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?
Best Answer
$\newcommand\si\sigma\newcommand\R{\mathbb R}$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 $y$.