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This question is actually a follow up to a question I asked earlier here, but I'll supply here all the necessary details.

Let $X$ be a continuous random variable (rv) equipped with the CDF $$F_X(x)= \begin{cases} \frac12(2-e^{-x}) & x\ge0 \\ \frac12e^x & x\le0 \end{cases}$$

Now, we define a truncated rv Y such that $$Y= \begin{cases} 0 & X\leq 0 \\ X & 0 < X < 5 \\ 5 & X\ge 5 \end{cases}$$

Now, I know that $Y$ has a mixed type distribution and that its CDF has a decomposition $$F_Y(y)=\alpha F^d(y)+(1-\alpha) F^c(y) $$

where $F^d(y)$ and $F^c(y)$ are the CDFs of the discrete and continuous distributions, respectively.

I'm trying to find $F^d(y)$, $F^c(y)$ and $\alpha$.

My try:

$$F_Y(y)=P(Y\le y|Y \in \{0,5 \})P(Y \in \{0,5 \})+P(Y\le y|Y \notin \{0,5 \})P(Y \notin \{0,5 \})$$

$$P(Y=0)=P(X\le0)=F_X(0)=\frac12$$

$$P(Y=5)=P(X\ge5)=1-F_X(5)=\frac12 e^{-5}$$

Therfore

$$\alpha=P(Y \in \{0,5 \})=\frac12 + \frac12 e^{-5}$$

$$F_Y(y)= \begin{cases} 0 & y< 0 \\ \frac12 & 0 \le y < 5 \\ \frac12 + \frac12 e^{-5} & y\ge 5 \end{cases}$$

However, $F_Y(y)$ is clearly not a CDF (since its limit at $\infty$ is not $1$).

What I'm doing wrong?

Additionally, how can I find $F^c(y)$?

Any help will be greatly appreciated.

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Brassican
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1 Answers1

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First, your specification of the CDF is obviously in error. If $x \ge 0$, you state $F_X(x) = e^x/2$, thus $\lim_{x \to \infty} F_X(x) = \infty$, which is absurd. I interpreted the intended CDF as $$F_X(x) = \begin{cases} \frac{1}{2}(2-e^{-x}), & x \ge 0 \\ \frac{1}{2}e^x, & x < 0. \end{cases}$$ With this in mind, and $Y$ defined as above, we clearly have $$\Pr[Y = 0] = \Pr[X \le 0] + \Pr[X \ge 5] = F_X(0) + (1 - F_X(5)) = \frac{1}{2}(1 + e^{-5}) \approx 0.503369.$$ Now note that $Y$ cannot be negative, nor can it be $5$ or greater: either $Y = 0$ or $Y = X$ if $X \in (0,5)$, and in no case is $Y < 0$ or $Y \ge 5$. So suppose $y \in (0,5)$: we now have $$F_Y(y) = F_X(y) + (1 - F_X(5)) = \frac{1}{2}(2 - e^{-y} + e^{-5}), \quad 0 < y < 5.$$ If $y \ge 5$, then $F_Y(y) = 1$. If $y < 0$, then $F_Y(y) = 0$. So the CDF of $Y$ is $$F_Y(y) = \begin{cases} 0, & y < 0 \\ \frac{1}{2}(2 - e^{-y} + e^{-5}), & 0 \le y < 5 \\ 1 & 5 \le y. \end{cases}$$

While I was composing the above response, you edited your definition of $Y$ so that $Y = 5$ if $X \ge 5$. In this case, we would have $$\Pr[Y = 0] = \Pr[X \le 0] = F_X(0) = \frac{1}{2}.$$ Similarly, $$\Pr[Y = 5] = \Pr[X \ge 5] = 1 - F_X(5) = e^{-5}/2 \approx 0.00336897.$$ And again, for $y \in (0,5)$, the same reasoning as above would give $$F_Y(y) = F_X(y),$$ so we would simply have $$F_Y(y) = \begin{cases} 0, & y < 0 \\ \frac{1}{2}(2 - e^{-y}), & 0 \le y < 5 \\ 1, & y \ge 5. \end{cases}$$ The easiest way to see this is to observe that the CDF of $Y$ is discontinuous at $y = 0$ and $y = 5$, and the size of the discontinuities are precisely the probabilities $\Pr[Y = 0]$ and $\Pr[Y = 5]$.

To test your understanding, what if instead we defined $$W = \begin{cases} -4, & X \le -4 \\ X, & -4 < X < 0 \\ 0 & X \ge 0? \end{cases}$$ What is the CDF of $W$? And what if we defined in general $$T(m,n) = \begin{cases} m, & X \le m \\ X, & m < X < n \\ n, & X \ge n,\end{cases}$$ for $m < n$?

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