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.


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 $\{X > \psi\}$.


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.


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: $F_{X|X>\psi}(x)= 1 - \left( \frac{\psi}{x} \right)^{\lambda}1_{(\psi,+\infty)}(x)$.

  • $\begingroup$ Your calculation of the cdf, $F_X(x)$ is wrong; there's at least two errors on the right hand side of the first equation. I didn't look beyond that. $\endgroup$
    – Glen_b
    Jan 22 at 15:38
  • $\begingroup$ @Glen_b Thank you for your report. I'll fix the issue. $\endgroup$
    – iStats7238
    Jan 22 at 15:41
  • $\begingroup$ @Glen_b I have seen the equation set for point 1 of the conditional distribution and I don't notice any errors in the CDF calculation. Could you point me to what you are referring to? $\endgroup$
    – iStats7238
    Jan 22 at 16:04
  • $\begingroup$ Your very first formula for the CDF is wrong (the right hand side doesn't even depend on $x$!) and the result you get is nonsensical: no CDF can be constant, nor can it have negative values (its values are probabilities, after all). $\endgroup$
    – whuber
    Jan 22 at 16:37
  • 1
    $\begingroup$ $F$ must increase to an upper limit of $1,$ not $0,$ as $x$ grows large. In your calculations you overlooked the fact that $f$ is zero for all $x\lt \theta.$ $\endgroup$
    – whuber
    Jan 22 at 20:45

1 Answer 1


$ 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 $$ P(X \leq x \mid X > \psi) = \frac{P(X \leq x \cap X > \psi)}{P(X > \psi)} = \frac{F(x) - F(\psi)}{1 - F(\psi)} = \frac{(\tfrac{\theta}{\psi})^\lambda - (\tfrac{\theta}{x})^\lambda}{(\tfrac{\theta}{\psi})^\lambda}= 1 -(\tfrac{\psi}{x})^\lambda.$$


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