I am studying Markov processes with exponential wait times. The following is said:
Assume there are $k$ point events, denoted $w_1, \dots, w_k$, that the waiting time for $w_i$ to occur is $T_i \sim \text{Exp}(\lambda_i) \ (i = 1, \dots, k)$, and that the $T_i$s are independent. Let $T = \min_{1 \le i \le k} T_i$ be the time to the first occurrence, and let $Z = i$ if $T = T_i$, i.e., if $w_i$ is the first event to occur.
The author then makes the following remark:
Since the occurrence times $T_i$ have a continuous law, the $w_i$s cannot occur simultaneously. It will follow that Markov process jumps to new states are always separated by a positive interval of time.
Why does the occurrence times $T_i$ having a continuous law imply that the $w_i$s cannot occur simultaneously? Since this mentions that it is because of a continuous law, I get the impression that this has something to do with real analysis, but I'm not sure.
If $X$ is a random variable with continuous law, then for any real number $y$, $P(X = y) = 0$.
If $X$ and $Y$ are independent random variables with continuous laws, then $P(X = Y) = 0$. To prove this, you can use the joint distribution and Fubini-Tonelli: \begin{align*} P(X=Y) &= \iint_{\mathbb R^2}1_{\{x = y\}}\mathcal L_X(dx)\mathcal L_Y(dy)\\ &= \int_{\mathbb R}\bigg(\int_{\mathbb R} 1_{\{x = y\}}\mathcal L_X(dx)\bigg)\mathcal L_Y(dy) \\ &= \int_{\mathbb R}P(X = y)\mathcal L_Y(dy) = 0. \end{align*}