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

Question

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?

Answer 1

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 $0$.

Answer 2

1. Yes.
2. Yes.
3. 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 $\Pr(X_1=X_2)=0.5$.