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Let $X_{n}$ denote a sequence of random variables. Then, $X_{n} = c + o_\text{P}(1)$ for some constant $c$ if, for all $\epsilon > 0$, $$\Pr\left(\left|X_{n} - c \right| \geq \epsilon \right) \to 0$$ for $n \to \infty$. Under this definition, $X_{n} \xrightarrow{P} c$.

I know the invariance property allows $f(X_{n}) \xrightarrow{P} f(c)$ for continuous functions $f$. I would like to see if this also holds for indicator functions.

Let $P_{n}$ denote a sequence of random variables separate from $X_{n}$, and let $\Delta\left(P_{n} \leq f\left(X_{n}\right) \right)$ denote an indicator function that returns one if the condition inside its argument is true, zero otherwise.

My question is, assuming $f$ is a smooth function and $X_{n} \xrightarrow {P} c$, does $$\Delta\left(P_{n} \leq f\left(X_{n}\right)\right) = \Delta\left(P_{n} \leq f\left(c\right) + o_{P}(1) \right)$$ imply $$\Delta\left(P_{n} \leq f\left(X_{n}\right)\right) = \Delta\left(P_{n} \leq f\left(c\right) \right) + o_{P}(1)?$$

I'm writing this post in a busy restaurant on my phone, but I still want to try to explain the avenues I've taken. My apologies if some details are omitted. I started by using the definition of $o_{P}$ and got nowhere due to the other $o_{P}$ within the indicator function's argument. Then, desperately, I wanted to see if $\Delta$ was a Bernoulli distributed random variable, which surprisingly (/s) didn't work either.

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  • $\begingroup$ @henry, sorry, I corrected the post. Thank you for pointing that out! $\endgroup$
    – JerBear
    Commented 2 days ago
  • $\begingroup$ Now I do not understand $\Delta\left(P_{n} \leq f\left(X_{n}\right)\right) = \Delta\left(P_{n} \leq f\left(c\right) + o_{P}(1) \right)$ is saying. In any case, what happens if you remove the randomness with $f(x)=x$, $X_n=c+\frac1n$, $P_n=c+\frac2n$ or with $f(x)=x$, $X_n=c+\frac2n$, $P_n=c+\frac1n$ ? $\endgroup$
    – Henry
    Commented 2 days ago
  • $\begingroup$ For this line, I'm replacing $f(X_{n})$ with $f(c) + o_{P}(1)$, assuming $f(c)$ is continuous at $c$. This I can do assuming that the sequence of random variables $X_{n}$ converge in probability to a constant. I think this comment actually helped me get closer to an answer, which I can post tomorrow God's will (still waiting on the shrimp scampi 😭) $\endgroup$
    – JerBear
    Commented 2 days ago

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