Suppose there is a parameter $\theta$, that we do not know. What we do know is some random variable $$ Y=\theta + \epsilon $$ where $\epsilon \sim N(0,1)$ is independent of $\theta$. Then $f_{Y\vert\theta}(y\vert \theta) \sim N(\theta,1)$, where $f$ denotes a PDF.
Since we do not know $\theta$, suppose that that $\theta$ is drawn from an (improper) uniform prior over the real line (but we still don't know what it is)
The definition of the posterior distribution is $$\tag{1} f_{\theta\vert Y}(x\vert y)=\frac{f_{\theta}(x)\cdot f_{Y\vert\theta}(y\vert x)}{\int_{-\infty}^{\infty}f_{\theta}(x)\cdot f_{Y\vert\theta}(y\vert x)dx} $$ because $f_\theta(x) =1,\ \forall x$ (since improper prior uniform over reals) we have that the integrand is $$ \int_{-\infty}^{\infty}f_{Y\vert\theta}(y\vert x)dx = \infty $$ Therefore the posterior is improper? Have I made a mistake?
To me the above seems incorrect, because I believe that $E[\theta\vert Y=y] =y$, and seeing as the mean is a property of a distribution, doesn't this mean that the conditional distribution exists?
I realize that if I have the prior be uniform over a large interval like $(-1000000,1000000)$, then the denominator is not $\infty$, but then the prior is also not improper...
To clarify the question: If we have an prior that is uniform over the reals, and $Y$ as above, does the conditional distribution $(1)$ exist, and if so what is it?
The unconditional distribution of $Y$--that is to say, your denominator--is $$\int_{x=-\infty}^\infty f_\theta(x) f_{Y \mid \theta}(y \mid x) \, dx = \int_{x=-\infty}^\infty 1 \cdot \frac{1}{\sqrt{2\pi}} e^{-(y - x)^2/2} \, dx = 1.$$ Thus, it too is improper, but the posterior distribution of $\theta$ is not: it is $$f_{\theta \mid Y}(x \mid y) = f_{Y \mid \theta}(y \mid x) = \frac{1}{\sqrt{2\pi}} e^{-(y-x)^2/2},$$ hence is normal with mean $y$ and variance $1$.