The distribution function is $F(x)=\begin{cases}1-e^{-x}, & \text{$x$>0} \\ 0, & \text{$x$≤0 } \end{cases}$ But I don't know why

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Let the random variables $X$ and $Y$ be independent,$X$ follows the exponential distribution with parameter $1$, and the probability distribution of $Y$ is

$$\mathcal{P}\{Y=-1\}=p,\;\mathcal{P}\{Y=1\}=1-p;\;Z=XY$$

(1) Find the probability density of $Z$.

(2) Find $p$ value,when the $X$ and $Z$ are not relevant.

(3) $X$ and $Z$ are independent of each other?

This topic is a graduate entry topic. The progress now is that I know the probability density function of $X$.

$$f(x)=\begin{cases}e^{-x}, & \text{$x$>0} \\ 0, & \text{$x$≤0 } \end{cases}$$

Some people say its distribution function is

$$F(x)=\begin{cases}1-e^{-x}, & \text{$x$>0} \\ 0, & \text{$x$≤0 } \end{cases}$$

But I don't know why. I really don’t understand this basic question.

The possible strategy is to find the distribution function of $Z$ and then derive it. This problem is obviously a problem of the probability. It is a graduate entrance examination, which is very important to me.

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Just integrate the pdf with $t>0$:

$$F(t) = \int\limits_{-\infty}^t f(x)dx = \int\limits_{-\infty}^0 0dx + \int\limits_{0}^t e^{-x}dx = 0 + -e^{-x}|_0^t = -e^{-t} + 1 = 1-e^{-t}$$

and if $t\leq 0$:

$$F(t) = \int\limits_{-\infty}^t f(x)dx = \int\limits_{-\infty}^t 0dx = 0$$