0
$\begingroup$

1) Does sum of two independent folded normal distribution always yield a folded normal distribution?

(if independent random variables X and Y follow normal distributions, then does $|X| +|Y|$ equal to some folded normal distribution?)

2) If so, then given the mean and variances of those two independent folded normal distribution, what are the mean and variance of the sum of those distributions?

(Suppose $|X|$ ~ $FN(\mu_{x},\sigma_{x})$, $|Y|$ ~ $FN(\mu_{y},\sigma_{y})$, then what are the mean and variance of $|X| + |Y|$ in terms of $\mu_{x},\mu_{y},\sigma_{y},\sigma_{x} $ if possible?)

I looked at other sources but it seems folded normal distribution is something not talked about a lot.

$\endgroup$
0
$\begingroup$

To answer (2), we still have linearity of expected value and variance, i.e. $$ \mathbb{E}[|X| + |Y|] = \mathbb{E}[|X|] + \mathbb{E}[|Y|] $$ and same for variance. You should be able to find the mean and variance of the folded normal from the parameters of the original normal using the definition, e.g. $$ \mathbb{E}[|X|] = \int_{-\infty}^\infty |x| \phi(x; \mu, \sigma) dx... $$

$\endgroup$
  • $\begingroup$ By the way, do you know how that expectation would change if I were to square root that? i.e. E[sqrt(|X|)]? and thanks for the confirmation on the answer. $\endgroup$ – Mike Chen Sep 27 '17 at 17:40
  • $\begingroup$ @MikeChen No earthly idea, but at least for the standard case you get $$ \mathbb{E}\left[\sqrt{|X|}\right] = \int_{-infty}^\infty \sqrt{|x|} \phi(x) dx = 2 \int_)^\infty \sqrt{x} e^{-x^2/2}dx = \frac{\Gamma(3/4)}{\sqrt[4]{2}} $$ and the indefinite integral is obtained from Wolfram Alpha at [wolframalpha.com/input/?i=Integrate%5BSqrt%5Bx%5D+*+Exp%5B-x%5E2%2F2%5D,+%7Bx,0,Infinity%7D%5D] $\endgroup$ – gt6989b Sep 27 '17 at 18:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.