Why is $\frac X2$ a standard normal variable if it's density is $\frac 12$ the density of the standard normal distribution?

Question

I don't understand the algebra of scaling a normal variable. Say I have $X\sim N(0,2^2)$ and an $ Y\sim N(0,1)$ , then for the density function I get: $$p_Y(t) = P(Y=t) = \frac 1{\sqrt{2\pi}} \exp{-\frac {t^2} 2} = 2P(X=2t) = 2p_{X/2}(t)$$

Now everyone knows that $X/2\sim N(0,1)$, by why is it just allowed to "correct" the density function with the factor $2$? Where does this factor come from? It seems "unmathematical" to say that we need the factor $2$ simply because otherwise, we wouldn't have a probability density. Could someone shed some light, please?

Answer 1

So if you want to derive the density of $Y=\frac{X}{2}$ you have just to apply the Fundamental Transformation Theorem

$$f_Y(y)=f_X[g^{-1}(y)]\Bigg|\frac{d}{dy}g^{-1}(y)\Bigg|$$

In your case you have

$$2\cdot \frac{1}{2\sqrt{2\pi}}e^{-\frac{(2y)^2}{2\cdot4}}=\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}$$

Answer 2

If $c >0$ then $P(cX \leq x)=P(X \leq \frac x c)$ or $F_{cX} (x)=F_X(\frac x c)$. To find density function you have to differentiate both sides. When you differentiate using chain rule you get $(F_X'(x)) \frac d {dx} (\frac x c )=\frac 1c F_X'(x)$ on the right side. That is how you get the factor $\frac 1 c$. Nothing special about normal distribution.