Limit of a integral whose integrand diverges under the limit

Question

I am trying to simplify the following limit of integral where $\mu$ is given:

$$p(y) = \lim_{\sigma \to 0} \int_{\mathbb R} |x| \cdot \frac{1}{\sqrt{2\pi\sigma^2} } e^{-\frac{1}{2\sigma^2} (xy - \mu)^2} f(x) dx,$$

however I am not sure if there is a way to simplify it, as the integrand does not converge under the limit $\sigma \to 0$ and the interchangeablity of limit and integral fails here:

$$\lim_{\sigma \to 0} |x| \cdot \frac{1}{\sqrt{2\pi\sigma^2} } e^{-\frac{1}{2\sigma^2} (xy - \mu)^2} f(x) = \begin{cases} \inf, & x \ne \mu /y; \\ 0, & x = \mu /y \end{cases}.$$

However, I DO know the answer of the limit of the integral, as naturally it is the probablity density function of $Y = T/X$, where $X$ has PDF $f(x)$ and $T \sim N(\mu, \sigma^2)$ is a normal distribution. The limit of integral at $\sigma \to 0$ then essentially means the PDF of $\mu / X$ and can be easily calculated.

So how do I calculate this limit of integral analytically, without the help of the intuition from probablity?

Answer 1

$\newcommand\si\sigma\newcommand\R{\mathbb R}$We have to find $$\lim_{\si\downarrow0}p_\si(y)$$ for all real $y$ such that the limit exists, where $$p_\si(y):=\int_\R\frac{|x|\,dx}{\si\sqrt{2\pi}}e^{-(xy-\mu)^2/(2\si^2)}\,f(x).$$ Assuming that $f$ is bounded, and using the substitution $x=(\mu+\si z)/y$ and dominated convergence, we get $$p_\si(y)=\frac1{y^2}\,\int_\R\frac{|\mu+\si z|\,dz}{\sqrt{2\pi}}e^{-z^2/2} f\Big(\frac{\mu+\si z}y\Big) \to p(y):=\frac{|\mu|}{y^2}\,f\Big(\frac\mu y\Big) \tag{1}\label{1}$$ (as $\si\downarrow0$) for all real $y\ne0$ such that $\mu/y$ is a point of continuity of $f$.

If $y=0$, then $\lim_{\si\downarrow0}p_\si(y)=\infty$. If $y\ne0$ and $\mu/y$ is not a point of continuity of $f$, then $\lim_{\si\downarrow0}p_\si(y)$ does not exist in general.

However, \eqref{1} holds for every real $y\ne0$ such $\mu/y$ is a Lebesgue point of continuity of $f$. So, \eqref{1} holds for almost all real $y$.

Answer 2

The Gaussian tends to a delta function in the $\sigma\rightarrow 0$ limit,

$$ \lim_{\sigma \to 0} \frac{1}{\sqrt{2\pi\sigma^2} } e^{-\frac{1}{2\sigma^2} (xy - \mu)^2} =\delta(xy-\mu)=|y|^{-1}\delta(x-\mu/y),$$ so the integral evaluates to $$p(y) = |y|^{-1}\int_{-\infty}^\infty |x| f(x)\delta(x-\mu/y) \,dx=|y|^{-1}|\mu/y|f(\mu/y).$$