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In statistical inference by casella, it provides a nice example that shows how convergence in probability does not imply almost surely convergence. However, there is no good counterexample showing how convergence in distribution does not imply convergence in probability.

After spending some time, it seems like if I can find two CDFs that are identical but have different random variables, it should work out. But the most I spend time on it, this seems impossible. Convergence in distribution says that $$\lim_{n \to \infty} F_{X_n}(x) = F_X(x)$$ at all points x, and $F_X(x)$ be continuous. But if the cdfs are the same, then the pdfs are also the same. How can we have convergence in distribution and not convergence in prob?

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  • $\begingroup$ For example if the variables lives in different spaces of probability $\endgroup$ May 2 '18 at 2:59
  • $\begingroup$ what do you mean by different space? it says CDFs must be the same at all points $x$ and I assumed that meant CDF must lie in the same space $\endgroup$
    – MoneyBall
    May 2 '18 at 3:01
  • $\begingroup$ Convergence in prob states that $\lim_n \mathbb{P}(|X_n-X|>\epsilon)=0$. For this, X_n and X must live in the same space. (If not, this probability no have sense!) $\endgroup$ May 2 '18 at 3:07
  • $\begingroup$ could you provide an example where $X_n$ and $X$ don't live in the same space but have the same CDFs? $\endgroup$
    – MoneyBall
    May 2 '18 at 3:08
  • $\begingroup$ If $A$ is a set with 1 element, define a random variable in the space $(A,\mathcal{P}(A),\mathbb{P})$ as 0 in the element of A. Then the CDF is 0 if x<0 and 1 if x\geq1. Now, the element of A is arbitrary $\endgroup$ May 2 '18 at 3:19
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Take the sample space $\Omega = \{0,1\}$ with $P(\omega= 0) = P(\omega = 1) = 1/2$.

Let $X_n$ for all $n \in \mathbb{N}$ and $X$ be random variables defined on the same sample space $\Omega$ such that $X_n(0) = 0 = X(1)$ and $X_n(1) = 1 = X(0)$.

Since $|X_n(\omega) - X(\omega)| = 1$, $X_n$ fails to converge to $X$ in probability.

However, you can show that $X_n$ and $X$ have the same distribution function,

$$F_X(x) = F_{X_n}(x) = \begin{cases}1/2, \,\,\, 0 \leqslant x < 1\\ 1, \,\,\,\,\,\,\,\,\, x \geqslant 1 \end{cases}$$

and, hence, $X_n \to X$ in distribution.

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