A possible misconception on the weak law of large numbers

Question

Let $\{X_n\}_{n\geq1}$ be a sequence of i.i.d. random variables having common probability density function $f(x)=xe^{-x}, x\geq0$

Let $\bar{X_n}=\frac{1}{n}\sum_{i=1}^nX_i$

Then, $\lim_{n\to\infty}P(\bar{X_n}=2)$ equals??

$\because \bar{X_n}\xrightarrow[n\to\infty]{P} 2=E[X_i].$ (hence, $\bar{X_n}$ converges in probabilty to a discrete distribution, that is equal to $2$ with probability $1$)

$\therefore \forall \epsilon>0, \lim_{n\to\infty}P(|\bar{X_n}-2|<\epsilon)=1$

$\therefore \lim_{n\to\infty}P(\bar{X_n}=2)=1 $ (here I derive intuition from the result in real analysis that says, that if two numbers are arbitrarily close to each other, then those numbers are the same.)

But, the given answer is that $\lim_{n\to\infty}P(\bar{X_n}=2)=0$.

So, what did i do wrong? and how to prove that $\lim_{n\to\infty}P(\bar{X_n}=2)=0$

Also, looking at the answer, I get the feeling that even as $n\to\infty, \bar{X_n}$ still remains a continuous random variable. But this is very counterintuitive, because all the probability of $\bar{X_n}$ is getting concentrated inside smaller and smaller intervals around $2$, but still in the limit, $\bar{X_n}$ does not become a discrete random variable. How is this possible?

Note: I do know that $\lim_{n\to\infty}P(|\bar{X}_n-\mu|\leq\epsilon)=1$ does not imply that, for large $n$, $|\bar{X}_n-\mu|\leq\epsilon$

Answer 1

As you correctly pointed out $\bar{X}_n \xrightarrow{p} 2$ but this is a slightly different statement than $\lim_{n\to\infty}P(\bar{X}_n =2) =1$.

Let $\bar{X}_n$ be the sequence of random variables representing the sample mean of $n$ iid samples of $X$. Being averages of continuous variables, each of these has a differentiable cumulative distribution function $F_n$ on $(0,\infty)$ (i.e., density function) denoted $f_n$.

Note that $\bar{X}_n \xrightarrow{p} 2 \implies F_n(x) \to \mathbf{1}_{\geq2}(x):=F(x)$, where $\mathbf{1}_A(x)$ is the indicator function.

Using the properties of CDFs, we define

$$P_n(x) := P(\bar{X}_n=x) = \lim_{\delta \to 0^+} \left[F_n(x+\delta)-F_n(x-\delta)\right]$$

Since each $F_n$ is differentiable at $x=2$, it is also continuous at $x=2$, hence

$$P_n(2)= 0 \;\;\forall n>0 \implies \lim_{n\to\infty} P_n(2) = 0 \neq \lim_{\delta \to 0^+} \left[F(2+\delta)-F(2-\delta)\right]$$

Put a different way

$$\lim_{n\to \infty}\lim_{\delta \to 0^+} \left[F_n(2+\delta)-F_n(2-\delta)\right] \neq \lim_{\delta \to 0^+}\lim_{n\to \infty} \left[F_n(2+\delta)-F_n(2-\delta)\right]$$

Your confusion comes from assuming these limits can be interchanged.

Answer 2

''here I derive intuition from the result in real analysis that says, that if two numbers are arbitrarily close to each other, then those numbers are the same''. The numbers here are no arbitrarily close to each other as you said in the note, becasue $X_n(\omega)$ is a different number for each $n$.

To prove that the limit is in fact zero, notice that the sum of i.i.d variables with density is a random variable with density (therefore for each fixed $n, \ P(\overline{X}_n = a)=0$ for every $a \in \mathbb{R}$).

To visualize this example, imagine that for each $n$ the density of $\overline{X}_n$ is a curve in $\mathbb{R}$ which as you said is getting closer to the constant $2$. However for each fixed $n$ the random variable has density.