Example 1.4 Karatzas & Shreve

Question

This example shows that two processes $X,Y$ can be a modification of one another, but they don't necessarily need to be indistinguishable.

Let $T$ be a continuous random variable with continuous distribution.
Let $X_t\equiv 0$ and $Y_t = \begin{cases} 0,& t \neq T\\ 1,& t =T\end{cases}$.
It is clear to me that $Y$ is a modification of $X$.
It is clear to me that given a fixed $t\geq 0$ we have that $P(X_t = Y_t) = P(T\neq t)$.
It is not clear to me why $P(T \neq t) = 1$
It is not clear to me why $P(X_s = Y_s \hspace{0.2cm};\forall s \geq 0) =0$

Given that $T$ is an arbitrary continuous random variable with continuous distribution, why does that set $\{ \omega \in \Omega: T(\omega) \neq t\}$ need to have probability 1?

Moreover, on the last point, I have been trying to prove that using $\{X_s = Y_s; \forall s \geq 0\} = \cap_{s\geq 0} \{ \omega \in \Omega : X_s(\omega) = Y_s(\omega)\}$, is this the right way to do that?

I am new to probabilistic reasoning from a measure theory point of view! Thanks for your patience.

score 0 · Accepted Answer · answered Jul 04 '21 at 05:20

0

For any continuous r.v. $X$ we have $P(X=t)=0$ so $P(X \neq t)=1-0=1$.

If $X_s=Y_s$ for all $s$ then $X_T=Y_T$ which gives $0=1$. Hence the event $(X_s=Y_s\, \forall s \geq 0)$ is empty.

answered Jul 04 '21 at 05:20

Kavi Rama Murthy

311,013

Why does $P(X=t)=0$? Why cant we have $X(\omega) = t$ for all $\omega \in \Omega$? – Jul 05 '21 at 00:56
That is the definition of a continuous distrbution. What is your definition of a continuous distrbution? @jcm – Kavi Rama Murthy Jul 05 '21 at 05:00

Example 1.4 Karatzas & Shreve

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