Confusion when using continuity of the $\Phi$ function for convergence in probability

Question

Suppose that $X\sim N(0, 1)$. Then $P(X<2)=\Phi(2)$. Suppose now that $Y_n$ is a sequence of random variables so that $Y_n\xrightarrow{p} 2$. I think that the following statement holds $$P(X<Y_n)\rightarrow P(X<2)$$

Proof: Since $\Phi$ is continuous, apply it to two sides of $Y_n\xrightarrow{p} 2$, leading $$\Phi(Y_n) \xrightarrow{p} \Phi(2)$$ By definition of $\Phi$, it follows that $$P(X<Y_n)\xrightarrow{p} P(X<2)$$ Note: convergence of a sequence of real numbers.

I am now confused.

1. Can I write $P(X<Y_n)=\Phi(Y_n)$?
2. Can I just simply throw $p$ away in the last statement because it involves only real numbers?

Answer 1

1. No. the left side is a number and right side is a random variable.

2. $P(X<Y_n)\xrightarrow{p} P(X<2)$ does not even make sense. $\xrightarrow{p} $ is for random variables, not for constants. But if you treat constants as r.v's, yes, you can drop the $p$ above the arrow.

Here is a valid proof of the statement: $$P(X<Y_n) \leq P(X<2+\epsilon)+P(Y_n >2+\epsilon) \to \Phi (2+\epsilon)+0$$ and

$$P(X<2-\epsilon) \leq P(Y_n<2-\epsilon)+P(X<Y_n) .$$ So $$P(X<Y_n)\geq P(X<2-\epsilon)-P(Y_n<2-\epsilon) \to \Phi (2-\epsilon)-0.$$ Now finish the proof using continuity of $\Phi$.

Answer 2

Based on the excellent proof given by @Kavi, I present here a complete version.

Take arbitrarily an $\varepsilon>0$, we need to show that there exist $n_0$ so that $\forall n>n_0$, we have $|P(X<Y_n)- P(X<2)|<\varepsilon$.

Let $f(y)=\Phi(2+y)-\Phi(2-y)$. Since $\Phi()$ is continuous, $f(y)$ is continuous. Hence, there exist $\delta>0$ so that $|f({\delta})-f(0)|<\varepsilon/2$, which leads to \begin{equation} \begin{aligned} |\Phi(2+\delta)-\Phi(2-\delta)-\left(\Phi(2+0)-\Phi(2-0)\right)|<\varepsilon/2\Rightarrow |\Phi(2+\delta)-\Phi(2-\delta)| <\varepsilon/2 \end{aligned} \end{equation} Note that $\Phi$ is monotonically increasing, we then have $$0< \Phi(2+\delta)-\Phi(2-\delta) <\varepsilon/2$$

Because $Y_n\xrightarrow{p} 2$, it follows that $\mathop {\lim }\limits_{n \to \infty }P(|Y_n-2|\geq \gamma)=0$ $\forall \gamma>0$. In particular, $\mathop {\lim }\limits_{n \to \infty }P(|Y_n-2|\geq \delta)=0$. Thus, there exists $n_0$ so that for all $n>n_0$, $P(|Y_n-2|\geq \delta)<\varepsilon/4$.

I now prove that $\forall n>n_0$, we have $|P(X<Y_n)-P(X<2)|<\varepsilon$.

Take arbitrarily an $n>n_0$. Note that if event $\{X<Y_n\}$ occurs then at least one of the two following events occurs: $\{X<2+\delta\}$ and $\{Y_n>2+\delta\}$ (Otherwise, their complements, i.e., $\{X\geq 2+\delta\}$ and $\{Y_n\leq 2+\delta\}$, occur, which leads to the concurrence of $\{X\geq Y_n\}$, a contradiction). Therefore, \begin{equation} \begin{aligned} P\left(X<Y_n\right)&\leq P\left(\{X<2+\delta\}\cup\{Y_n>2+\delta\}\right)\leq P\left(X<2+\delta\right) + P\left(Y_n>2+\delta\right)\\ &<\Phi\left(2+\delta\right)+ \varepsilon/4 \end{aligned} \end{equation} (Since $P\left(Y_n>2+\delta\right) + P\left(Y_n<2-\delta\right)=P\left(|Y_n-2|>\delta\right)<\varepsilon/4$)

Using a similar argument, we have: \begin{equation} \begin{aligned} &P\left(X<2-\delta\right)\leq P\left(\{X<Y_n\}\cup\{Y_n<2-\delta\}\right)\leq P\left(X<Y_n\right) + P\left(Y_n<2-\delta\right)\\ &\Rightarrow P\left(X<Y_n\right) \geq P\left(X<2-\delta\right) - P\left(Y_n<2-\delta\right)>\Phi(2-\delta)-\varepsilon/4\\ \end{aligned} \end{equation} Thus, \begin{equation} \Phi(2-\delta)-\varepsilon/4 < P\left(X<Y_n\right)< \Phi\left(2+\delta\right)+ \varepsilon/4 \end{equation} We also have \begin{equation} \Phi(2-\delta)-\varepsilon/4 < \Phi(2)=P\left(X<2\right) < \Phi\left(2+\delta\right)+ \varepsilon/4 \end{equation} Hence, \begin{equation} |P(X<Y_n)- P(X<2)|< \Phi\left(2+\delta\right)+ \varepsilon/4 - \left(\Phi(2-\delta)-\varepsilon/4\right)<\varepsilon/2+\varepsilon/4+\varepsilon/4=\varepsilon. \end{equation}