2
$\begingroup$

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

$\endgroup$
11
  • $\begingroup$ 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
    Commented Jan 22, 2023 at 15:38
  • $\begingroup$ @Glen_b Thank you for your report. I'll fix the issue. $\endgroup$
    – iStats7238
    Commented Jan 22, 2023 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
    Commented Jan 22, 2023 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
    Commented Jan 22, 2023 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
    Commented Jan 22, 2023 at 20:45

1 Answer 1

1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.