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Let $X$ be a random variable having PDF $f(x) = \frac{1}{\lambda} e^{-\frac{\lambda}{x}} , x > 0 , \lambda > 0$.

And I was trying to find out the $p$th Quantile, for which we have to set $\int_{-\infty}^{x} f(x) dx = p$

so,the $x$ is our $p$th quantile.

$\int_{-\infty}^{x} \frac{1}{\lambda}e^{-\frac{\lambda}{t}}dt = p$

$\int_{-\infty}^{x} e^{-\frac{\lambda}{t}}dt = \int_{0}^{x} e^{-\frac{\lambda}{t}}dt$ as $x > 0$

so,$\int_{0}^{x} e^{-\frac{\lambda}{t}}dt = p\lambda$.

How do I solve this integration next in order to get the value of $x$?

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    $\begingroup$ I think you want $e^{-x/\lambda}$. That integral is far easier to evaluate, leaving you to find $1-e^{-x/\lambda}=p$ which may be solved using logarithms. Your given PDF is not a PDF and actually is not even normalizable because its integral does not converge. $\endgroup$ – Ian Aug 20 '17 at 14:19
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    $\begingroup$ As written the integral does not converge, so it is not a PDF. $\endgroup$ – Ross Millikan Aug 20 '17 at 14:22
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    $\begingroup$ actually we want thisas the limit is from $0$ to $x$ ,it's showing some values at least , does this converge? $\endgroup$ – BAYMAX Aug 20 '17 at 16:00
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    $\begingroup$ Unless you put in a maximum for $x$, there is no way to force $e^{-1/x}$ to be a PDF. Now something like $f(x)=e^{-1/x}$ for $x \in (0,1)$, otherwise $0$, that can be normalized to be a PDF. $\endgroup$ – Ian Aug 20 '17 at 23:00

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