Consider the process $\{X(t),t\geq0\}$ defined by: $$X(t)= \begin{cases} 1, &\text{if t$\le$ Y}\\ 0, &\text{if t$\gt$ Y} \end{cases}$$ where $Y$ is a uniformly distributed random variable on the interval $(0,1)$.

1. Compute, for $t\in[0,1]$, the first-order probability mass of $X$, $f(t,x)$.

2. What is the first-order probability mass of $X$ for $t>1$? Could someone explain what is a first-order probability mass?

  • $\begingroup$ I haven't heard this term, but my guess would be that they are referring to the one-dimensional distribution, i.e. the distribution of $X$ at just one time. This is as opposed to, say, the joint distribution of $X$ at some number of times considered together. $\endgroup$ – Ian Oct 5 '17 at 20:45
  • $\begingroup$ @Ian, Hi! Nice to see you again! I think your interpretation is right! But I don't know how to solve this problem, does it mean we just need to compute the probability of t<=Y, but I don't know how to do this. $\endgroup$ – Qing Oct 5 '17 at 20:49
  • $\begingroup$ I think they actually want the density but you can start from the CDF, which has $F_X(x)=0$ for $x<0$, $P(t>Y)$ for $0 \leq x<1$, and $F_X(x)=1$ for $x \geq 1$. $P(t>Y)$ should be easy to calculate. $\endgroup$ – Ian Oct 5 '17 at 20:54
  • $\begingroup$ Could you elaborate a bit about P(t>Y), I think that's what I don't know how to solve, I mean P(t>Y)=1-P(t$\le$Y)=1-t? $\endgroup$ – Qing Oct 5 '17 at 21:01
  • 1
    $\begingroup$ What we just discussed may be summarized by saying $F_{X(t)}(x)=\begin{cases} 0 & x<0 \\ t & 0 \leq x<1 \\ 1 & x \geq 1 \end{cases}$. They ask you for $f_{X(t)}(x)$. $\endgroup$ – Ian Oct 5 '17 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.