I am trying to obtain the correct posterior distribution for a Bayes model with a uniform distribution for the likelihood, and a Pareto distribution for the prior (the Pareto is the conjugate prior for the Uniform). So far I have the following:

$p(a)$ is the prior

$p(D|a)$ is the likelihood of the data set D

$D = \{x_{1}, ... , x_{n}\}$ where all data points $x_1$ through $x_n$ are real valued, and $n$ is the total number of points

$p(a|D)$ is the posterior distribution (what I am trying to derive)

Now, the Uniform Distribution is defined like so:

$Unif(a|c) = \dfrac{1}{2c} \times I(a \in [-c, c])$

where $I(true) = 1$ and $I(false) = 0$.

And the Pareto distribution is defined like so:

$Pareto(a|m,k) = k \times m^{k} \times a^{-(k+1)}$

The likelihood is a slight variation of the uniform distribution:

$p(D|a) = (\dfrac{1}{2a})^{n} \times I(a \ge max(\{|x_{1}|, ... , |x_{n}|\}))$

The prior distribution is derived from the Pareto distribution:

$p(a) = k \times m^{k} \times a^{-(k+1)} \times I(a \ge m)$

The posterior distribution is the likelihood times the prior, normalized:

$p(a|D) = \dfrac{p(a) \ p(D|a)}{\int_{a' \in A} p(a') \ p(D|a') \ da'}$

Plug in the distribution formulas, and we get this:

$p(a|D) = \dfrac{k \ m^{k} \ a^{-(k+1)} \ (\dfrac{1}{2a})^{n} \times I(a \ge m) \times I(a \ge max(\{|x_{1}|, ... , |x_{n}|\}))}{\int_{a' \in A} k \ m^{k} \ a^{-(k+1)} \ (\dfrac{1}{2a})^{n} \times I(a \ge m) \times I(a \ge max(\{|x_{1}|, ... , |x_{n}|\})) \ da'}$

which we can re-arrange like so:

$p(a|D) = \dfrac{k \ m^{k} \ (\dfrac{1}{2})^{n} \times a^{-(n+k+1)} \times I(a \ge m) \times I(a \ge max(\{|x_{1}|, ... , |x_{n}|\}))}{k \ m^{k} \ (\dfrac{1}{2})^{n} \times \int_{a' \in A} \ a'^{-(n+k+1)} \times I(a' \ge m) \times I(a' \ge max(\{|x_{1}|, ... , |x_{n}|\})) \ da'}$

which we can simplify to:

$p(a|D) = \dfrac{a^{-(n+k+1)} \times I(a \ge m) \times I(a \ge max(\{|x_{1}|, ... , |x_{n}|\}))}{\int_{a' \in A} \ a'^{-(n+k+1)} \times I(a' \ge m) \times I(a' \ge max(\{|x_{1}|, ... , |x_{n}|\})) \ da'}$

if we simply set $m = max(\{|x_{1}, ... , |x_{n}|\})$, we can simplify to:

$p(a|D) = \dfrac{a^{-(n+k+1)} \times I(a \ge m) \times I(a \ge max(\{|x_{1}|, ... , |x_{n}|\}))}{\int_{a' = m}^{a' = +\infty} \ a'^{-(n+k+1)} \ da'}$

work out the integral on the bottom, we get the final answer for our posterior distribution:

$p(a|D) = \dfrac{a^{-(n+k+1)} \times I(a \ge m) \times I(a \ge max(\{|x_{1}|, ... , |x_{n}|\}))}{\dfrac{m^{-(n+k)}}{(n+k+1)}}$

where $n$ must be an integer greater than zero, and $k$ must be a real value greater than zero, and $m$ still satisfies this: $m = max(\{|x_{1}|, ... , |x_{n}|\})$

My question is this: our final answer for the posterior distribution (just like any other probability distribution) when integrated over its entire domain must equal 1 ... regardless of the values we pick for the hyper parameters (m, k, n, etc). But my answer does not. In fact, as the hyperparameters change, the value of the distribution when integrated also changes. Could someone please tell me where I went wrong? I've worked it out several times and keep arriving at the same answer.

Thanks in advance

  • 1
    $\begingroup$ If you have a Pareto prior and conjugate max-unif likelihood, then you should be able to deduce the parameters of the Pareto posterior just looking at the numerator: PRIOR $\times$ LIKELIHOOD $\propto$ POSTERIOR. Try that first to straighten out the notation, then try to show that the posterior integrates to 1 over the correct support. // However, the point of using conjugate prior and likelihood is usually to avoid having to do the integration. $\endgroup$ – BruceET Jul 26 '17 at 7:28
  • $\begingroup$ @BruceET I thought that the whole point of normalizing it (using whats in the denominator) was so that you could use any values for m, n, k, etc within their boundaries and still have the whole thing integrate to 1. Is this not true? Where is the mistake? Is setting m=max({|x1|, ... , |xn|}) the problem? $\endgroup$ – Josh Jul 26 '17 at 14:11
  • $\begingroup$ @BruceET and if not, then what would the relationship be between these hyperparameters that will cause the whole thing to integrate to 1? $\endgroup$ – Josh Jul 26 '17 at 14:12

Here is a practical application of a Pareto/Uniform Bayesian problem. For the benefit of a more general audience, I changed the observed uniform distribution to $\mathsf{Unif}(0,\theta)$ in order to have a development that more nearly matches treatments of the uniform estimation problem in Wikipedia and in many statistics texts.

Problem. Let $X_1, X_2, \dots, X_n$ be a random sample from $\mathsf{UNIF}(0, \theta),$ where $\theta$ is unknown. We wish to find a 95% interval estimate for $\theta.$ We explore both traditional frequentist and Bayesian methods.

Frequentist approach. The maximum likelihood estimator of $\theta$ is the maximum value $M = X_{(n)}$ of the sample. The MLE is biased with $E(M) = \frac{n}{n+1}\theta$, so that $M_u = \frac{n+1}{n}M$ is unbiased with $E(M_u) = \theta).$ An intuitive rationale is that "on average" the $n$ observations divide $(0, \theta)$ into $n + 1$ equal subintervals with the maximum falling at $\frac{n}{n+1}\theta.$ A formal proof is not difficult.

More generally, the PDF of the distribution of $M$ is $f_M(M) = nM^{n–1}/\theta^n,$ for $0 < M < \theta.$ Hence a 95% CI for $\theta$ is $(M, M/(.05)^{1/n}).$ [See Wikipedia on ‘continuous uniform distribution’ under Estimation.]

The simulation below illustrates some key assertions in the case with $n = 25$ and $\theta = 10.$

B = 10^6;  n = 25;  th = 10
M = replicate(B, max(runif(n, 0, th)) )
mean(M);  (n/(n+1))*th
[1] 9.615483                  # aprx E(M)
[1] 9.615385                  # exact E(M)
M.u = ((n+1)/n)*M;  mean(M.u)
[1] 10.0001                   # aprx E(M.u) = 10                              
hist(M, prob=T, col="skyblue2")
curve(n*x^(n-1)/th^n, lwd=2, col="red", add=T) # arg of ‘curve’ must be ‘x’

enter image description here

If we have a sample of size $n = 25$ from $\mathsf{UNIF}(0, \theta),\; \theta$ unknown, with $M = 9.234,$ we know $\theta > M$ and the 95% CI $(M,\, M/.05^{1/25})$ becomes $(9.234, 10.410).$

Bayesian approach.

Prior distribution. A Pareto distribution has PDF $f(\theta) = kL^k/\theta^{k+1},$ for $k > 0, \theta > L > 0.$ The Pareto is a heavy-tailed distribution. Its CDF is $F(\theta) = 1 – (L/\theta)^k.$

In particular, we choose the relatively noninformative prior $\mathsf{Par}(k = 1,\,L=2)$ with kernel $f(\theta) \propto 1/\theta^{1 + 1},$ for $\theta > L=2 .$ A consequence of this prior is that we believe $$P(2 < \theta < 12) = 1 – (2/12) – [1 – (2/2)] = 5/6.$$

Likelihood function. If $M$ is the maximum of $n = 25$ observations from $\mathsf{UNIF}(0, \theta),$ then the likelihood function has kernel $f(x|\theta) \propto 1/\theta^n,$ for $\theta > M.$

Notice that the prior and likelihood are conjugate (of compatible mathematical form).

Posterior distribution. According to Bayes’ Theorem, the posterior density has kernel $$f(\theta|x) \propto 1/\theta^2 × 1/\theta^n \propto 1/\theta^{27},\;\; \text{for} \;\; \theta > \max(L,M).$$

which we recognize as the kernel of $\mathsf{Par}(26, M),$ where the first argument is $k+n$ and second argument is $\max(L,M),$ determining the support of the distribution.

Application. If we have $n = 25$ independent observations from a uniform distribution with unknown $\theta$ and maximum observation $M = 9.234,$ then one way to get a 95% Bayesian posterior probability interval is to cut 5% from the upper tail of $\mathsf{Par}(26, 9.234).$

Because we know $\theta > M,$ this amounts to $(9.234, 10.362),$ which is numerically about the same as the frequentist 95% CI.

Note: We used the 25 observations simulated by the R code below:

set.seed(1234); x = round(runif(25, 0, 10),3);  M = max(x);  M
## 9.234  

Ok, I found it. Just a typo. One of the expressions should be "$(n+k)$" and not "$(n+k+1)$". Fix this, and it will integrate to 1 every time, without fail, regardless of the values for n, k, m, etc (as long as they are all within their limitations).

The last expression we arrived at for our posterior distribution was **this:

$p(a|D) = \dfrac{a^{-(n+k+1)} \times I(a \ge m) \times I(a \ge max(\{|x_{1}|, ... , |x_{n}|\}))}{\dfrac{m^{-(n+k)}}{(n+k+1)}}$

But it should have been:

$p(a|D) = \dfrac{a^{-(n+k+1)} \times I(a \ge m) \times I(a \ge max(\{|x_{1}|, ... , |x_{n}|\}))}{\dfrac{m^{-(n+k)}}{(n+k)}}$

See the change?

The error had to have occurred when the normalization integral in the denominator was worked out.

Now we can integrate the distribution itself:

$\int_{a \in A} \ p(a|D) = \int_{a \in A} \ \dfrac{a^{-(n+k+1)} \times I(a \ge m) \times I(a \ge max(\{|x_{1}|, ... , |x_{n}|\}))}{\dfrac{m^{-(n+k)}}{(n+k)}}$

$\int_{a \in A} \ p(a|D) = \int_{a = m}^{a = +\infty} \ \dfrac{a^{-(n+k+1)} \times I(a \ge m)}{\dfrac{m^{-(n+k)}}{(n+k)}}$

$\int_{a \in A} \ p(a|D) = \int_{a = m}^{a = +\infty} \ \dfrac{a^{-(n+k+1)}}{\dfrac{m^{-(n+k)}}{(n+k)}} = 1$

where $n$ is an integer representing the number of data points in the set of training data $D$, and $k$ is a relatively small real positive number, and $m$ is a real number *greater* than zero.

  • 1
    $\begingroup$ Checked back to see how this is going. Almost there (+1). But I don't think the $m$ of the Pareto prior can be the same as $M =max(|x_i|)$ of the likelihood. The former is posited before you see data and the latter is based on data. Can you express posterior Pareto parameters in terms of prior $m, k$ and likelihood $M, n$? $\endgroup$ – BruceET Jul 26 '17 at 23:26
  • $\begingroup$ @BruceET you're right .... it shouldn't have been m=max(|xi|) .... i just did that so that i wouldn't have to bother splitting the integral into a sum of two integrals (just being lazy) .... but m cannot be zero (look at the formula for Pareto) ... so what should it be set to? $\endgroup$ – Josh Jul 27 '17 at 14:51
  • $\begingroup$ In a real application, Pareto prior would be based on some actual knowledge or opinion. For this theoretical exercise, I'd probably just use m=k=1.m=k=1. // Perhaps a little easier to think about real application if data come from Unif(0,a) instead of Unif(-a,a). $\endgroup$ – BruceET Jul 29 '17 at 17:48

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