Complicated Demonstration - Violation of the theorem that converges in probability and not in distribution

Question

I was thinking that if a sequence of random variables $Y_n$ with c.d.f. $H_n$ which converges to $c$ in probability, such that $H_n(c)$ does not converge to $H(c)=1$.

How could I make an example where the above is fulfilled?

Answer 1

I'm not sure this is what you want, but I though about the sigmoid function for the cumulative distribution function: $$ H_n(x) = \frac{1}{1 + e^{nx}} $$

For larger values of $n$, the function will be more steep near $x=0$.

For all $n$, $H_n(0)=1/2$. However, in the limit: $$ H(x) = \lim_{n \to +\infty} H_n(x) = \left\{ \begin{align} 0 & \textrm{ if } x < 0 \\ 1 & \textrm{ if } x \geq 0 \end{align} \right. $$ And, so, $H(0) = 1 \neq 1/2 = H_n(0)$.

This type of approach can be done when the limit of a continuous variable will behave like a discrete variable, like in Dirac $\delta$ function.