Why is the expected length of an interval containing a point $\mathbb{E}\left[X^2\right]/\mathbb{E}[X]$ if interval length is IID? 
In general, for processes, in which the interarrival intervals with distribution $F_X(x)$ are IID, the expected length of an arbitrary chosed interval is $\mathbb{E}\left[X^2\right]/\mathbb{E}[X]$. We see that for the above parts, the formula is certainly valid.

Why is this statement true?
For context:
Suppose there existed an infinitely long line that is divided into IID intervals with lengths following some distribution. Let the random variable $X$ denote the length of an interval. If I arbitrarily picked a point on this line, the expected length of the interval containing this point is $\mathbb{E}\left[X^2\right]/\mathbb{E}[X]$.
Let $\mathbb{E}[X_s]$ be the expected length from start of interval to the point.
Let $\mathbb{E}[X_e]$ be the expected length from the point to end of interval.
I understand that if such distribution is exponential, then the expected length is $\mathbb{E}[X_s] + \mathbb{E}[X_e]$, which happens to be $\mathbb{E}\left[X^2\right]/\mathbb{E}[X]$. However, I'm not sure why this holds for the general case.
Edit:
For full context, this is taken from problem 4 on this pset. The quote is from the 4.b) answer
Edit 2:
The interval choice strategy is to first pick a point, then choose the interval containing this point. As @user6247850 mentions, larger intervals are favored.
 A: I will first take some care to make sure everything is set up rigorously. Let $X_1,X_2,\dots$ be the iid sequence of waiting times, all distributed like $X$, and let $T_n=X_1+X_2+\dots+X_n$ denote the $n^\text{th}$ arrival time.
We are interested in the length of the interval containing some particular point $t$. I will denote this by $L_t$. In mathematical terms,
$$
L_t=T_{N(t)+1}-T_{N(t)},\qquad \text{where }N(t)= \sup\{n:T_n\le t\}
$$
Finally, let
$$
\ell(t)=E[L_t]
$$
We can then prove the following theorem, which gives a rigorous interpretation to your question:

Theorem: $\lim_{t\to\infty} \ell(t)=E[X^2]/E[X]$.

Proof: I will assume that $X$ has a pdf of $f_X(x)$ for simplicity, but this is actually not necessary. Using the law of total expectation with respect to $X_1$, we get
$$
\ell(t) = \int_0^t f_X(x) \ell(t-x)\,dx+\int_t^\infty x f_X(x)\,dx
$$
Note that $\int_t^\infty x f_X(x)\,dx=E[X\cdot {\bf 1}(X\ge t)]$ is just some function of $t$, which I will denote $g(t)$ for convenience. It is well known that $E[X]<\infty$ implies $\lim_{t\to\infty} g(t)=0$ (proof: dominated convergence theorem).The above equation can be written succinctly using a convolution:
$$
\ell = f*\ell + g
$$
This convolution equation has a solution in terms of the renewal function $m(t)$, defined as follows:
$$
m(t)=E[N(t)]=E\big[\text{number of arrivals in $[0,t]$}\big]
$$
Using the methods at this page, (see box numbered $24$), we can prove that
$$
\ell(t) = g(t)+\int_0^tg(s) m'(t-s)\,ds,
$$
Now that we have "solved" for $\ell(t)$, we just need to determine its long term behavior. The key renewal theorem* is exactly suited to this purpose, and says that
$$
\lim_{t\to\infty} \int_0^t g(s)m'(t-s)\,ds=\frac1{E[X]}\int_0^\infty g(s)\,ds
$$
All that remains is to prove $\int_0^\infty g(s)\,ds=E[X^2]$, which is done as follows:
$$
\int_0^\infty g(s)\,ds=\int_0^\infty E[X\cdot {\bf 1}(X>s)]\,ds=E\left[X\int_0^\infty {\bf 1}(X>s)\,ds\right]=E[X\cdot X]
$$
*A condition for applying the key renewal theorem is that $\int_0^\infty g(s)\,ds <\infty$. Fortunately, the very last computation shows this is true as long as $E[X^2]<\infty$.
