# Convergence in probability and iid sequence

Let $X_n$ and $Y_n$ two sequences of real random variables.

Assume $Y_n$ iid with law $\mu$ non degenerated, and $X_n - Y_n \xrightarrow[n\infty]{\mathbb{P}} 0$. Can we say $X_n \xrightarrow[n\infty]{\mathbb{P}} Y_1$ ?

• You need to converge in probability to a random variable, not a distribution. – Batman Oct 17 '16 at 1:03
• @Batman, i edited my post. – anonymus Oct 17 '16 at 1:07

## 2 Answers

No, e.g., let $X_n=Y_n$ and you don't expect $Y_n\to_pY_1$, given the independence assumption. More concretely, take the iid fair coin flipping space, the unit interval with lebesgue measure with $Y_n$ the indicator of those $\omega\in [0,1]$ with $1$ in the $nth$ place of their binary expansion. Then $Y_1$ is concentrated on $[1/2,1]$ but for $n>1$ $Y_n$ is half concentrated on $[0,1/2]$.

• but you can say that $X_n$ converges in distribution to $Y_1$, right ? Since $X_n-Y_n$ converges in probability to $0$ which is a constant and $Y_n$ converges in distribution to $Y_1$, Slutsky lemma can be applied. – anonymus Oct 17 '16 at 11:47

If $Y_n$ is iid with law $\mu$, then clearly, $Y_n$ converges in distribution to $\mu$. But $Y_n$ may not converge to $Y_1$ (or any $Y_k$) in probability.

To see this, assume $Y_n \sim \mathcal{N}(0,1)$ iid, and let $Z\sim \mathcal{N}(0,2)$ then : $$P (|Y_n - Y_1|>\epsilon) = P(|Z|>\epsilon) > C_\epsilon > 0.$$

Furthermore, we cannot even say that $X_n$ converges in distribution to $\mu$.