verifying a posterior is proper There's a homework problem in a textbook that asks to verify propriety of a certain posterior distribution, and I'm having a little trouble with it. The setup is you have a logistic regression model with one predictor, and you have an improper uniform prior over $\mathbb{R}^2$. 
Specifically, we assume for $i=1,\ldots,k$ that 
$$
y_i \mid \alpha, \beta,x_i \sim \text{Binomial}(n,\text{invlogit}(\alpha + \beta x_i)),
$$ 
so the likelihood is
$$
p(y \mid \alpha, \beta, x ) = \prod_{i=1}^k [\text{invlogit}(\alpha + \beta x_i)]^{y_i}[1-\text{invlogit}(\alpha + \beta x_i)]^{n-y_i}.
$$
The trouble is that I suspect that this posterior is actually improper. 
For the specific situation where $k=1$, if we use the change of variables $s_1 = \text{invlogit}(\alpha + \beta x)$ and $s_2 = \beta$, then we can see that
\begin{align*}
\iint_{\mathbb{R}^2}p(y \mid \alpha, \beta, x ) \text{d}\alpha \text{d}\beta &= \iint_{\mathbb{R}^2}[\text{invlogit}(\alpha + \beta x)]^{y}[1-\text{invlogit}(\alpha + \beta x)]^{n-y} \text{d}\alpha \text{d}\beta \\
&= \int_{-\infty}^{\infty}\int_0^1 s_1^{y-1}(1-s_1)^{n-y-1}  \text{d}s_1 \text{d}s_2 \\
&= B(y,n-y)  \int_{-\infty}^{\infty} 1 \text{d}s_2 \tag{*}\\
&= \infty.
\end{align*}
In the line with the asterisk we assume that $0 < y < n$, but if it doesn't, then we end up with the same thing.
Am I doing something silly here? Or is this an improper posterior?
 A: For $=1$, you can see directly from your equation for the likelihood that the posterior density $(,|_1,_1)$ will be constant along parallell lines at which $+_$ takes constant values. So the posterior is indeed improper and has the shape of a ridge for $=1$.  Basically, any regression line fitting the observed response at $_1$ will do equally well.
Next, suppose we have $k=2$ observations.  Consider the reparameterization given by
\begin{align}
\eta_1 &= \alpha + \beta x_1 \\
\eta_2 &= \alpha + \beta x_2
\end{align}
Since this is a linear transformation of $\alpha,\beta$ with a constant determinant the prior for $\eta_1,\eta_2$ is also uniform over $\mathbb{R}^2$, provided that $x_1\neq x_2$.  Consider the further reparameterization, the inverse logit transformation
\begin{align}
p_i = \frac1{1+e^{-\eta_i}},
\end{align}
for $i=1,2$. Clearly, $p_1,p_2$ are also a priori independent with densities given by
$$
\pi(p_i)=\pi(\eta_i)\Big|\frac{d\eta_i}{dp_i}\Big|\propto \frac d{dp_i}\ln\frac{p_i}{1-p_i} = \frac1{(1-p_i)p_i}
$$
These are so called improper Haldane priors, that can be interpreted as a certain form of limit of the density of a Beta distribution with both parameters approaching zero.  Conditional on the data $y_1,y_2$, provided that $0<y_i<n$, the posterior marginal density for each $p_i$ are proper Beta distributions with parameters $y_i,n-y_i$.  Backtransforming, the posterior distributions of $(\eta_1,\eta_2)$ and $(\alpha,\beta)$ must also be proper.  This holds except in special cases such as one $y_i$ taking a value of 0 or $n$ in which case the normalising beta function $B(y_i,n-y_i)$ is infinite and the posterior of $p_i$ (and hence the posterior of $\alpha$ and $\beta$) is improper.
For $k>2$ observations, the posterior must also be proper since the non-normalized posterior density of $\alpha,\beta$ is bounded by the posterior based on the first $k=2$ observations.
A: It's improper, I believe. I only need to prove that $$\int\limits_{\alpha \in \mathbb{R} \\ \beta>0} p(y \mid \alpha, \beta, x) = +\infty.$$
Denote function $$\sigma = \mathrm{invlogit}$$
Now that $\sigma$ is a monotonically increasing function, when $\beta > 0$, we have $$\mathrm{\sigma}(\alpha + \beta x_i) > \mathrm{\sigma}(\alpha - \beta \max |x_i|) > 0,$$ $$1 - \mathrm{\sigma}(\alpha + \beta x_i) > 1 - \mathrm{\sigma}(\alpha + \beta \max |x_i|) > 0.$$
Thus the the integral 
$$\begin{aligned}
\int\limits_{\alpha \in \mathbb{R} \\ \beta>0} p(y \mid \alpha, \beta, x) >& \int\limits_{\alpha \in \mathbb{R} \\ \beta>0} \prod \left[ \mathrm{\sigma}(\alpha - \beta \max |x_i|) \right]^{y_i} \left[ 1 - \mathrm{\sigma}(\alpha + \beta \max |x_i|) \right]^{n_i - y_i} \\
>& \int\limits_{\alpha \in \mathbb{R} \\ \beta>0} \prod \left[ \mathrm{\sigma}(\alpha - \beta \max |x_i|) \right]^{\max n_i} \left[ 1 - \mathrm{\sigma}(\alpha + \beta \max |x_i|) \right]^{\max n_i} \\
>& \int\limits_{\alpha \in \mathbb{R} \\ \beta>0} \left[ \mathrm{\sigma}(\alpha - \beta \max |x_i|) \right]^{k\max n_i} \left[ 1 - \mathrm{\sigma}(\alpha + \beta \max |x_i|) \right]^{k\max n_i} \\
\end{aligned}$$
More properties about $\sigma$ are needed: 
$$\left( \sigma(x) \right)^N = \frac{1}{(1+e^{-x})^N} > \dfrac{1}{2^N (\max\{1,e^{-x}\}) ^N} = \dfrac{1}{2^N (\max\{1,e^{-Nx}\})} > \dfrac{1}{2^N}\sigma(Nx)$$
Let $\xi = \alpha - \beta \max |x_i|$, $\eta = \alpha + \beta \max |x_i|, N=k\max n_i$, then
\begin{aligned}
\int\limits_{\alpha \in \mathbb{R} \\ \beta>0} p(y \mid \alpha, \beta, x)
>& \int\limits_{\alpha \in \mathbb{R} \\ \beta>0} \left[ \mathrm{\sigma}(\alpha - \beta \max |x_i|) \right]^{N} \left[ 1 - \mathrm{\sigma}(\alpha + \beta \max |x_i|) \right]^{N} \\
\propto& \int\limits_{-\infty < \xi < \eta < +\infty} \left[ \mathrm{\sigma}(\xi) \right]^{N} \left[ \mathrm{\sigma}(-\eta) \right]^{N}\\
>& \frac{1}{2^{2N}}\int\limits_{\xi}^{+\infty} \Big( \int\limits_{-\infty}^{+\infty}  \mathrm{\sigma}(N\xi) \mathrm{d}\xi \Big) ~ \mathrm{\sigma}(- N\eta) \mathrm{d}\eta \\
=& +\infty
\end{aligned}
A: I've already accepted an answer, but I did want to point out that the posterior isn't proper for all possible data sets. The posterior is proportional to the likelihood, which is
$$\prod_{i=1}^k [\text{invlogit}(\alpha + \beta x_i)]^{y_i}[1-\text{invlogit}(\alpha + \beta x_i)]^{n-y_i}.
$$
If $y_1 = y_2 = \cdots = y_k = n$, then this simplifies to 
$$
\prod_{i=1}^k [\text{invlogit}(\alpha + \beta x_i)]^n,
$$
and we can see that
\begin{align*}
&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \prod_{i=1}^k [\text{invlogit}(\alpha + \beta x_i)]^n \text{d}\alpha \text{d}\beta\\
&\ge \int_0^{\infty}\int_{-\infty}^{\infty} \prod_{i=1}^k [\text{invlogit}(\alpha + \beta x_i)]^n \text{d}\alpha \text{d}\beta  \\
&\ge \int_0^{\infty}\int_{-\infty}^{\infty}  [\text{invlogit}(\alpha + \beta x_{(1)})]^{nk} \text{d}\alpha \text{d}\beta  \\
&\ge \int_0^{\infty}\int_{-\infty}^{\infty}  [\text{invlogit}(r_1)]^{nk} \text{d}r_1 \text{d}r_2  \\
&= \infty.
\end{align*}
