To be specific, suppose $X\_1$ and $X\_2$ are independent exponential random variables with parameters $\lambda\_1$ and $\lambda\_2$; what is $P(X\_1 = X\_2)$?

According to section 5.2.3 of the book "Introduction to Probability Models" by Sheldon M.Ross (the 10th edition), $P(X\_1 < X\_2) = \frac{\lambda\_1}{\lambda\_1 + \lambda\_2}$. Symmetrically, $P(X\_1 > X\_2) = \frac{\lambda\_2}{\lambda\_1 + \lambda\_2}$.

$$P(X\_1 < X\_2) = \int\_{0}^{\infty} P(X\_1 < X\_2 \mid X\_1 = x) \lambda\_1 e^{-\lambda\_1 x}dx \\ = \int\_{0}^{\infty} P(x < X\_2) \lambda\_1 e^{-\lambda\_1 x}dx \\ = \int\_{0}^{\infty} e^{-\lambda\_2 x} \lambda\_1 e^{-\lambda\_1 x}dx \\ = \int\_{0}^{\infty} \lambda\_1 e^{-(\lambda\_1 + \lambda\_2) x}dx = \frac{\lambda\_1}{\lambda\_1 + \lambda\_2}.$$

**My Problems:**

1. Can I now conclude that $P(X\_1 = X\_2) = 1 - P(X\_1 < X\_2) - P(X\_1 > X\_2) = 0$?
2. I am confused about the argument because of the fact that the probability that a continuous random variable will take on any *particular* value is zero. Is it reasonable to calculate the probability that the two variables *simultaneously* take on the same particular value?
3. If so, how to calculate $P(X\_1 = X\_2)$ directly, in the same way used in the calculation of $P(X\_1 < X\_2)$? Furthermore, can this calculation be applied to all continuous random variables?