Why isn't this property preserved going from pmfs to pdfs? $\Pr(cX=x)=\Pr(X=x/c)$, but $f_X(x/c)\neq f_Y(x),\ Y=cX$ I've tried searching for this but I think the question is a bit too specific. I was working through a problem and along the way I had to write the probability density of some variable $cX$ at the point $x$. In my case $X\sim N(0,1)$. Then I wrote
"The event $cX=x$ is equivalent to the event $X=\frac{x}{c}$, so $f_{Y}(x)=f_X\left(\frac{x}{c}\right)$, where $f_Y(\cdot)$ is the density function for the variable $Y=cX$."
But then I thought about it for a while and realised it wasn't true. We have
\begin{equation*}
    f_X\left(\frac{x}{c}\right)=\frac{1}{\sqrt{2\pi}}e^{-\frac{\left(\frac{x}{c}\right)^2}{2}}=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2c^2}}
\end{equation*}
whereas we have
\begin{equation*}
    f_Y(x)=\frac{1}{\sqrt{2\pi c^2}}e^{-\frac{x^2}{2c^2}}
\end{equation*}
So my statement above was false. However, I gather that it would be true if $X$ was discrete.
I am aware of the transformation-of-variables formula $Y=g(X)\implies f_Y(y)=f_X(g^{-1}(y))\bigg|\frac{\mathrm{d}}{\mathrm{d}y}g^{-1}(y)\bigg|$, and I just need a bit of help understanding which step in my logic was faulty. Is it because probability mass functions for discrete variables are not trivially analogous to probability density functions for continuous variables?
Thank you!
 A: When viewing as a probability, the property is conserved:
For continuous random variables $X$ and $Y=cX$, $c>0$, $a<b$,
$$\begin{align*}
\Pr\left(\frac ac<X < \frac bc\right) &= \int_{a/c}^{b/c} f_X(x)\ dx\\
\Pr(a< Y < b) &= \int _{a}^{b} f_Y(y)\ dy\\
&= \int_{a/c}^{b/c}f_Y(cx)(c\ dx)\\
f_X(x) &= c\  f_Y(cx)
\end{align*}$$
So in some sense, to maintain the same integral (area), stretching the domain (width) of $f_X$ means compressing the function return (height).
Like how stretching rubber band makes it thinnner.

While for the discrete case, probability masses are added, not integrated. Like
$$\begin{align*}
\Pr(X\in\{1,2,4\}) &= \sum_{x\in\{1,2,4\}} p_X(x)\\
\Pr(Y\in\{c,2c,4c\}) &= \sum_{y\in\{c,2c,4c\}} p_Y(y)\\
\end{align*}$$
If it helps, imagine there is a rectangle of unit width and height $p_X(x)$ around a probability mass at each $x$, then the summation is like the area of these rectangles.
But because it is a summation, when stretching to $Y=cX$, those rectangles are still rectangle of unit width and the same height, but at a different location $y=cx$.
Like if you stick some solid dots (that doesn't stretch, like the rectangles) on the rubber band, and stretch that rubber band, the dots just get farther apart.
A: 
Why isn't this property preserved going from discrete to continuous variables?

Wrong, it is, namely:
$\mathbb{P}(c\cdot X \in [a,b])=\mathbb{P}(X\in[\frac{a}{c},\frac{b}{c}]),$
for any $a\le b$. Start by anderstending why this is actually correct. Probability is a measure , that is a function from a previosly prescribed sigma-field into interval $[0,1]$,
which satisfy some axioms. And our formula is just a shortcut notation for:
$\mathbb{P}(\{\omega \in \Omega: c\cdot X \in [a,b]\}) = \mathbb{P}(\{\omega\in \Omega: X\in [\frac{a}{c},\frac{b}{c}]\})$.
Notice, that we are really measuring the same set:
$\{\omega \in \Omega: c\cdot X \in [a,b]\}=\{\omega\in \Omega: X\in [\frac{a}{c},\frac{b}{c}]\}$,
and hence equality of probabilities is an obvious consequence. This formula is a bit more general then Yours, but idea is the same.

Is it because probability mass functions for discrete variables are not trivially analogous to probability density functions for continuous variables?

Yes, they are not. Density is not o continuous analogue of probability. It is a different mathematical object, heavily linked but not a continuous counterpart. 
So transformation-of-variables formula is the basically the right way. Maybe it is not so intuitive at first, but it encodes what is know as integration by substitution theorem. 
