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

• 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. Jan 22 at 15:38
• @Glen_b Thank you for your report. I'll fix the issue. Jan 22 at 15:41
• @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? Jan 22 at 16:04
• 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).
– whuber
Jan 22 at 16:37
• $F$ 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 $x\lt \theta.$
– whuber
Jan 22 at 20:45

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