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Post Closed as "Needs details or clarity" by Snoop, Leucippus, Harish Chandra Rajpoot, Kurt G., amWhy
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JerBear
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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 $\delta, \epsilon > 0$$\epsilon > 0$, there exists some $n_{0}$ such that,$$\Pr\left(\left|X_{n} - c \right| \geq \epsilon \right) \to 0$$ for all $n > n_{0}$, $$\Pr\left(\left|\frac{X_{n}}{c_{n}}\right| > \delta \right) \leq \epsilon.$$$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.

Let $X_{n}$ denote a sequence of random variables. Then, $X_{n} = c + o_\text{P}(1)$ for some constant $c$ if, for all $\delta, \epsilon > 0$, there exists some $n_{0}$ such that, for all $n > n_{0}$, $$\Pr\left(\left|\frac{X_{n}}{c_{n}}\right| > \delta \right) \leq \epsilon.$$ 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.

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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JerBear
  • 185
  • 7

Does the invariance property hold for consistent estimators within an indicator function?

Let $X_{n}$ denote a sequence of random variables. Then, $X_{n} = c + o_\text{P}(1)$ for some constant $c$ if, for all $\delta, \epsilon > 0$, there exists some $n_{0}$ such that, for all $n > n_{0}$, $$\Pr\left(\left|\frac{X_{n}}{c_{n}}\right| > \delta \right) \leq \epsilon.$$ 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.