# Solved – Examples of when confidence interval and credible interval coincide

In the wikipedia article on Credible Interval, it says:

For the case of a single parameter and
data that can be summarised in a
single sufficient statistic, it can be
shown that the credible interval and
the confidence interval will coincide
if the unknown parameter is a location
parameter (i.e. the forward
probability function has the form Pr(x
| μ) = f(x − μ) ), with a prior that
is a uniform flat distribution; and
also if the unknown parameter is a
scale parameter (i.e. the forward
probability function has the form Pr(x
| s) = f(x / s) ), with a Jeffreys'
prior  — the latter following
because taking the logarithm of such a
scale parameter turns it into a
location parameter with a uniform
distribution. But these are distinctly
special (albeit important) cases; in
general no such equivalence can be

Could people give specific examples of this? When does the 95% CI actually correspond to "95% chance", thus "violating" the general definition of CI?

Contents

## normal distribution:

Take a normal distribution with known variance. We can take this variance to be 1 without losing generality (by simply dividing each observation by the square root of the variance). This has sampling distribution:

\$\$p(X_{1}…X_{N}|mu)=left(2piright)^{-frac{N}{2}}expleft(-frac{1}{2}sum_{i=1}^{N}(X_{i}-mu)^{2}right)=Aexpleft(-frac{N}{2}(overline{X}-mu)^{2}right)\$\$

Where \$A\$ is a constant which depends only on the data. This shows that the sample mean is a sufficient statistic for the population mean. If we use a uniform prior, then the posterior distribution for \$mu\$ will be:

\$\$(mu|X_{1}…X_{N})sim Normalleft(overline{X},frac{1}{N}right)implies left(sqrt{N}(mu-overline{X})|X_{1}…X_{N}right)sim Normal(0,1)\$\$

So a \$1-alpha\$ credible interval will be of the form:

\$\$left(overline{X}+frac{1}{sqrt{N}}L_{alpha},overline{X}+frac{1}{sqrt{N}}U_{alpha}right)\$\$

Where \$L_{alpha}\$ and \$U_{alpha}\$ are chosen such that a standard normal random variable \$Z\$ satisfies:

\$\$Prleft(L_{alpha}<Z<U_{alpha}right)=1-alpha\$\$

Now we can start from this "pivotal quantity" for constructing a confidence interval. The sampling distribution of \$sqrt{N}(mu-overline{X})\$ for fixed \$mu\$ is a standard normal distribution, so we can substitute this into the above probability:

\$\$Prleft(L_{alpha}<sqrt{N}(mu-overline{X})<U_{alpha}right)=1-alpha\$\$

Then re-arrange to solve for \$mu\$, and the confidence interval will be the same as the credible interval.

## Scale parameters:

For scale parameters, the pdfs have the form \$p(X_{i}|s)=frac{1}{s}fleft(frac{X_{i}}{s}right)\$. We can take the \$(X_{i}|s)sim Uniform(0,s)\$, which corresponds to \$f(t)=1\$. The joint sampling distribution is:

\$\$p(X_{1}…X_{N}|s)=s^{-N};;;;;;;0<X_{1}…X_{N}<s\$\$

From which we find the sufficient statistic to be equal to \$X_{max}\$ (the maximum of the observations). We now find its sampling distribution:

\$\$Pr(X_{max}<y|s)=Pr(X_{1}<y,X_{2}<y…X_{N}<y|s)=left(frac{y}{s}right)^{N}\$\$

Now we can make this independent of the parameter by taking \$y=qs\$. This means our "pivotal quantity" is given by \$Q=s^{-1}X_{max}\$ with \$Pr(Q<q)=q^{N}\$ which is the \$beta(N,1)\$ distribution. So, we can choose \$L_{alpha},U_{alpha}\$ using the beta quantiles such that:

\$\$Pr(L_{alpha}<Q<U_{alpha})=1-alpha=U_{alpha}^{N}-L_{alpha}^{N}\$\$

And we substitute the pivotal quantity:

\$\$Pr(L_{alpha}<s^{-1}X_{max}<U_{alpha})=1-alpha=Pr(X_{max}L_{alpha}^{-1}>s>X_{max}U_{alpha}^{-1})\$\$

And there is our confidence interval. For the Bayesian solution with jeffreys prior we have:

\$\$p(s|X_{1}…X_{N})=frac{s^{-N-1}}{int_{X_{max}}^{infty}r^{-N-1}dr}=N (X_{max})^{N}s^{-N-1}\$\$ \$\$implies Pr(s>t|X_{1}…X_{N})=N (X_{max})^{N}int_{t}^{infty}s^{-N-1}ds=left(frac{X_{max}}{t}right)^{N}\$\$

We now plug in the confidence interval, and calculate its credibility

\$\$Pr(X_{max}L_{alpha}^{-1}>s>X_{max}U_{alpha}^{-1}|X_{1}…X_{N})=left(frac{X_{max}}{X_{max}U_{alpha}^{-1}}right)^{N}-left(frac{X_{max}}{X_{max}L_{alpha}^{-1}}right)^{N}\$\$

\$\$=U_{alpha}^{N}-L_{alpha}^{N}=Pr(L_{alpha}<Q<U_{alpha})\$\$

And presto, we have \$1-alpha\$ credibility and coverage.

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