Is it possible to do the following approximation $\Pr({\frac{f}{{XY}}<Z<\frac{f}{a}})\approx\Pr({Z<\frac{f}{a}})-\Pr({\frac{f}{{XY}}<z})$?

Question

Is it possible to do the following approximation ?

$\Pr \left( {\frac{f}{{XY}} < Z < \frac{f}{a}} \right) \approx \Pr \left( {Z < \frac{f}{a}} \right) - \Pr \left( {\frac{f}{{XY}} < z} \right)$

Where $X,Y,Z$ are exponential random variable and $f,a>0$.

Thank you very much !

Answer 1

These are two different values. Is it true that $0.42 \approx 0.55$? In order to speak about approximation, we must have limit.

We have $X \sim \exp(\lambda_x)$, $Y \sim \exp(\lambda_y)$, $Z \sim \exp(\lambda_z)$. Hence $\lambda_x X \sim \exp(1)$ and $\lambda_y Y \sim \exp(1)$.

Suppose that $\lambda_x \lambda_y \to 0$, $f$, $a$ and $\lambda_Z$ i are fixed. Then $\frac{f}{XY} \to 0$ in distribution.

As $P \left( {\frac{f}{{XY}} < Z < \frac{f}{a}} \right) \approx \Pr \left( 0 < Z < \frac{f}{a} \right) =\Pr \left( Z \le \frac{f}{a} \right) $

$$ \Pr \left( {Z < \frac{f}{a}} \right) - \Pr \left( Z < {\frac{f}{{XY}} } \right) \approx \Pr \left( {Z < \frac{f}{a}} \right) - \Pr \left( {Z < 0} \right) = \Pr \left( Z \le \frac{f}{a} \right). $$ Thus $$P \left( {\frac{f}{{XY}} < Z < \frac{f}{a}} \right) \approx \Pr \left( {Z < \frac{f}{a}} \right) - \Pr \left( Z < {\frac{f}{{XY}} } \right) $$ $$ \approx \Pr \left( Z \le \frac{f}{a} \right) = 1 - e^{-\lambda_z\frac{f}a}$$