Warning
I have divided the question into two subquestions: the first concerns the conditional distribution, which I discuss here; the second part is found in this question.
Exercise
Let $X \thicksim Pa(\lambda, \theta)$ with density function
$
f(x; \theta, \lambda) = \frac{\lambda \theta^{\lambda}}{x^{\lambda+1}}
$
where $x \geq \theta$, $\lambda > 0$ and $\theta > 0$.
Fixed $\psi > 0$, find the distribution of $X$ conditional on the event $\{X > \psi\}$.
Try
First, I have found the CDF of $X$
$$
F_X (x) = P(X \leq x)
= \int^{x}_{\theta} \frac{\lambda \theta^{\lambda}}{t^{\lambda+1}} dt
= \lambda \theta^{\lambda} \int^{x}_{\theta} \frac{1}{t^{\lambda+1}} dt
= \lambda \theta^{\lambda} \int^{x}_{\theta} t^{-(\lambda+1)} dt
$$
$$
F_X (x) = \lambda \theta^{\lambda} \left[ \frac{ t^{-\lambda}}{-\lambda} \right]^{x}_{\theta}
= \lambda \theta^{\lambda} \left[ \frac{ x^{-\lambda}}{-\lambda} – \frac{ \theta^{-\lambda}}{-\lambda} \right]
= – \theta^{\lambda}x^{-\lambda} + 1 = 1 – \left( \frac{\theta}{x} \right)^{\lambda}
$$
So far I'm in, but after that I wouldn't know how to proceed.
UPDATE
I have understood because when I calculate the integral of the density doesn't provide me the CDF. I didn't insert $\theta$ as lower bound in the integral. Sorry. Now I have fixed it.
Meanwhile, I have found this question in which the calculation is given by:
$$
P(X \leq x | X > \psi) = \frac{F(x) – F(\psi)}{1 – F(\psi)} 1_{(\psi,+\infty)}(x)
$$
where $1_{(\psi,+\infty)}(x)$ is the indicator function for the set $(\psi,+\infty)$. So, let's proceed:
$$
\begin{align*}
F_{X|X>\psi}(x) & = P(X \leq x | X > \psi) = \frac{F(x) – F(\psi)}{1 – F(\psi)} 1_{(\psi,+\infty)}(x) \\
& = \frac{1 – \left( \frac{\theta}{x} \right)^{\lambda} – 1 + \left( \frac{\theta}{\psi} \right)^{\lambda}}{1 – 1 + \left( \frac{\theta}{\psi} \right)^{\lambda}}
= \frac{\left( \frac{\theta}{\psi} \right)^{\lambda} – \left( \frac{\theta}{x} \right)^{\lambda} }{\left( \frac{\theta}{\psi} \right)^{\lambda}} \\
& = 1 – \frac{\left( \frac{\theta}{x} \right)^{\lambda} }{\left( \frac{\theta}{\psi} \right)^{\lambda}}
= 1 – \frac{\theta^{\lambda}}{x^{\lambda}} \cdot \frac{\psi^{\lambda}}{\theta^{\lambda}} \\
& = 1 – \left( \frac{\psi}{x} \right)^{\lambda}
\end{align*}
$$
Finally, I have the CDF: $F_{X|X>\psi}(x)= 1 – \left( \frac{\psi}{x} \right)^{\lambda}1_{(\psi,+\infty)}(x)$.
Best Answer
$ f(x; \theta, \lambda) = \frac{\lambda \theta^{\lambda}}{x^{\lambda+1}} $ So, $$ F(x) = \int_\theta^x f(t)\,dt = \lambda \theta^\lambda \left[\frac{-1}{\lambda}t^{-\lambda}\right]_\theta^x = \theta^\lambda (\theta^{-\lambda} - x^{-\lambda}) = 1 - (\tfrac{\theta}{x})^\lambda $$ and $$ P(X \leq x \mid X > \psi) = \frac{P(X \leq x \cap X > \psi)}{P(X > \psi)} = \frac{F(x) - F(\psi)}{1 - F(\psi)} = \frac{(\tfrac{\theta}{\psi})^\lambda - (\tfrac{\theta}{x})^\lambda}{(\tfrac{\theta}{\psi})^\lambda}= 1 -(\tfrac{\psi}{x})^\lambda.$$