Example 5. Find the distribution function of a random variable distributed according to the exponential law.
Solution. We will use formula (2.3.2). For $x \leq 0$

$$
F(x)=\int_{-\infty}^{x} p(t) d t=\int_{-\infty}^{x} 0 d t=0
$$

If $x>0$, then

$$
\begin{aligned}
& F(x)=\int_{-\infty}^{x} p(t) d t=\int_{-\infty}^{0} p(t) d t+\int_{0}^{x} p(t) d t=\int_{-\infty}^{0} 0 d t+ \\
& +\int_{0}^{x} \alpha e^{-\alpha t} d t=-\left.e^{-\alpha t}\right|_{0} ^{x}=-\left(e^{-\alpha x}-e^{0}\right)=1-e^{-\alpha x}
\end{aligned}
$$

Therefore, the distribution function is

$$
F(x)= \begin{cases}0 & \text { for } x<0 \\ 1-e^{-\alpha x} & \text { for } x \geq 0\end{cases}
$$

The graph of the function $F(x)$ is shown in Fig. 3.7.

![](https://cdn.mathpix.com/cropped/2024_05_22_720b3cbdc89c57e2b629g-228.jpg?height=375&width=518&top_left_y=981&top_left_x=308)

## Fig. 3.7