I encountered the following statement in some lecture notes and it does not seem right to me. I just wanted to have my thoughts verified or falsified.

$(X_n)_{n\in\mathbb{N}}$ is a sequence of random variables and $X_n \to X$ almost surely for some r.v. $X$. The claim is as follows. $$P\{X_n > b\} \to P\{X > b\}$$ for any $b \in \mathbb{R}$.

I have a counter example for this claim. Take a probability space with two elements: $\omega_1,\omega_2$. Set $P\{\omega_1\} = P\{\omega_2\} = 1/2$.

Define $X_n(\omega_1) = 1 + \frac{1}{n}$ and $X_n(\omega_2) = 0$. Then define $X(\omega_1) = 1$ and $X(\omega_2) = 0$ so that the convergence hypothesis holds.

Take $b = 1$. $P\{X_n > b\} = 1/2$ for every $n$ but $P\{X > b\} = 0$. So the assertion in the lecture notes seems to be false. The reason is as follows.

While it is correct by the dominated convergence theorem that

$$\lim_{n\to\infty}P\{X_n > b\} = E[ \lim_{n\to\infty} 1_{X_n > b}]$$

it is not true that

$$E[ \lim_{n\to\infty} 1_{X_n > b}] = E[ 1_{\underbrace{\lim_{n\to\infty} X_n}_X > b}]$$ since the indicator function is not continuous.

  • 1
    $\begingroup$ Your counterexample looks convincing to me. $\endgroup$ – Giuseppe Negro Sep 23 '18 at 12:25
  • $\begingroup$ @GiuseppeNegro Thanks. $\endgroup$ – Calculon Sep 23 '18 at 13:30

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