Warum enthält ein 95% -Konfidenzintervall aus der Perspektive der Bayes'schen Wahrscheinlichkeit nicht den wahren Parameter mit einer Wahrscheinlichkeit von 95%?

14

Auf der Wikipedia-Seite zu Konfidenzintervallen :

... Wenn Konfidenzintervalle aus mehreren separaten Datenanalysen von wiederholten (und möglicherweise unterschiedlichen) Experimenten erstellt werden, entspricht der Anteil solcher Intervalle, die den wahren Wert des Parameters enthalten, dem Konfidenzniveau ...

Und von derselben Seite:

Ein Konfidenzintervall sagt nicht voraus, dass der wahre Wert des Parameters mit einer bestimmten Wahrscheinlichkeit in dem Konfidenzintervall liegt, wenn die tatsächlich erhaltenen Daten vorliegen.

Wenn ich es richtig verstanden habe, wird diese letzte Aussage unter Berücksichtigung der frequentistischen Interpretation der Wahrscheinlichkeit getroffen. Allerdings aus einer Bayes - Wahrscheinlichkeit Perspektive, warum nicht ein 95% Konfidenzintervall die wahren Parameter mit 95% Wahrscheinlichkeit enthält? Und wenn nicht, was ist dann falsch an der folgenden Argumentation?

Wenn ich weiß, dass ein Prozess in 95% der Fälle eine korrekte Antwort liefert, beträgt die Wahrscheinlichkeit, dass die nächste Antwort korrekt ist, 0,95 (vorausgesetzt, ich habe keine zusätzlichen Informationen zum Prozess). Wenn mir jemand ein Konfidenzintervall anzeigt, das von einem Prozess erstellt wird, der in 95% der Fälle den wahren Parameter enthält, sollte ich dann nicht zu Recht sagen, dass er den wahren Parameter mit einer Wahrscheinlichkeit von 0,95 enthält, wenn ich weiß, was ich weiß?

Diese Frage ist ähnlich, aber nicht gleichbedeutend mit: Warum impliziert ein 95% CI keine 95% ige Chance, den Mittelwert zu enthalten? Die Antworten auf diese Frage konzentrierten sich darauf, warum ein 95% -KI keine 95% ige Chance impliziert, den Mittelwert aus einer häufigeren Perspektive zu enthalten. Meine Frage ist dieselbe, aber aus der Perspektive der Bayes'schen Wahrscheinlichkeit.

bayesian confidence-interval Rasmus Bååth
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Ein Weg, dies zu sehen, ist, dass 95% CI ein "langfristiger Durchschnitt" ist. Nun gibt es viele Möglichkeiten, Ihre "kurzfristigen" Fälle so aufzuteilen, dass eine ziemlich willkürliche Abdeckung erzielt wird - aber wenn der Durchschnitt herausgerechnet wird, erhalten Sie insgesamt 95%. Ein andere, mehr abstrakte Weise wird erzeugen

x_{i} \sim B e r n o u l l i (p_{i})

$x_i\sim Bernoulli (p_i)$ für

i = 1, 2, \dots

$i=1, 2,\dots$ , so dass

\sum_{i = 1}^{\infty} p_{i} = 0.95

$\sum_{i=1}^{\infty} p_i=0.95$ . Es gibt unendlich viele Möglichkeiten, dies zu tun. Hier

x_{i}

$x_i$ gibt an, ob das mit dem i-ten Datensatz erstellte CI den Parameter enthielt, und

p_{i}

$p_i$ ist die Überdeckungswahrscheinlichkeit für diesen Fall.

Wahrscheinlichkeitislogic

11

Update : Nach ein paar Jahren habe ich eine präzisere Behandlung des im Wesentlichen gleichen Materials als Antwort auf eine ähnliche Frage verfasst.

So erstellen Sie eine Vertrauensregion

Beginnen wir mit einer allgemeinen Methode zum Aufbau von Vertrauensbereichen. Sie kann auf einen einzelnen Parameter angewendet werden, um ein Konfidenzintervall oder eine Reihe von Intervallen zu erhalten. und es kann auf zwei oder mehr Parameter angewendet werden, um Bereiche mit höherem Maßvertrauen zu erhalten.

Wir behaupten, dass die beobachteten Statistiken $D$ aus einer Verteilung mit Parametern $\theta$ , nämlich der Stichprobenverteilung $s(d|\theta)$ über mögliche Statistiken $d$ , und suchen einen Vertrauensbereich für $\theta$ in der Menge möglicher Werte $\Theta$ . Definieren eines HDR (Highest Density Region): Der $h$ HDR eines PDF ist die kleinste Teilmenge seiner Domäne, die die Wahrscheinlichkeit $h$ unterstützt . Bezeichne das $h$ HDR von $s(d|\psi)$ als $H_\psi$ für jedes $\psi \in \Theta$ . Dann wird der $h$ Vertrauensbereich für $\theta$ , Daten gegeben $D$ ist,der Satz $C_D = \{ \phi : D \in H_\phi \}$ . Ein typischer Wert von $h$ wäre 0,95.

Eine häufige Interpretation

Aus der vorstehenden Definition eines Vertrauensbereich folgt

d \in H_{ψ} ⟷ ψ \in C_{d}

$d \in H_\psi \longleftrightarrow \psi \in C_d$ mit

C_{d} = {ϕ : d \in H_{ϕ}}

$C_d = \{ \phi : d \in H_\phi \}$ . Nun stellen Sie eine große Menge von ( imaginären ) Beobachtungen

{D_{i}}

$\{D_i\}$ , genommen unter ähnlichen Umständen zu

D

$D$ . dh sie sind Abtastwerte von

s (d | θ)

$s(d|\theta)$ . Da

H_{θ}

$H_\theta$ Abstützungen Wahrscheinlichkeitsmasse

h

$h$ der PDF

s (d | θ)

$s(d|\theta)$ ,

P (D_{i} \in H_{θ}) = h

$P(D_i \in H_\theta) = h$ für alle

i

$i$ . Daher ist der Bruchteil von

{D_{i}}

$\{D_i\}$ für den

D_{i} \in H_{θ}

$D_i \in H_\theta$ ist,

h

$h$ . Und so ist unter Verwendung der obigen Äquivalenz der Bruchteil von

{D_{i}}

$\{D_i\}$ für den

θ \in C_{D_{i}}

$\theta \in C_{D_i}$ ist, ebenfalls

h

$h$ .

Dies ist es also, was die häufigste Behauptung für den $h$ Vertrauensbereich für $\theta$ beträgt:

Nehmen Sie eine große Anzahl von imaginären Beobachtungen $\{D_i\}$ aus der Stichprobenverteilung $s(d|\theta)$ , die zu der beobachteten Statistik $D$ . Dann liegt $\theta$ innerhalb eines Bruchteils $h$ der analogen, aber imaginären Vertrauensbereiche $\{C_{D_i}\}$ .

Der Konfidenzbereich $C_D$ erhebt daher keinen Anspruch auf die Wahrscheinlichkeit, dass $\theta$ irgendwo liegt! Der Grund ist einfach, dass es in der Formulierung nichts gibt, das es uns erlaubt, von einer Wahrscheinlichkeitsverteilung über $\theta$ zu sprechen . Die Interpretation ist nur aufwendiger Überbau, der die Basis nicht verbessert. Die Basis ist nur $s(d | \theta)$ und $D$ , wobei $\theta$ thgr; nicht als verteilte Größe erscheint, und es gibt keine Informationen, mit denen wir das ansprechen können. Grundsätzlich gibt es zwei Möglichkeiten, eine Verteilung über $\theta$ :

Ordnen Sie eine Verteilung direkt aus den vorliegenden Informationen zu: $p(\theta | I)$ .
Beziehen $\theta$ zu einer anderen verteilten Menge: $p(\theta | I) = \int p(\theta x | I) dx = \int p(\theta | x I) p(x | I) dx$ .

In beiden Fällen muss $\theta$ irgendwo links stehen. Frequentisten können keine der beiden Methoden anwenden, da sie beide einen ketzerischen Prior benötigen.

Bayesianische Sicht

Das Beste, was ein Bayesianer aus dem $h$ Konfidenzbereich $C_D$ ohne Einschränkung machen kann , ist einfach die direkte Interpretation: Es ist die Menge von $\phi$ für die $D$ in den $h$ HDR $H_\phi$ der Stichprobenverteilung $s(d|\phi)$ . Es sagt uns nicht unbedingt viel über $\theta$ , und hier ist der Grund dafür.

The probability that $\theta \in C_D$ , given $D$ and the background information $I$ , is:

\begin{aligned} P (θ \in C_{D} | D I) & = \int_{C_{D}} p (θ | D I) d θ \\ = \int_{C_{D}} \frac{p (D | θ I) p (θ | I)}{p (D | I)} d θ \end{aligned}

$\begin{align*} P(\theta \in C_D | DI) &= \int_{C_D} p(\theta | DI) d\theta \\ &= \int_{C_D} \frac{p(D | \theta I) p(\theta | I)}{p(D | I)} d\theta \end{align*}$ Notice that, unlike the frequentist interpretation, we have immediately demanded a distribution over

θ

$\theta$ . The background information

I

$I$ tells us, as before, that the sampling distribution is

s (d | θ)

$s(d | \theta)$ :

\begin{aligned} P (θ \in C_{D} | D I) & = \int_{C_{D}} \frac{s (D | θ) p (θ | I)}{p (D | I)} d θ \\ = \frac{\int_{C_{D}} s (D | θ) p (θ | I) d θ}{p (D | I)} \\ i.e. P (θ \in C_{D} | D I) & = \frac{\int_{C_{D}} s (D | θ) p (θ | I) d θ}{\int s (D | θ) p (θ | I) d θ} \end{aligned}

$\begin{align*} P(\theta \in C_D | DI) &= \int_{C_D} \frac{s(D | \theta) p(\theta | I)}{p(D | I)} d \theta \\ &= \frac{\int_{C_D} s(D | \theta) p(\theta | I) d\theta}{p(D | I)} \\ \text{i.e.} \quad\quad P(\theta \in C_D | DI) &= \frac{\int_{C_D} s(D | \theta) p(\theta | I) d\theta}{\int s(D | \theta) p(\theta | I) d\theta} \end{align*}$ Now this expression does not in general evaluate to

h

$h$ , which is to say, the

h

$h$ confidence region

C_{D}

$C_D$ does not always contain

θ

$\theta$ with probability

h

$h$ . In fact it can be starkly different from

h

$h$ . There are, however, many common situations in which it does evaluate to

h

$h$ , which is why confidence regions are often consistent with our probabilistic intuitions.

For example, suppose that the prior joint PDF of $d$ and $\theta$ is symmetric in that $p_{d,\theta}(d,\theta | I) = p_{d,\theta}(\theta,d | I)$ . (Clearly this involves an assumption that the PDF ranges over the same domain in $d$ and $\theta$ .) Then, if the prior is $p(\theta | I) = f(\theta)$ , we have $s(D | \theta) p(\theta | I) = s(D | \theta) f(\theta) = s(\theta | D) f(D)$ . Hence

\begin{aligned} P (θ \in C_{D} | D I) & = \frac{\int_{C_{D}} s (θ | D) d θ}{\int s (θ | D) d θ} \\ i.e. P (θ \in C_{D} | D I) & = \int_{C_{D}} s (θ | D) d θ \end{aligned}

$\begin{align*} P(\theta \in C_D | DI) &= \frac{\int_{C_D} s(\theta | D) d\theta}{\int s(\theta | D) d\theta} \\ \text{i.e.} \quad\quad P(\theta \in C_D | DI) &= \int_{C_D} s(\theta | D) d\theta \end{align*}$ From the definition of an HDR we know that for any

ψ \in Θ

$\psi \in \Theta$

\begin{aligned} \int_{H_{ψ}} s (d | ψ) d d & = h \\ and therefore that \int_{H_{D}} s (d | D) d d & = h \\ or equivalently \int_{H_{D}} s (θ | D) d θ & = h \end{aligned}

$\begin{align*} \int_{H_\psi} s(d | \psi) dd &= h \\ \text{and therefore that} \quad\quad \int_{H_D} s(d | D) dd &= h \\ \text{or equivalently} \quad\quad \int_{H_D} s(\theta | D) d\theta &= h \end{align*}$ Therefore, given that

s (d | θ) f (θ) = s (θ | d) f (d)

$s(d | \theta) f(\theta) = s(\theta | d) f(d)$ ,

C_{D} = H_{D}

$C_D = H_D$ implies

P (θ \in C_{D} | D I) = h

$P(\theta \in C_D | DI) = h$ . The antecedent satisfies

C_{D} = H_{D} ⟷ \forall ψ [ψ \in C_{D} \leftrightarrow ψ \in H_{D}]

$C_D = H_D \longleftrightarrow \forall \psi \; [ \psi \in C_D \leftrightarrow \psi \in H_D ]$ Applying the equivalence near the top:

C_{D} = H_{D} ⟷ \forall ψ [D \in H_{ψ} \leftrightarrow ψ \in H_{D}]

$C_D = H_D \longleftrightarrow \forall \psi \; [ D \in H_\psi \leftrightarrow \psi \in H_D ]$ Thus, the confidence region

C_{D}

$C_D$ contains

θ

$\theta$ with probability

h

$h$ if for all possible values

ψ

$\psi$ of

θ

$\theta$ , the

h

$h$ -HDR of

s (d | ψ)

$s(d | \psi)$ contains

D

$D$ if and only if the

h

$h$ -HDR of

s (d | D)

$s(d | D)$ contains

ψ

$\psi$ .

Now the symmetric relation $D \in H_\psi \leftrightarrow \psi \in H_D$ is satisfied for all $\psi$ when $s(\psi + \delta | \psi) = s(D - \delta | D)$ for all $\delta$ that span the support of $s(d | D)$ and $s(d | \psi)$ . We can therefore form the following argument:

$s(d | \theta) f(\theta) = s(\theta | d) f(d)$ (premise)
$\forall \psi \; \forall \delta \; [ s(\psi + \delta | \psi) = s(D - \delta | D) ]$ (premise)
$\forall \psi \; \forall \delta \; [ s(\psi + \delta | \psi) = s(D - \delta | D) ] \longrightarrow \forall \psi \; [ D \in H_\psi \leftrightarrow \psi \in H_D ]$
$\therefore \quad \forall \psi \; [ D \in H_\psi \leftrightarrow \psi \in H_D ]$
$\forall \psi \; [ D \in H_\psi \leftrightarrow \psi \in H_D ] \longrightarrow C_D = H_D$
$\therefore \quad C_D = H_D$
$[s(d | \theta) f(\theta) = s(\theta | d) f(d) \wedge C_D = H_D] \longrightarrow P(\theta \in C_D | DI) = h$
$\therefore \quad P(\theta \in C_D | DI) = h$

Let's apply the argument to a confidence interval on the mean of a 1-D normal distribution $(\mu, \sigma)$ , given a sample mean $\bar{x}$ from $n$ measurements. We have $\theta = \mu$ and $d = \bar{x}$ , so that the sampling distribution is

s (d | θ) = \frac{\sqrt{n}}{σ \sqrt{2 π}} e^{- \frac{n}{2 σ^{2}} {(d - θ)}^{2}}

$s(d | \theta) = \frac{\sqrt{n}}{\sigma \sqrt{2 \pi}} e^{-\frac{n}{2 \sigma^2} { \left( d - \theta \right) }^2 }$ Suppose also that we know nothing about

θ

$\theta$ before taking the data (except that it's a location parameter) and therefore assign a uniform prior:

f (θ) = k

$f(\theta) = k$ . Clearly we now have

s (d | θ) f (θ) = s (θ | d) f (d)

$s(d | \theta) f(\theta) = s(\theta | d) f(d)$ , so the first premise is satisfied. Let

s (d | θ) = g ((d - θ)^{2})

$s(d | \theta) = g\left( (d - \theta)^2 \right)$ . (i.e. It can be written in that form.) Then

\begin{matrix} s (ψ + δ | ψ) = g ((ψ + δ - ψ)^{2}) = g (δ^{2}) \\ and s (D - δ | D) = g ((D - δ - D)^{2}) = g (δ^{2}) \\ so that \forall ψ \forall δ [s (ψ + δ | ψ) = s (D - δ | D)] \end{matrix}

$\begin{gather*} s(\psi + \delta | \psi) = g \left( (\psi + \delta - \psi)^2 \right) = g(\delta^2) \\ \text{and} \quad\quad s(D - \delta | D) = g \left( (D - \delta - D)^2 \right) = g(\delta^2) \\ \text{so that} \quad\quad \forall \psi \; \forall \delta \; [s(\psi + \delta | \psi) = s(D - \delta | D)] \end{gather*}$ whereupon the second premise is satisfied. Both premises being true, the eight-point argument leads us to conclude that the probability that

θ

$\theta$ lies in the confidence interval

C_{D}

$C_D$ is

h

$h$ !

We therefore have an amusing irony:

The frequentist who assigns the $h$ confidence interval cannot say that $P(\theta \in C_D) = h$ , no matter how innocently uniform $\theta$ looks before incorporating the data.
The Bayesian who would not assign an $h$ confidence interval in that way knows anyhow that $P(\theta \in C_D | DI) = h$ .

Final Remarks

We have identified conditions (i.e. the two premises) under which the $h$ confidence region does indeed yield probability $h$ that $\theta \in C_D$ . A frequentist will baulk at the first premise, because it involves a prior on $\theta$ , and this sort of deal-breaker is inescapable on the route to a probability. But for a Bayesian, it is acceptable---nay, essential. These conditions are sufficient but not necessary, so there are many other circumstances under which the Bayesian $P(\theta \in C_D | DI)$ equals $h$ . Equally though, there are many circumstances in which $P(\theta \in C_D | DI) \ne h$ , especially when the prior information is significant.

We have applied a Bayesian analysis just as a consistent Bayesian would, given the information at hand, including statistics $D$ . But a Bayesian, if he possibly can, will apply his methods to the raw measurements instead---to the $\{x_i\}$ , rather than $\bar{x}$ . Oftentimes, collapsing the raw data into summary statistics $D$ destroys information in the data; and then the summary statistics are incapable of speaking as eloquently as the original data about the parameters $\theta$ .

CarbonFlambe--Reinstate Monica
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Would it be correct to say that a Bayesian is committed to take all the available information into account, while interpretation given in the question ignored D in some sense?

qbolec

Is it a good mental picture to illustrate the situation: imagine a grayscale image, where intensity of pixel x,y is the joint ppb of real param being y and observed stat being x. In each row y, we mark pixels which have 95% mass of the row. For each observed stat x, we define CI(x) to be the set of rows which have marked pixels in column x. Now, if we choose x,y randomly then CI(x) will contain y iff x,y was marked, and mass of marked pixels is 95% for each y. So, frequentists say that keeping y fixed, chance is 95%, OP says, that not fixing y also gives 95%, and bayesians fix y and don't know

qbolec

@qbolec It is correct to say that in the Bayesian method one cannot arbitrarily ignore some information while taking account of the rest. Frequentists say that for all

y

$y$ the expectation of

y \in C I (x)

$y \in \mathrm{CI}(x)$ (as a Boolean integer) under the sampling distribution

p r o b (x | y, I)

$\mathrm{prob}(x\, |\, y, I)$ is 0.95. The frequentist 0.95 is not a probability but an expectation.

CarbonFlambe--Reinstate Monica

6

from a Bayesian probability perspective, why doesn't a 95% confidence interval contain the true parameter with 95% probability?

Two answers to this, the first being less helpful than the second

There are no confidence intervals in Bayesian statistics, so the question doesn't pertain.
In Bayesian statistics, there are however credible intervals, which play a similar role to confidence intervals. If you view priors and posteriors in Bayesian statistics as quantifying the reasonable belief that a parameter takes on certain values, then the answer to your question is yes, a 95% credible interval represents an interval within which a parameter is believed to lie with 95% probability.

If I have a process that I know produces a correct answer 95% of the time then the probability of the next answer being correct is 0.95 (given that I don't have any extra information regarding the process).

yes, the process guesses a right answer with 95% probability

Similarly if someone shows me a confidence interval that is created by a process that will contain the true parameter 95% of the time, should I not be right in saying that it contains the true parameter with 0.95 probability, given what I know?

Just the same as your process, the confidence interval guesses the correct answer with 95% probability. We're back in the world of classical statistics here: before you gather the data you can say there's a 95% probability of randomly gathered data determining the bounds of the confidence interval such that the mean is within the bounds.

With your process, after you've gotten your answer, you can't say based on whatever your guess was, that the true answer is the same as your guess with 95% probability. The guess is either right or wrong.

And just the same as your process, in the confidence interval case, after you've gotten the data and have an actual lower and upper bound, the mean is either within those bounds or it isn't, i.e. the chance of the mean being within those particular bounds is either 1 or 0. (Having skimmed the question you refer to it seems this is covered in much more detail there.)

How to interpret a confidence interval given to you if you subscribe to a Bayesian view of probability.

There are a couple of ways of looking at this

Technically, the confidence interval hasn't been produced using a prior and Bayes theorem, so if you had a prior belief about the parameter concerned, there would be no way you could interpret the confidence interval in the Bayesian framework.
Another widely used and respected interpretation of confidence intervals is that they provide a "plausible range" of values for the parameter (see, e.g., here). This de-emphasises the "repeated experiments" interpretation.

Moreover, under certain circumstances, notably when the prior is uninformative (doesn't tell you anything, e.g. flat), confidence intervals can produce exactly the same interval as a credible interval. In these circumstances, as a Bayesianist you could argue that had you taken the Bayesian route you would have gotten exactly the same results and you could interpret the confidence interval in the same way as a credible interval.

TooTone
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but for sure confidence intervals exist even if I subscribe to a bayesian view of probability, they just wont dissapear, right? :)The situation I was asking about was how to interpret a confidence interval given to you if you subscribe to a Bayesian view of probability.

Rasmus Bååth

The problem is that confidence intervals aren't produced using a Bayesian methodology. You don't start with a prior. I'll edit the post to add something which might help.

TooTone

2

I'll give you an extreme example where they are different.

Suppose I create my 95% confidence interval for a parameter $\theta$ as follows. Start by sampling the data. Then generate a random number between $0$ and $1$ . Call this number $u$ . If $u$ is less than $0.95$ then return the interval $(-\infty,\infty)$ . Otherwise return the "null" interval.

Now over continued repititions, 95% of the CIs will be "all numbers" and hence contain the true value. The other 5% contain no values, hence have zero coverage. Overall, this is a useless, but technically correct 95% CI.

The Bayesian credible interval will be either 100% or 0%. Not 95%.

probabilityislogic
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So is it correct to say that before seeing a confidence interval there is a 95% probability that it will contain the true parameter, but for any given confidence interval the probability that it covers the true parameter depends on the data (and our prior)? To be honest, what I'm really struggling with is how useless confidence intervals sounds (credible intervals I like on the other hand) and the fact that I never the less will have to teach them to our students next week... :/

Rasmus Bååth

This question has some more examples, plus a very good paper comparing the two approaches

probabilityislogic

1

"from a Bayesian probability perspective, why doesn't a 95% confidence interval contain the true parameter with 95% probability? "

In Bayesian Statistics the parameter is not a unknown value, it is a Distribution. There is no interval containing the "true value", for a Bayesian point of view it does not even make sense. The parameter it's a random variable, you can perfectly know the probability of that value to be between x_inf an x_max if you know the distribuition. It's just a diferent mindset about the parameters, usually Bayesians used the median or average value of the distribuition of the parameter as a "estimate". There is not a confidence interval in Bayesian Statistics, something similar is called credibility interval.

Now from a frequencist point of view, the parameter is a "Fixed Value", not a random variable, can you really obtain probability interval (a 95% one) ? Remember that it's a fixed value not a random variable with a known distribution. Thats why you past the text :"A confidence interval does not predict that the true value of the parameter has a particular probability of being in the confidence interval given the data actually obtained."

The idea of repeating the experience over and over... is not Bayesian reasoning it's a Frequencist one. Imagine a real live experiment that you can only do once in your life time, can you/should you built that confidence interval (from the classical point of view )?.

But... in real life the results could get pretty close ( Bayesian vs Frequencist), maybe thats why It could be confusing.

blew
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Warum enthält ein 95% -Konfidenzintervall aus der Perspektive der Bayes'schen Wahrscheinlichkeit nicht den wahren Parameter mit einer Wahrscheinlichkeit von 95%?

Antworten:

So erstellen Sie eine Vertrauensregion

Eine häufige Interpretation

Bayesianische Sicht

Final Remarks