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Say, $R_i$ $(i=1,2,\ldots,n)$ is a binary variable.

I have an estimator $n^{-1}\sum_{i=1}^{n}(1-R_i)$.

How can I show $n^{-1}\sum_{i=1}^{n}(1-R_i)$ converges in probability to $P(R_i=0)$?

$\lim_{n\to \infty}n^{-1}\sum_{i=1}^{n}(1-R_i)=?$

Say, $R_i$ $(i=1,2,\ldots,n)$ is a binary variable.

I have an estimator $n^{-1}\sum_{i=1}^{n}(1-R_i)$.

How can I show $n^{-1}\sum_{i=1}^{n}(1-R_i)$ converges in probability to $P(R_i=0)$? That is,

$\lim_{n\to \infty}n^{-1}\sum_{i=1}^{n}(1-R_i)=?$

I know $\sum_{i=1}^{n}(1-R_i)P(1-R_i)=\mathbb E(1-R_i)$.

But is $\lim_{n\to \infty}n^{-1}\sum_{i=1}^{n}(1-R_i)=\mathbb E(1-R_i)$?

Say, $R_i$ $(i=1,2,\ldots,n)$ is a binary variable.

What will be the expectation, $\mathbb E[1-R_i]$?

Is it $\mathbb E[1-R_i]=\sum_{i=1}^{n}(1-R_i)P(R_i=0)$?

But if anyone show that $n^{-1}\sum_{i=1}^{n}(1-R_i)$ converges in probability to $P(R_i=0)$, does he assume $P(R_i=0)=\lim_{n\to \infty}\frac{1}{n}$ ?

That is, to show $n^{-1}\sum_{i=1}^{n}(1-R_i)\overset{p}{\to}P(R_i=0)$, is the following way correct:

$\lim_{n\to \infty}n^{-1}\sum_{i=1}^{n}(1-R_i)=\mathbb E[1-R_i]=P(R_i=0)$?

Say, $R_i$ $(i=1,2,\ldots,n)$ is a binary variable.

I have an estimator $n^{-1}\sum_{i=1}^{n}(1-R_i)$.

How can I show $n^{-1}\sum_{i=1}^{n}(1-R_i)$ converges in probability to $P(R_i=0)$?

$\lim_{n\to \infty}n^{-1}\sum_{i=1}^{n}(1-R_i)=?$

Say, $R_i$ $(i=1,2,\ldots,n)$ is a binary variable.

What will be the expectation, $\mathbb E[1-R_i]$?

Is it $\mathbb E[1-R_i]=\sum_{i=1}^{n}(1-R_i)P(R_i=0)$?

But if anyone show that $n^{-1}\sum_{i=1}^{n}(1-R_i)$ converges in probability to $P(R_i=0)$, does he assume $P(R_i=0)=\lim_{n\to \infty}\frac{1}{n}$ ?

That is, to show $n^{-1}\sum_{i=1}^{n}(1-R_i)\overset{p}{\to}P(R_i=0)$, is the following way correct:

$\lim_{n\to \infty}n^{-1}\sum_{i=1}^{n}(1-R_i)=\mathbb E[1-R_i]=P(R_i=0)$.

Say, $R_i$ $(i=1,2,\ldots,n)$ is a binary variable.

What will be the expectation, $\mathbb E[1-R_i]$?

Is it $\mathbb E[1-R_i]=\sum_{i=1}^{n}(1-R_i)P(R_i=0)$?

But if anyone show that $n^{-1}\sum_{i=1}^{n}(1-R_i)$ converges in probability to $P(R_i=0)$, does he assume $P(R_i=0)=\lim_{n\to \infty}\frac{1}{n}$ ?

That is, to show $n^{-1}\sum_{i=1}^{n}(1-R_i)\overset{p}{\to}P(R_i=0)$, is the following way correct:

$\lim_{n\to \infty}n^{-1}\sum_{i=1}^{n}(1-R_i)=\mathbb E[1-R_i]=P(R_i=0)$?