Application WLLN in a convergence related problem

Question

So, I have stumbled upon a question like this:

${X_n}_{\{n\geq1\}}$ be a sequence of iid random variables having common pdf $f_X(x)=xe^{-x}I_{x>0}$. Define $\overline{X_n}=\frac{1}{n}\sum_{i=1}^{n}{X_i},n=1,2,...$. Then $\lim{n\to \infty}P(\overline{X_n}=2)=$

1. $0$
2. $1$
3. $1/2$
4. $1/4$

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Now,as $X_i$ is a continuous ranfom variable $\overline{X_n}$ is also continuous. So,$P(\overline{X_n}=2)=0$ for any $n$. But Weak of of Large Numbers says that $\overline{X_n}$ converges in probability to $2$ ( as $E(\overline{X_n})=2)$ which means(?) $\overline{X_n}=2$ with probability $1$. So, I must be missing something here. Which option is correct? Any help would be very much appreciated!

Answer 1

We say that a sequence $(X_n)_n$ of random variables converges in probability to the random variable $X$ (notation: $X_n \stackrel{\mathbb{P}}{\to} X)$ if

$$\forall \epsilon > 0: \lim_n\mathbb{P}(|X_n-X| \geq \epsilon)=0$$

If $\overline{X_n} = 2$ with probability $1$ then for all $\epsilon > 0$ we have $\mathbb{P}(|\overline{X_n}-2| \geq \epsilon) = 0$ and thus we have $\overline{X}_n \stackrel{\mathbb{P}}{\to} 2$ trivially.

There is no contradiction. Of course $\overline{X}_n = 2$ with probability $1$ is stronger than convergence in probability.

Answer 2

Let $S_n = \sum_{i=1}^n X_i$. Then the density of $S_1$ is $f_1(x)= \frac{x^{2\cdot1-1}}{(2\cdot1-1)!}e^{-x}$. Assume that the density of $S_n$ is $f_n(x) = \frac{x^{2n-1}}{(2n+1)!}e^{-x}$ for some positive integer $n$. Then the density of $S_{n+1}$ is given by \begin{align} f_{n+1}(x) &= f_{S_n}\star f_1(x)\\ &= \int_0^x \frac{y^{2n-1}}{(2n+1)!}e^{-y} (x-y)e^{-(x-y)}\ \mathsf dy\\ &= \frac{x^{2(n+1)}}{(2n+3)!}e^{-x}, \end{align} so by induction the density of $S_n$ is $f_n(x) = \frac{x^{2n-1}}{(2n+1)!}e^{-x}$, $n=1,2,\ldots$. The mean of $S_n$ may be computed by $$ \int_0^\infty \frac{x^{2n}}{(2n-1)!}e^{-x}\ \mathsf dx = 2n, $$ and hence $\mathbb E[\overline X_n] = 2$. By the weak law of large numbers, $$ \lim_{n\to\infty} \mathbb P\left(\overline X_n=2\right) = 1. $$