I am trying to show that a set of numbers distributed from a Gaussian distribution {g1, g2,...,gn} with a standard deviation of 1 and a mean of zero can be transformed to a set of Gaussian numbers {G1, G2,...,Gn} that have a standard deviation of s and a mean of m by the formula:
Gi = s * gi + m (ignoring any normalization constants)
Despite my best efforts I cannot seem to find a way to prove this well known relationship true.
On
Let $E g_i = 0$ and $\text{Var } g_i = 1$, i.e. $g_i \sim \mathcal{N}(0,1)$. Now, let $$G_i = sg_i + m$$ $$E G_i = E(sg_i + m) = sEg_i +m= s(0) + m = m$$ and $$\text{Var } G_i = \text{Var }(sg_i + m) = s^2\text{Var }g_i + 0= s^2(1) = s^2 = \sigma^2$$ So $G_i$ has mean $m$ and standard deviation $s$. But scaling and shifting a Normal distribution is also normal, hence $$G_i \sim \mathcal{N}(m,s^2)$$ If you're interested why a shifted and scaled version of a normal is also normal, well let $$G = sg + m$$ where $g \sim \mathcal{N}(0,1)$. Please see below.
PROVING SIMULTANEOUSLY THAT $G_i = sg_i + m$ is normal with the above mean and variance
Let's get: $$F_{G}(y) = Pr(G \leq y) = P(sg + m \leq y) = Pr(s \leq \frac{y-m}{s}) = \int\limits_{-\infty}^{\frac{y-m}{s}} \Phi(x) \ dx$$ where $\Phi(x)$ is the PDF of a standard normal distribution (i.e. mean $0$ and variance $1$). Now take a change of variable, $t = sx + m$, that is $dt = s dx$. Then the above becomes $$F_{G}(y) = \int\limits_{-\infty}^{y} \Phi(\frac{t-m}{s})\frac{1}{s} \ dt$$, hence the PDF of $g$ is $$f_g(t) = \Phi(\frac{t-m}{s})\frac{1}{s} \sim \mathcal{N}(m,s^2)$$
Since $g_i$ has cdf $\Phi(x)$, the cdf of $G_i$ is $P(G_i\le x)=P(g_i\le \frac{x-m}{s})=\Phi(\frac{x-m}{s})$. Differentiating gives the cdf of $G_i$, viz. $\frac{1}{s}\phi(\frac{x-m}{s})=\frac{1}{s\sqrt{2\pi}}\exp -\frac{1}{2}(\frac{x-m}{s})^2$ as expected.