$X_1, X_2, ..., X_n$ are random sample with pdf $f(x;\theta)=\theta(1-\theta)^x, x=0,1,2,...$
based on that I see that it is the same with $X_i$ ~ $GEO(\theta)$ because the pdf of geometric distribution with parameter $\theta$ is $f(x;\theta)=\theta(1-\theta)^{x-1}, x=1,2,...$
and get that $E[X]=1/\theta$
However, from the solution it used transformation such this
If $X$ has pdf $f(x;\theta)=\theta(1-\theta)^x,$ for $x=0,1,2,...$ then $Y=1+X$ has the $GEO(\theta)$ distribution
I don't understand why i need to transform it? Isn't the first pdf already equal to the second pdf?