Say we now a Random Variable X is normal $X \sim N(\mu, \sigma) $

Then we know that :

$Z= (X-\mu)/\sigma $ ~$N(0,1)$

I answered in another question that we know Y is normal if it can be obtained through the inverse transform $Z^{-1}$ , inverse to above, i.e., X is normal iff:

$X== \sigma Z + \mu $ ,

for some Real $\sigma >0, \mu $

Is this correct?


2 Answers 2


If I've understood your question...

If $Z$ is a standard normal random variable $Z \sim \mathcal{N}(0,1)$

Then $X = \mu + \sigma Z $ is a normal random variable $X \sim \mathcal{N}(\mu, \sigma^2)$.


Based on OP's "iff", I think we need to prove that:

Given $Z$ follows standard normal distribution,

  1. For real numbers $(\mu, \sigma) \in R^2$, $Y=\mu + \sigma Z$ follows normal distribution, as given by Demetri Pananos.

  2. If $Y$ follows normal distribution, then there exist real numbers $(\mu, \sigma) \in R^2$ such that $Y=\mu + \sigma Z$. It is true because $\left(E(Y), \sqrt{Var(Y)}\right)$ meet the requirement.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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