# If $X_n \to X$ in probability and the law of $X_n$ and $Y_n$ is the same, then $Y_n \to X$ in probability?

If $X_n \to X$ in probability and the law of $X_n$ and $Y_n$ is the same for all $n$, then $Y_n \to X$ in probability?

My reasoning is as follows...

I know that $X_n \to X$ in probability is equivalent to $\lim_{n \to\infty}E\left[\frac{\vert X_n-X\vert}{1+\vert X_n-X\vert}\right]=0$ But since $X_n$ and $Y_n$ have the distribution we can replace $X_n$ in the above expression to $Y_n$ to conclude $\lim_{n \to\infty}E\left[\frac{\vert Y_n-X\vert}{1+\vert Y_n-X\vert}\right]=0$ and therefore $Y_n \to X$ in probability.

I was reading a book by Skorokhod and he says that the limiting random variable might not be the same! But that implies my reasoning is wrong. Can somebody help?

No, $X_n \to X$ in probability refers to the r.v.s $X_n-X$. This at the very least requires them to be defined on the same probability space. $X_n$ and $Y_n$ in this case needn't even be defined on the same probability space.
Even if all the relevant random variables are defined on the same probability space, counterexamples still abound. Consider a fixed random variable $X$ with a nondegenerate symmetric distribution. Take $X_n=X$ and $Y_n=-X$. By the symmetry, $Y_n$ has the same distribution as $X_n$. But $|X_n-X| \equiv 0$ whereas $|Y_n-X|=2|X|$. Thus $X_n \to X$ in probability and $Y_n \not \to X$ in probability.
An alternative answer requiring no measure theory: the error in your reasoning is that just because $X_n$ and $Y_n$ have the same distribution does not mean that $(X_n,X)$ and $(Y_n,X)$ have the same joint distribution. If they did, then that would be sufficient to carry your reasoning through.
• But if the joint distributions are the same then it will converge to the same rv $X$ right? Nov 15 '17 at 20:00