I was given the following stochastic process:
$ X(t) = V h(t-Z)$ , $t > 0$
Where $f_V (v) = 1/2, 0\leq v\leq 2$ , $f_Z(z) = 2 e^{-2z}, z\geq 0$ , and:
$
h(t) = \begin{cases}
1, & 0 \leq t\leq 1 \\
0, & \text{otherwise}
\end{cases}
\
$
I was asked to find the CDF of $X(1/2)$. What I did is to write:
$ F_{X(1/2)} (k)= P({X(1/2)}\leq k) = P(0 \leq z \leq 1/2)P(V \leq k) + P(Z > 1/2)P(k \geq 0)$
Solving this I get:
$
F_{X(1/2)} (k) = \begin{cases}
1, & k \geq 2 \\
(1-e^{-1})\frac{k}{2} + e^{-1}, & 0 \leq k \leq 2 \\
0 & k < 0
\end{cases}
\
$
I'm not sure about this result because this function is discontinuous at $k=0$. If I wanted to get, for example, the PDF by derivation I would get: $f_{X(1/2)}(k)=(1-e^{-1})/2$ which has an area $1-e^{-1} \neq 1$.
Is something wrong with my reasoning ?
Best Answer
Yes, it is possible to have a discontinuous CDF. In fact, every discrete distribution has a discontinuous CDF; however, your question relates to a mixed distribution with a combination of density and discrete probability mass. Consider the example
$$W \mid X = \min(X, 2), \\ X \sim \operatorname{Exponential}(\lambda = 1).$$ Then $W$ is what we call a right-censored exponential random variable: $W = X$ if $X < 2$, otherwise $W = 2$. Hence $$\Pr[W \le w] = \begin{cases} 0, & w < 0 \\ 1 - e^{-w}, & 0 \le w < 2 \\ 1, & w \ge 2. \end{cases}$$ There is a nonzero density for $W$ in the interval $(0,2)$, and then a discrete probability mass at $W = 2$ with value $\Pr[W = 2] = \Pr[X > 2] = e^{-2}$.
That said, we need to take a closer look at your computations. I will use the indicator function $\mathbb 1$ in place of your $h(t)$.
I get $$\begin{align} X(1/2) &= V \mathbb 1 (0 \le 1/2 - Z \le 1) \\ &= \begin{cases}V, & |Z| \le 1/2 \\ 0, & |Z| > 1/2 \end{cases} \\ &= \begin{cases} V, & 0 \le Z \le 1/2 \\ 0, & Z > 1/2. \end{cases} \end{align}$$ This piecewise definition is the origin of the discontinuous nature of the CDF. Proceeding with our computations, $$\begin{align} F_{X(1/2)}(x) &= \Pr[X(1/2) \le x] \\ &= \Pr[V \le x \mid 0 \le Z \le 1/2] \Pr[0 \le Z \le 1/2] + \Pr[0 \le x \mid Z > 1/2] \Pr[Z > 1/2] \\ &= \left( \frac{x}{2} \mathbb 1 (0 \le x \le 2) + \mathbb 1 (x > 2) \right) (1 - e^{-1}) + \mathbb 1(x \ge 0) e^{-1} \\ &= \begin{cases} 0, & x < 0 \\ \frac{x}{2} (1 - e^{-1}) + e^{-1} , & 0 \le x \le 2 \\ 1, & x > 2.\end{cases} \end{align}$$ This is equivalent to your solution. While it is true that the density equals $$f_{X(1/2)}(x) = \frac{1 - e^{-1}}{2}$$ for $0 < x < 2$, there is also a discrete probability mass at $x = 0$: $$\Pr[X = 0] = e^{-1},$$ which we can see by going back to the original definitions: $$\Pr[X(1/2) = 0] = \Pr[|Z| > 1/2] = \Pr[Z > 1/2] = e^{-2/2} = e^{-1}.$$