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From my understanding, to prove a statement "something is an A if and only if it is a B" we start with B which will lead to A, then we start with A which will lead to B. In this case (please see the screenshot), A is: $$f_X(x) = \frac1{\sigma}f((x-\mu)/\sigma))$$ and B are $$f_Z(z)$$ and $$X = \sigma Z + \mu$$ If so, why didn't the "only if" part of the attached proof start with A, that is, $f_X(x) = \frac1{\sigma}f((x-\mu)/\sigma))$? However, that would lead to $f_X(x) = \frac1{\sigma}f(z)$ instead of $f(z)$. Note: I derived the last equality ($f_X(x) = \frac1{\sigma}f(z)$) based on the premise that $z = g(x) = (x - \mu)/\sigma$. I'm so confused with the second part (only if) of the attached screenshot. Could you please point it out what I did wrong?

enter image description here

enter image description here

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  • $\begingroup$ Thanks for the editing, @StubbornAtom. This will help me know how to post better in the future. $\endgroup$
    – Nemo
    Commented Jan 30, 2020 at 10:50

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