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

iStats7238
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  • 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. – Glen_b Jan 22 '23 at 15:38
  • @Glen_b Thank you for your report. I'll fix the issue. – iStats7238 Jan 22 '23 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? – iStats7238 Jan 22 '23 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 '23 at 16:37
  • @whuber because $F_X(x) = P(X \leq x) = \int^{x}_{-\infty} f(t)dt$. Right? If I was true, maybe, I really made some mistakes. – iStats7238 Jan 22 '23 at 16:49
  • @whuber I have calculated just now the CDF and the result is $F_X(x) = - (\theta/x)^{\lambda}$. This was the CDF I was supposed to be looking for, right? – iStats7238 Jan 22 '23 at 16:57
  • @whuber Besides I have done the control with the derivative and it provides me the density function. OMG I made some wrongs... Thank you very much for your report, I was doing crazy calculations without consciousness. – iStats7238 Jan 22 '23 at 17:10
  • Again: any function that returns a negative value for even one argument cannot possibly be a CDF. You're close, so consider checking your result by differentiating it and comparing to the original density. Also consider the limiting value of $F(x)$ as $x$ grows large: the axiom of total probability implies this limit must be $1.$ Plotting your solution is always a good idea because you can easily see whether it has these two fundamental properties (plus the others that make it a CDF, like left continuity). – whuber Jan 22 '23 at 17:35
  • @whuber I have studied $F(x)$ when $x$ grows large and I find that the limit is $0^{-}$. $F(-x)$ has as limit $0^{+}$. $F(x)$ has right continuity. The integral of density on $\mathbb{R}$ is 1. I have plotted the function and it seems an hyperbolic function. The behaviuor of $F(x)$ seems ok according to plot. Have I done something wrong? – iStats7238 Jan 22 '23 at 18:53
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    $F$ 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 '23 at 20:45
  • Let us continue this discussion in chat. – iStats7238 Jan 22 '23 at 23:15

1 Answers1

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

Hunaphu
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