I'm dealing with the following problem. Given $X_1,\ldots,X_n \sim^{iid}$ with density function:
$$ f_{X}(x) = \dfrac{1}{8} x e^{-\dfrac{|x|}{4}} \; \mathbb{I}_{(-\infty,\infty)} (x) $$
Define: $ U_{n} = \dfrac{\sum_{i=1}^{n} |x_i| }{n-2}$. Prove
$$U_n \overset{P}{\to} c$$
And find the value of $c$.
My attemp: Rearranging,
$$ U_{n} = \dfrac{n}{n-2} \dfrac{\sum_{i=1}^{n} |x_i| }{n} $$.
Using the LLN: (I don't know if I can do this)
$$ \dfrac{\sum_{i=1}^{n} |x_i| }{n} \overset{P}{\to} \mathbb{E}(|X|) $$.
So If that's correct, I need to compute $\mathbb{E}(|X|) $.
$$ \mathbb{E}(|X|) = \int_{-\infty}^{\infty} |x| \dfrac{1}{8} x e^{-\dfrac{|x|}{4}} dx = \int_{-\infty}^{0} - \dfrac{1}{8} x^2 e^{\dfrac{x}{4}} dx + \int_{0}^{\infty} \dfrac{1}{8} x^2 e^{-\dfrac{x}{4}} dx = 0$$
So I think, $U_n \overset{P}{\to} 0 $ because $\dfrac{n}{n-2} \to 1$ when $n \to \infty$
But I'm not sure.
The density should be $\displaystyle\frac{|x|}{8}\exp(-|x|/4)$ and now notice that $\displaystyle\int_{-\infty}^{\infty}\frac{|x|}{8}\exp(-|x|/4)\,dx=\frac{2}{8}\int_{0}^{\infty}xe^{-x/4}\,dx=1$ and is indeed a valid density.
The initial definition that you were using is faulty as $P(X\leq 0)=\frac{1}{8}\int_{-\infty}^{0}xe^{-x/4}\,dx=-\frac{1}{2}$ .
So you are getting negative probability. In particular, a density function must be non-negative (or if you know measure theory, then it is non-negative outside a set of measure $0$).
Using this, you now have that $E(|X|)=\int_{-\infty}^{\infty}\frac{1}{8}|x|^{2}e^{-|x|/4}\,dx=\frac{1}{4}\int_{0}^{\infty}x^{2}e^{-x/4}\,dx=32$
Also note that you should never get $E(|X|)=0$. If you do get, then it means that $X=0$ (or if you know measure theory, then $X=0$ almost surely).
Now you can apply the Strong Law of large numbers and conclude that $\frac{\sum_{k=1}^{n}|X_{i}|}{n}\xrightarrow{a.s.} 32$ (or if you want convergence in probability, then fine, use the weak law and say that $\frac{\sum_{k=1}^{n}|X_{i}|}{n}\xrightarrow{P} 32$ )
(I don't know why at all you are unsure if you can use the weak law/strong law. They both only require iid-ness and that $E(|X_{1}|)<\infty$ as assumptions.)
And as $\frac{n-2}{n}\to 1$, you also have $\frac{\sum_{k=1}^{n}|X_{i}|}{n-2}\xrightarrow{a.s.} 32$