# Proof of posterior consistency

Given the model $x_1,...,x_n$ ~ $Poi(\theta)$ (iid), and the prior $\theta$~$Gamma(a,b)$. How do we prove the consistency of posterior distribution? Thanks in advance.

Modification: The definition of posterior consistency given on my lecture slides is:

“given the data $x_1,...,x_n$ iid and the model f(x;$\theta$), if we have posterior consistency at $\theta_0$, then $\forall \epsilon$, $\lim_{n\rightarrow +\infty}\pi(|\theta-\theta_0|>\epsilon|X)$=0.” $\theta_0$ is not specified but I think it is the true value of $\theta$.

• Could you explain what you mean for a distribution to be consistent?
– whuber
Commented Dec 22, 2017 at 19:34
• I have added this in the original question. Commented Dec 22, 2017 at 20:06
• Thank you. That was important, because consistency is not a property of any distribution: it's a property of the estimator and the model. (Presumably, "$\pi$" means "probability.") As far as your question goes, could you please indicate what attempts you have made towards a solution? For instance, have you worked out what the posterior distribution actually is in terms of $a$, $b$, and $X$?
– whuber
Commented Dec 22, 2017 at 20:48

Writing $$X^{(n)}=(X_1,\dots,X_n)$$, Bayes theorem tells us that $$\Theta\mid X^{(n)}\sim\mathrm{Ga}\!\left(a+\sum_{i=1}^n X_i,\,b+n\right).$$ Define $$M=\mathrm{E}[\Theta\mid X^{(n)}]=\frac{a+\sum_{i=1}^n X_i}{b+n}$$ and $$V=\mathrm{Var}[\Theta\mid X^{(n)}]=\frac{a+\sum_{i=1}^n X_i}{(b+n)^2}.$$ For any given $$\epsilon>0$$, if $$|\Theta-M|<\epsilon/2$$ and $$|M-\theta_0|<\epsilon/2$$, then, using the triangle inequality, we have $$|\Theta-\theta_0| = |\Theta-M+M-\theta_0|\leq |\Theta-M|+|M-\theta_0| < \epsilon.$$ Considering the contrapositive of the previous statement, we see that $$\{|\Theta-\theta_0|\geq\epsilon\} \subset \{|\Theta-M|\geq\epsilon/2\} \cup \{|M-\theta_0|\geq\epsilon/2\}.$$ Hence, monotonicity and Chebyshev's (Чебышёв) inequality give us $$\Pr(|\Theta-\theta_0|\geq\epsilon\mid X^{(n)}) \leq \frac{4V}{\epsilon^2} + \Pr(|M-\theta_0|\geq\epsilon/2\mid X^{(n)}).$$ Let $$\mu_0$$ be a $$\mathrm{Poi}(\theta_0)$$ distribution and note that $$V\to 0$$ a.s. $$[\mu_0]$$, and $$M\to \theta_0$$ a.s. $$[\mu_0]$$, as $$n\to\infty$$.
This is essentially what you need to see that this posterior is "simulation consistent". That is, if you're simulating iid $$X_i$$'s from a $$\mathrm{Poi}(\theta_0)$$ distribution, then your posterior distribution will concentrate all its mass at $$\theta_0$$, as your sample size goes to infinity.