Crest Factor of OFDM signal
In statistical case OFDM (without specific PAPR reduction precoding and with most of subcarriers used) signal could be modelled as white Gaussian process. So PDF of OFDM signal is well-known one for normal distribution:

$f(x) = 1/(\sigma\cdot\sqrt{2\cdot\pi}) \cdot \exp(-(x-\mu)^2/(2\sigma^2))$

For OFDM $\mu=0$. When we calculate crest factor we need to define the number of samples to be used for measurement. We say measurement for $1e+6$ samples is valid since crest factor doesn't increase significantly, when we take more values. In another words if we have e.g. $1e+2...1e+4$ samples, crest factor will vary from one measurement to another. But if we have a million of samples or more, crest factor of white noise or OFDM will converge to one specific value, it is about 13.5 dB. You can check it easily with MATLAB. 

It can be derived analitically with aid of PDF of normal distribution. Look, we need to estimate how much peak value exceeds $\sigma$ to estimate crest factor. If for example $1e+6$ samples are used the probability of peak value is $1e-6$. It is rought approximation. For normal distribution large peaks have low probabilty and it decreases with peak amplitude increase (see plot of normal distr. PDF). We need to estimate maximum peak for million samples, so it is one peak for million and its probabilty is $1e-6$. 

The equation to be solved is

$P(x)=\int_x^\infty f(x)dx = 1e-6$,

where $f(x)$ is presented above.

To avoid infinite integration we can roughtly simplify (forgive me, mathematicians!) this equation to something like this

$f(x) \le P(x) = 1e-6$. 

By usign $f(x)$ instead of $P(x)$ and solving this equation for different $\sigma$ we can find out that $4.5\sigma \le x \le 5\sigma$ in general. The creast factor could be find as $20\cdot\log_{10}(x/\sigma)\approx 13...14$. The result is very similar to what we can observe in practice. It can be more accurate if someone suggest the way to approximate $\int_x^\infty f(x)dx$ term in the original equation (I believe it could be done since $f(x)$ is decreasing in the interval from $x$ to $\infty$).