Notation and Transformation Rules of the Normal Distribution So, I'm wondering. So far I have seen both notations:
(1) $X \sim \mathcal{N}(\mu, \sigma^2)$ and
(2) $X \sim \mathcal{N}(\mu, \sigma)$
where $\mu$ is the mean.
My first questions is directed to $\sigma$.  
In (1), does it resemble the variance and in (2) the standard deviation?
I assume there is no 'correct' notation, but it is rather a choice of preference, isn't it?
I also saw the following notation
(3) $\mu + \sigma \cdot \mathcal{N}(0, 1)$
Which of the aforementioned notations, i.e.,  (1) and (2),  can be transformed to (3)? Or can both be transformed to (3) without violating any fundamental rules? And what does $\sigma$ in (3) refer to? Is it the standard deviation of $\mathcal{N}$?
Thanks in advance!
 A: The easiest way to understand that the choice of parameter results in equal outcome can be found in the probability density function (PDF). The PDF of the Normal distribution is given by:
$\mathcal{N}(\mu, \sigma^2) = \frac{1}{\sqrt{2\pi\sigma^2} } e^{ -\frac{(x-\mu)^2}{2\sigma^2}}$
The function is parametrized by $\mu$ (mean) and $\sigma$ (standard deviation) or $\sigma^2$ (variance). Both are equivalent, so is merely at the user's convenience and convention. For completeness I note that there is another parameterization people sometimes choose: its precision, given by $\frac{1}{\sigma^2}$.
In the third notation, you find the standard Normal distribution $\mathcal{N}(0, 1)$. The parameters are used to shift and scale the distribution. Here it does matter that we use the variance (as scaling by $\sigma$ multiplies the variance by $\sigma^2$)!
To illustrate this, let's say we have variable $x$ that is distributed as a standard Gaussian:
$x \sim N(0,1)$
The mean and variance are given by:
$\mu = \mathbb{E}[x] = \int x p(x) dx = 0$
$\sigma^2 = \text{var}[x] = \mathbb{E}[(x - \mu)^2] = \mathbb{E}[x^2] - \mathbb{E}[x]^2 = 1$
When we add $c$ to $x$, the mean shifts by $c$:
$\mathbb{E}[x + c] = \mathbb{E}[x] + \mathbb{E}[c] = \mathbb{E}[x] + c$
When we multiply $x$ by $c$, the variance multiplies with $c$:
$\text{var}[cx] = \mathbb{E}[(cx)^2] -\mathbb{E}[cx]^2 = \mathbb{E}[c^2x^2] - (c\mathbb{E}[x])^2 = c^2(\mathbb{E}[x^2] - \mathbb{E}[x]^2) = c^2\text{var}[x]$
These notes on shifting and scaling and on expectations might be helpfull to develop a deeper understanding. 
