I come across the following question:
Let $X_1,...,X_n$ be a random sample from a distribution with density $f(x; β) = \frac{\beta}{(1+x)^{\beta+1}}$ when $x > 0$ and zero elsewhere. The parameter $β > 0$ is an unknown parameter.
Show that $Y_i=\log(1+X_i) \sim EXP(1/\beta)$
So I did the following:
$$f(Y_i=y)=f(\log(1+X_i)=y)=f(1+X_i=e^y)=f(X_i=e^y-1)=\frac{\beta}{e^{y(\beta+1)}}=\frac{1}{(1/\beta)}e^{\frac{-y}{1/(\beta+1)}} \neq \frac{1}{(1/\beta)}e^{\frac{-y}{1/\beta}}$$
So I do not find the pdf of $EXP(1/\beta)$, which is the last expression.
Now apparently the method you should use is first find the cdf of $X_i$ from its pdf and then apply the above method, like so:
$$F(y)=P(Y_i \leq y)=P(X_i \leq e^y-1)=\int^{e^y-1}_{0}f(x)dx=...=1-e^{\frac{-y}{1/\beta}}$$
Which is the cdf of $EXP(1/\beta)$. So this method indeed finds the right answer. My question is: why is the first method (that I used) invalid?
The first method is not valid because there is a confusion about the meaning of the pdf. An expression such as $f(Y_i=y)$ is meaningless. You could do a valid heuristic reasoning with infinitesimal quantities such as \begin{equation} f(y) d y \approx P(Y_i \in [y, y+d y]) = P(X_i \in [e^y-1, e^{y+dy}-1]) \approx P(X_i \in [e^y-1, e^y-1+e^y d y]) = \frac{\beta}{(e^y-1+1)^{\beta+1}} e^y d y \end{equation} And then one finds the correct result. Now some people could question the mathematical rigor of this reasoning.