Let $X$ be a random variable having density function $$f(x;\vartheta,\lambda)=\lambda\,\vartheta^\lambda\,x^{-(\lambda +1)}$$ $$\vartheta>0\qquad\lambda>0\qquad x\geq\vartheta$$ Ginen $\psi>\vartheta$, find the density funtion of $X$ conditioned to the event $\lbrace{ X>\psi\rbrace}$.
Intuitively, I'd say $$f(x;\vartheta,\lambda | \lbrace{ X>\psi\rbrace}) =\lambda\,\psi^\lambda\,x^{-(\lambda +1)}$$ $$\psi>0\qquad\lambda>0\qquad x\geq\psi$$
This can be found forcing $F(\psi)=1-\left(\frac{\vartheta}{\psi} \right)^\lambda = 0$, where $F(x)$ is the cumulative distribution function of the variable $X$.
Is this way of reasoning right or am I missing something? Thanks in advance for your help!
Your intuition is correct, but you do need to formalize it. To do this efficiently, it is easiest to work with the survival function $$S_X(x \mid \vartheta, \lambda) = \Pr[X > x] = \int_{t=x}^\infty \lambda \vartheta^\lambda t^{-(\lambda+1)} \, dt = (\vartheta/x)^\lambda, \quad x \ge \vartheta.$$ So for $x \ge \psi \ge \vartheta$, $$\Pr[X > x \mid X > \psi] = \frac{S_X(x \mid \vartheta, \lambda)}{S_X(\psi \mid \vartheta, \lambda)} = \frac{(\vartheta/x)^\lambda}{(\vartheta/\psi)^\lambda} = (\psi/x)^\lambda = S_X(x \mid \psi, \lambda).$$ This completes the proof.