}p_{\theta}(\theta) d\theta$$, $$ P(a \le \theta \le b) = P(h(a) \le \phi \le h(b))$$, Now, we need to show that a prior chosen as the square root of the Fisher Information admits this property. One thing I would like to note that if you look at the proof for this invariance, it is only important that we have the variance of a (differentiable) function of the density function of the sampling distribution. Is DAC used as stand-alone IC in a circuit? As I explained earlier in the comments, it is essential to understand how jacobians work (or differential forms). I made some edits, I think it explains clearly now why the Wikipedia link is not a real answer. $\mathsf P_\theta\in\mathrm M^1(\Omega,\mathcal A)$, $X\subset \mathrm M^1(\Omega,\mathcal A)^\Theta$, $(\mathsf P_\theta)_{\theta\in\Theta}\in X\implies (\mathsf P_{h(\theta)})_{\theta\in\Theta}\in X$, \begin{align*}\rho: X&\to \mathrm M^\sigma(\Theta, \mathcal B(\Theta))\\ (\mathsf P_\theta)_{\theta\in\Theta}&\mapsto\rho[(\mathsf P_\theta)_{\theta\in\Theta}]\end{align*}, $$h_\# \rho[(\mathsf P_{h(\theta)})_{\theta\in\Theta}] = \rho[(\mathsf P_\theta)_{\theta\in\Theta}]$$, $f_\theta=\frac{\mathrm d\mathsf P_\theta}{\mathrm d\nu}$, $\frac{\partial^2}{\partial\theta^2}\ln f_\theta\in L^1(\Omega,\mathcal A, \mathsf P_\theta)$, $\mathrm M^\sigma(\Theta, \mathcal B(\Theta))$, $\rho[(\mathsf P_\theta)_{\theta\in\Theta}]$, $\mathsf P_{\theta}=\mathsf P_{\vartheta}$, $\rho[(\mathsf P_\theta)_{\theta\in\Theta}]=0$, $p\in\mathrm M^\sigma(\Theta,\mathcal B(\Theta))$, $$\rho:X\to\mathrm M^{\sigma}(\Theta,\mathcal B(\Theta))$$, $$\rho[(\mathsf P_\theta)_{\theta\in\Theta}] =\begin{cases}h^{-1}_\# p, &\text{ if }(\mathsf P_{\theta})_{\theta\in\Theta}=(\mathrm Q_{h(\theta)})_{\theta\in\Theta} \text{ for some bijective }h\in C^\infty(\Theta;\Theta)\\0,&\text{otherwise}. The final line applies the definition of Jeffreys prior on $\varphi{(\theta)}$. Let $p_{\theta}(\theta)$ be the prior on $\theta$. Those equations (quoted from Wikipedia) omit the Jacobian because they refer to the case of a binomial trial, where there is only one variable and the Jacobian of $I$ is just $I$. Why not say ? But nonetheless, we can make sure that our priors are at least uninformative in some sense. zyx's answer is excellent but it uses differential forms. Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. The motivation for suggesting this By the transformation of variables formula, $$p_{\phi}(\phi) = p_{\theta}( h^{-1} (\phi)) \Bigg| \frac{d}{d\phi} h^{-1}(\phi) \Bigg| $$. Now how do we define a completely "uninformative" prior? Understanding why the Uniform distribution does not make a good prior. I like to understand things by approaching the simplest example first, so I'm interested in the case of a binomial trial, i.e. The clearest answer I have found (ie, the most blunt "definition" of invariance) was a comment in this Cross-Validated thread, which I combined with the discussion in "Bayesian Data Analysis" by Gelman et. What's the meaning of "Making demands on someone" in the following context? p (\varphi (\theta) |y) & = & \frac{1}{| \varphi' (\theta) |} p (\theta In the above case, the prior is telling us that "I don't want to give one value p$_1$ more preference than another value p$_2$" and it continues to say the same even on transforming the prior. Ask Question Asked 10 years, 10 months ago Modified 10 months ago Viewed 8k times 19 I've been trying to understand the motivation I'm fairly certain it's a logical point that I'm missing, rather than something to do with the formal details of the mathematics. A trivial choice is $X=\mathrm M^1(\Omega,\mathcal A)$ and $\rho=0$, because the measure assigning $0$ to all measurable sets is invariant under push-forward by any map. Asking for help, clarification, or responding to other answers. What Jeffreys provides is a prior construction method $M$ which has this property. This means if we rescale our variable, the prior will not I've been trying to understand the motivation for the use of the Jeffreys prior in Bayesian statistics. While his office is located in Flint, Michigan his reputation for obtaining outstanding results has led him to almost every county in the State of Michigan. Why do people generally discard the upper portion of leeks? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Computationally it is expressed by Jacobians but only the power-of-$A$ dependences matter and having those cancel out on multiplication. I agree with William Huber. University of Michigan $$\frac{\mathrm d\rho[(\mathsf P_\theta)_{\theta\in\Theta}]}{\mathrm d\lambda} $$. rule} \\ So they will use the $\lambda^{-1}d\lambda$ prior, the Jeffrey's prior (because it is the only general solution in the one-parameter case for scale-invariance). This explanation is of course restricted to the unidimensional case. M\{ f(x\mid h(\theta)) \} = M\{ f(x \mid \theta) \}\circ h, The comments on this question make no sense if you don't already know that @did's comment was originally an answer, which was deleted by a moderator and made into a comment, and that the following two comments were originally, $$ $$h_\# \rho[(\mathsf P_{h(\theta)})_{\theta\in\Theta}] = \rho[(\mathsf P_\theta)_{\theta\in\Theta}]$$ for all bijective $h\in C^\infty(\Theta;\Theta)$. (I will let you verify this by deriving the information from the likelihood. & = & \frac{1}{| \varphi' (\theta) |} \sqrt{I (\theta)} \\ Perhaps I can answer this myself now, but if you'd like to post a proper answer detailing it then I'd be happy to award you the bounty. and deriving & \propto & p (\varphi (\theta)) p (y| \theta) To answer your question, the missing bit is the bit where I said "I'd like to understand this sense [of invariance] form of a functional equation similar to (ii), so that I can see how it's satisfied by (i)." How to cut team building from retrospective meetings? the case where the support is $\{1,2\}$. When in {country}, do as the {countrians} do, Any difference between: "I am so excited." In fact the desired invariance is a property of $M$ itself, rather than of the priors it generates. $$ I still think that your problem is with jacobians and the fact that the formula (ii) is correct for the special case I does not make correct in general. How to combine uparrow and sim in Plain TeX? The property of "Invariance" does not necessarily mean that the prior distribution is Invariant under "any" transformation. RESPIRATORY AND CNS STIMULANTS. Invariance under parameter transformation with the My question is simply how Gelman reasoned from the first line of the equation to the second. for any smooth function $\varphi(\theta)$. but mostly it's because it's really unclear exactly what's being sought, which is why I wanted to express it as a functional equation in the first place. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Here $| \varphi' (\theta) |$ is the inverse of the jacobian of the transformation. \end{align*}, $\frac{d \log p(y|\theta(\phi))}{d \theta}$, Understanding the Proof for why Jeffreys' prior is invariant, Moderation strike: Results of negotiations, Our Design Vision for Stack Overflow and the Stack Exchange network, Parametrisation invariance/covariance of the Jeffreys prior, Compute $\pi(H_0|x)$ with Jeffreys prior for a family $N(\theta,1)$, Obtaining Jeffreys prior by taking the limit of a particular prior density on $(\mu, \Sigma)$, Understanding definition of informative and uninformative prior distribution, Significance of parameterisation invariance of Jeffreys prior, Jeffreys' prior invariance under reparametrization. My own party belittles me as a player, should I leave? However, I can't see how to express this invariance property in the form of a functional equation similar to $(ii)$, which is what I'm looking for as an answer to this question. Rufus settings default settings confusing. Is the following parametrizations identifiable? Thanks for the hints. If $h$ is increasing, then $h'$ is positive and we don't need the absolute value. To make sure that we are on the same page, let us take the example of the "Principle of Indifference" used in the problem of Birth rate analysis given by Laplace. p (\varphi (\theta) ) & = & \frac{1}{| \varphi' (\theta) |} p (\theta |y)\\ rev2023.8.22.43591. For the distribution $f_\theta (x) = \theta x^{\theta-1}$, what is the sufficient statistic corresponding to the Monotone Likelihood Ratio? AND "I am just so excited.". Hi! Understanding the Proof for why Jeffreys' prior is invariant What happens if you connect the same phase AC (from a generator) to both sides of an electrical panel? & \propto & \frac{1}{| \varphi' (\theta) |} p (\theta) p (y| \theta)\\ Say if the aliens used the same principle, they would definitely arrive at a different answer than ours. In (i), it is $\pi$. Then $\frac{d \log p(y|\theta(\phi))}{d \theta}$ (the "score function") is $0$ on average. &= \left(\frac{d^2 \log p(y|\theta(\phi))}{d \theta d\phi}\right)\left( \frac{d\theta}{d\phi}\right) + \left(\frac{d \log p(y|\theta(\phi))}{d \theta}\right) \left( \frac{d^2\theta}{d\phi^2}\right) \tag{prod. \begin{align*}\rho: X&\to \mathrm M^\sigma(\Theta, \mathcal B(\Theta))\\ (\mathsf P_\theta)_{\theta\in\Theta}&\mapsto\rho[(\mathsf P_\theta)_{\theta\in\Theta}]\end{align*} satisfying the equivariance property Invariance Property - University of South Carolina The preference for Jeffreys form of invariant prior is based on other considerations. The use of these "Uninformative priors" is completely problem-dependent and not a general method of forming priors. Having come back to this question and thought about it a bit more, I believe I have finally worked out how to formally express the sense of "invari This happens through the relationship $ \sqrt{I (\theta)} = \sqrt{I (\varphi (\theta))} | \varphi' (\theta) | $. In particular, I remember him arguing in favour of an "uninformative" prior for a binomial distribution that's an improper prior proportional to $1/(p(1-p))$. I was reviewing the section of Andrew Gelman's "Bayesian Data Analysis" on uninformative priors, and came across this explanation for why Jeffreys' prior is invariant to parameterization. WebANOREXIGENIC AGENTS, MISCELLANEOUS. - Mathematics Stack Exchange In what sense is the Jeffreys prior invariant? Fix now any "privileged" family of distributions $(\mathrm Q_\theta)_{\theta\in\Theta}$ (in the language of Bayesians this would be a "privileged parametrization") and the "privileged" prior $p\in\mathrm M^\sigma(\Theta,\mathcal B(\Theta))$ that you want to obtain. This paper considers a generalization of the connection between Jeffreys prior and the Kullback-Leibler divergence as a procedure for generating a wide class of Now, for the prior. &= \frac{d}{d\phi} \left( \frac{d \log p(y|\theta(\phi))}{d \theta} \frac{d\theta}{d\phi} \right) \tag{chain rule}\\ It is natural to ask for something local on the parameter space, so the invariant prior will be built from a finite number of derivatives of the likelihood evaluated at $\theta$. Because changes of coordinate alter $dV$, an invariant prior has to depend on more than $p(\theta)$. re the second comment, the distinction is between functions and differential forms. Invariant Properties of Probability Distributions? Your answer, @N. Virgo, has greatly improved my understanding of what the Jeffreys prior is and in what sense the word "invariant" is used. To use any other prior than this will have the consequence that a change in the time scale will lead to a change in the form of the prior, which would imply a different state of prior knowledge; but if we are completely ignorant of the time scale, then all time scales should appear equivalent. This is of course undesired (we want to generate any desired prior, not just $0$) and it doesn't seem very useful in practical problems to have multiple distinct parameters with the same probability distribution assigned to them. During law school, Clothier was the Judicial Law Clerk Note that by Definition of $X$, $\rho$ is well-defined, since $h$ in the first case is unique if it exists. for any (smooth, differentiable) function $\varphi$ -- but it's easy enough to see that this is not satisfied by the distribution $(i)$ above (and indeed, I doubt there can be any density function that does satisfy this kind of invariance for any transformation). This is simply because $0$ is the only $\sigma$-finite measure that remains unchanged when being pushforwarded by any smooth bijective map (actually I should prove this statement but I believe this is true). Your answer is really clear, but I think is not quite there yet. This proof is clearly laid out in these lecture notes. Is there a way to smoothly increase the density of points in a volume using the 'Distribute points in volume' node? In other words, on transforming the prior to a log-odds scale, the prior still says "See, I still consider no value of p1 to be preferable over another p2" and that is why the log-odds transform is not going to be flat. In the univariate case, does the expression in your first sentence reduce to $p(\theta) d\theta$? & \propto & \frac{1}{| \varphi' (\theta) |} \sqrt{I (\theta)} p (y| Amazing answer. You can see that the use of Jeffreys prior was essential for $\frac{1}{| \varphi' (\theta) |}$ to cancel out. =\sqrt{-\int_{\Omega} \frac{\partial^2}{\partial\theta^2}\ln f_\theta(x)\,\mathrm d\mathsf P_\theta(x)}.$$. Why do Airbus A220s manufactured in Mobile, AL have Canadian test registrations? Definition. On applying the (dv/v) rule on the positive semi-infinite interval, we get the 1/p(1-p) dependence which Jeffreys accepts only for the semi-infinite interval. Use MathJax to format equations. Here the argument used by Laplace was that he saw no difference in considering any value p$_1$ over p$_2$ for the probability of the birth of a girl. I'm not sure I understand what you mean in your other comment, though - could you spell your counterexample out in more detail? WebAttorney/Advisor. We then define Jeffreys prior (not-normalized) $\rho[(\mathsf P_\theta)_{\theta\in\Theta}]$ as the measure over $\Theta$ whose density with respect to the Lebesgue measure $\lambda$ is the square root of the Fisher information, i.e. But unfortunately, if their clocks were running at different speeds (say, t' = qt) then their results will definitely be conflicting if they did not consider this difference in time-scales. WebUnderstanding the Proof for why Jeffreys' prior is invariant Ask Question Asked 6 years, 6 months ago Modified 4 years, 7 months ago Viewed 2k times 5 I was reviewing the \int_{\theta_1}^{\theta_2} \rho(\theta) d \theta = So there must be some other sense intended by "invariant" in this context. The problem here is about the apparent "Principle of Indifference" considered by Laplace. This should be posted as a comment rather than an answer, since it is not an answer. The only difference is that the second line applies Bayes rule. By clicking Post Your Answer, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct. Also, it would help me a lot if you could expand on the distinction you make between "densities $p(x) dx$" and "the. the equations are between densities $p(x) dx$, but written as though for the density functions $p()$ that define the priors. The property of "Invariance" does not necessarily mean that the prior distribution is Invariant under " any " transformation. To make sure that we \int_{\varphi(\theta_1)}^{\varphi(\theta_2)} \rho(\varphi(\theta)) d \varphi \qquad\qquad(ii) I will add some clarifications to my answer regarding your question about the invariance depending on the likelihood to my answer. My answer is written as it is because yes, I believe. So using the chain rule, is this the correct reasoning? For the binomial regression model, Jeffreys's Let me know if you are stuck somewhere. Throughout this answer we fix a measurable space $(\Omega,\mathcal A)$, as well as a parameter space $\Theta\subset\mathbb R$, that, for simplicity, I assume to be an interval (the arguments here should also work for more general parameter spaces and the reader is invited to repeat them in a more general setting). Formula (ii) is not correct in either the special case or in general. & = & \sqrt{I (\varphi (\theta))} \\ Thanks both, I got myself confused because it was to the right of the conditional. Edit: The dependence on the likelihood is essential for the invariance to hold, because the information is a property of the likelihood and because the object of interest is ultimately the posterior. The goal of this answer is to provide a rigorous mathematical framework of the "invariance" property and to show that the prior obtained by Jeffreys method is not unique. The clearest answer I have found (ie, the most blunt "definition" of invariance) was a comment in this Cross-Validated thread , which I combined w $\rho$ satisfies the equivariance property by construction. WebThe Jereys Prior Uniform priors and invariance Recall that in his female birth rate analysis, Laplace used a uniform prior on the birth rate p2[0;1]. The key point is we want the following: If $\phi = h(\theta)$ for a monotone transformation $h$, then: $$P(a \le \theta \le b) = P(h(a) \le \phi \le h(b))$$. Perhaps I can, but it seems not at all trivial to me. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Henceforth I will use the word equivariant instead of invariant since it is a better fit in my opinion. \end{cases}$$. 1. The first line is only applying the formula for the jacobian when transforming between posteriors. Properties and Implementation of Jeffreys's Prior in Binomial But whatever we estimate from our priors and the data must necessarily lead to the same result. Why do Airbus A220s manufactured in Mobile, AL have Canadian test registrations? $$ rev2023.8.22.43591. First we show a probability density for which this is satisfied. What is the intuition or motivation about Translation-invariant priors? (Note that these equations omit taking the Jacobian of $I$ because they refer to a single-variable case.) United States Federal Court in the Eastern District of Michigan, 2000, United States Federal Court in the Western District of Michigan, 2007, TOP 10 Criminal Law Attorney for Michigan by American Jurist Institute, TOP 100 Trial Lawyer recognized by the National Trial Lawyers Association, 10 Best Attorney Client Satisfaction American Institute Criminal Attorneys, TOP 100 OWI Attorney recognized by National Advocacy for DUI Defense, Distinguished High Legal Ability and Ethical Standards-Martindale Hubbell. While other attorneys practice several areas of the law, Jeffrey E. Clothier does one thing, SOLVES YOUR PROBLEMS. Then, start with some simple examples of some monotonic transformations in order to see the invariance. $$ Determinants appear because there is a factor of $\det J$ to be killed from the change in $dV$, and because we will want the changes of the local quantities to multiply and cancel each other as is the case in Jeffreys prior, which practically requires a reduction to one dimension where the coordinate change can act on each factor by multiplication by a single number. &= \left(\frac{d^2 \log p(y|\theta(\phi))}{d \theta^2 }\right)\left( \frac{d\theta}{d\phi}\right)^2 + \left(\frac{d \log p(y|\theta(\phi))}{d \theta}\right) \left( \frac{d^2\theta}{d\phi^2}\right) \tag{chain rule} I suggest to start with $\varphi(\theta)=2\theta$ and $\varphi(\theta)=1-\theta$. \frac{d^2\log p(y | \phi)}{d\phi^2} \frac{d^2\log p(y \mid \phi)}{d\phi^2} However, regardless what likelihood you use, the invariance will hold through. \end{eqnarray*} In this case the Jeffreys prior is given by \begin{eqnarray*} It's not important that this function is the logarithm of this pdf, so indeed there are infinitely many of these kinds of methods. Also, to answer your question, the constants of integration do not matter here. The prior does not lose the information. ANOREXIGENICS;RESPIRATORY,CNS STIMULANTS. &= \int_{a}^{b} p_{\theta}(\theta) \Bigg| h'(\theta) \Bigg|^{-1} h'(\theta) d\theta, The problem is not that I don't understand those equations. I would like to understand this sense in the form of a functional equation similar to $(ii)$, so that I can see how it's satisfied by $(i)$. P(h(a)\le \phi \le h(b)) &= \int_{h(a)}^{h(b)} p_{\phi}(\phi) d\phi\\

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