We know that a Chi distribution is the distribution of the root of the sum of $k$ squared independent Gaussian random variables $Z_i \sim N(0, 1)$:
$$ Y = \sqrt{\sum_{i=0}^k Z_i^2} $$.
If we have another random variable $Z^R_I$ which is a rectified Gaussian, $$ Z_i^R = max(0, Z_i) $$,
Would it be possible to determine the Variance and Expected value for the distribution of the root of the sum of $k$ squared independent rectified Gaussian random variables $Z_i^R $ ?
$$ Y^R = \sqrt{\sum_{i=0}^k (Z_i^R)^2} $$ $$ E[Y^R]= ? $$ $$ Var[Y^R]= ? $$
We can give formulas, which are not totally closed.
First, the distribution of $Y^2$ is $\chi^2_k$, that is $\Gamma(d/2,1/2)$. The expectation of $Y^2$ is simple, $E[Y^2]=k$, but the expectation of $Y$ is not. $$E[Y] = \int_0^\infty x^{1/2} \frac{x^{k/2-1}e^{-x/2}}{2^{k/2}\Gamma(k/2)} dx = 2^{1/2}\frac{\Gamma(k/2+1/2)}{\Gamma(k/2)} =: m_k.$$
Next, for each $i$, $P[Z_i^R>0]=1/2$ and the distribution of $(Z_i^R)^2$ given $Z_i^R>0$ equals the distribution of $Z_i^2$. By using the independence and the equidistribution of the $Z_i$, we derive that the distribution of the number $N$ of indexes $i$ such that $Z_i>0$ is a binomial random variable with parameters $k$ and $1/2$. Moreover, the distribution of $(Y^R)^2$ given $N=n$ is $\chi^2_n$.
Therefore, $E[(Y^R)^2|N] = N$ and $E[Y^R|N] = m_N$. Taking expectations, we get $E[(Y^R)^2] = EN = k/2$ and $$E[Y^R] = E[m_N] = \sum_{n=0}^k m_n{k \choose n}/2^k.$$