Is it reasonable to calculate $P(X_1 = X_2)$ given that $X_1$ and $X_2$ are continuous random variables? 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:


*

*Can I now conclude that $P(X_1 = X_2) = 1 - P(X_1 < X_2) - P(X_1 > X_2) = 0$?

*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? 

*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?  

 A: *

*Yes.

*Yes.

*Your confusion appears to be the following.  If $X_1,X_2$ are continuously distributed random variables then $X_1-X_2$ may or may not be continuously distributed.  If $X_1-X_2$ is continuously distributed then $\Pr(X_1=X_2)=0$.  If it is not continuously distributed then it can be anything.  For instance, suppose that $X_1$ has a $N(0,1)$ distribution, $Z$ has a Bernoulli (0/1) distribution with $p=0.5$ and $X_2=X_1+Z$.  Then $X_1,X_2$ have continuous distributions but $\Pr(X_1=X_2)=0.5$.

A: It can be confusing to think of an event such as $\{X_1=X_2\}$, since both sides are random; perhaps it will be easier to think about it a little differently.
Instead of considering the event that $X_1=X_2$, let's consider the (clearly equivalent) event that $X_1-X_2=0$.  Note that $X_1-X_2$ is just another random variable -- and, in particular, since $X_1$ and $X_2$ both have densities, and are independent, it is not hard to show that $X_1-X_2$ also has a density.
We can compute that density, $f_{X_1-X_2}$, explicitly by considering cases:
If $x\geq0$, then
$$
f_{X_1-X_2}(x)=\int_x^{\infty}f_{X_1}(t)\cdot f_{X_2}(t-x)\,dt=\frac{\lambda_1\lambda_2e^{-\lambda_1x}}{\lambda_1+\lambda_2},
$$
while for $x<0$ we have
$$
f_{X_1-X_2}(x)=\int_{-\infty}^{x}f_{X_1}(x-t)\cdot f_{X_2}(-t)\,dt=\frac{\lambda_1\lambda_2e^{\lambda_2x}}{\lambda_1+\lambda_2}.
$$
where $f_{X_1}$ and $f_{X_2}$ are the densities for $X_1$ and $X_2$, respectively.
But, since $X_1-X_2$ has a density, it is a continuous random variable -- and so the probability that it takes any particular value is $0$.
