# Do general normal distribution and standard normal distribution have the same domain or not?

a random variable is understood as a measurable function defined on a probability space whose outcomes are typically real numbers.

every function has its domain.

I assume the domain of standard normal distribution is $$(-\infty, \infty)$$ (wiki does not claim this, so, more solid References may be needed)

what wiki does claim is

Every normal distribution is a version of the standard normal distribution whose domain has been stretched by a factor $$\sigma }$$ (the standard deviation) and then translated by $$\mu }$$ (the mean value)

considered the stretch and translation, Do general normal distribution and standard normal distribution have the same domain or not?

• The domain of the cumulative distribution function is the entire real line for any normally distributed RV. The domain of the RV itself is the sample space $\Omega$ from when we write: $X:\Omega \to \mathbb{R}$, which we sweep under the rug. The support of the probability density function is the set of reals where the density is non-zero, in the case of normal densities, it is again the entire real line. Aug 3, 2019 at 2:40
• Thanks for your comments. To be clear, your "entire real line" and my "$(-\infty, \infty)$" are the same thing, right? Aug 3, 2019 at 2:50
• Yes, of course. Aug 3, 2019 at 2:59
• @NapD.Lover consider the Bernoulli model, the $\Omega$ is {0,1}, how to apply your $X:\Omega \to \mathbb{R}$ in this particular case? What is the domain? Aug 7, 2019 at 1:46
• the sample space doesn’t have to be $\{0,1\}$ for Bernoulli trials, it could be an abstract categorical set of any two events like “Heads” and “Tails” but that can be encoded as binary events with the standard representation of $\{0,1\}$. If $\omega\in \Omega$ then $X(\omega)$ is equal to either $0$ or $1$. Under the Bernoulli distribution for a fixed chance $p$ it does not matter whether $\Omega=\{0,1\}$ or $\Omega=\{T,H\}$ because, provided $H$ is encoded as $1$ and $1$ is encoded as $1$, then $X(H)=X(1)=1$ with probability $p$, etc. The events are distributed identically. Aug 7, 2019 at 2:14

As explained in the comments, you are using slightly unusual terminology which is causing some of your confusion.

In fact, I would suggest you to forget about random variables for this question - it is not adding anything and only confusing things. Instead, your question is about the standard normal distribution, which can be rigorously defined as the object which assigns to each interval $$(a,b)\subseteq \mathbb R$$ the number $$\frac{1}{\sqrt{2\pi}}\int_a^b e^{-x^2/2}\ dx.$$ Notice there is no random variable involved in this definition. (The number assigned to the interval $$(a,b)$$ can be interpreted as the probability that a standard normal random variable lies in $$(a,b)$$ - but I would recommend thinking of this as a consequence of the definition.)

Saying that the "domain" (or more precisely, "support") of this distribution is $$\mathbb R$$ simply means that all non-empty intervals $$(a,b)$$ are assigned a strictly positive number by the above formula. In fact, this is easy to see since you can get a lower bound for the integral by using the minimum of the density (which always occurs at one of the endpoints), and integrating a positive number over an interval of positive size yields another positive number. Common examples of distributions where this is not the case would be the Uniform$$[a,b]$$ distribution, the Exponential$$(\lambda)$$ distribution, or any discrete distribution - in each of these cases, you can find an interval that gets assigned $$0$$ - for instance, $$(a-2,a-1)$$ in the first case, $$(-2,-1)$$ in the second case, and any interval that avoids the discrete values in the third case.

Now I will clarify what is meant when it is said that every normal distribution is obtained by translating and stretching a standard normal distribution. The normal distribution with mean $$\mu$$ and variance $$\sigma^2$$ can be defined rigorously as the object which assigns to the interval $$(a,b)$$ the number $$\frac{1}{\sqrt{2\pi}}\int_a^b e^{-(\frac{x-\mu}{\sigma})^2/2}\ dx.$$ You will notice that the formula is the same as before, except $$x$$ has been replaced by $$\frac{x-\mu}{\sigma}.$$ If you think about what the operation $$x\mapsto (x-\mu)/\sigma$$ does, you will see that it has the effect of shifting the real line $$\mathbb R$$ over by $$\mu$$, and then stretching by a factor $$\sigma$$. However, this does not mean that the domain itself is changing: all that has happened is that the domain has been relabeled, but as a whole it has not changed.

• Thank a lot. Your answer is very helpful. By "slightly unusual terminology", which terminology does that mean? Aug 7, 2019 at 1:28
• The "slightly unusual terminology" is what you cited from the wiki that refers to the domain of the standard normal distribution. The more common term for this concept is the support of a distribution, not the domain. Aug 7, 2019 at 3:20
• In this particular case, are "domain" and "support" the same thing? Aug 7, 2019 at 9:03
• The part of the wiki you are quoting from (where it refers to domain of the normal distribution) is using it to mean the support. If you ask most mathematicians if domain and support are the same thing, they would say no - which is why I described the terminology as slightly unusual. Aug 8, 2019 at 3:53
• "these cases" refers to Uniform$[a,b]$, Exponential$(\lambda)$, and any distribution supported on a discrete set - for instance, geometric distribution, dirac delta distribution, poisson distribution, etc etc Aug 9, 2019 at 2:45