1663. Due to various reasons, the TVs that are delivered to the repair shop for repair usually have all kinds of malfunctions. Therefore, the time (service time) required to eliminate the malfunctions is a random variable $T$. Find the mathematical expectation and variance of the service time of TVs, if the probability distribution function is as follows:

$$
F(t)= \begin{cases}0 & \text { for } t \leqslant 0 \\ 1-e^{-v t} & \text { for } t>0\end{cases}
$$

(exponential distribution of service time).
Solution. Since the formulas for the expected value and variance involve the probability density function, we first find the density $p(t)$:

$$
p(t)=F^{\prime}(t)=\left\{\begin{array}{ll}
0 & \text { for } t \leqslant 0, \\
\left(1-e^{-v t}\right)^{\prime} & \text { for } t>0
\end{array}= \begin{cases}0 & \text { for } t \leqslant 0 \\
v e^{-v t} & \text { for } t>0\end{cases}\right.
$$

Then the average service time for the television is

$$
\begin{gathered}
E T=\int_{-\infty}^{+\infty} t p(t) d t=\int_{0}^{+\infty} t v e^{-v t} d t=-\int_{0}^{+\infty} t d e^{-v t}=-\left.t e^{-v t}\right|_{0} ^{+\infty}+ \\
+\int_{0}^{+\infty} e^{-v t} d t=-\left.\frac{1}{v} \cdot e^{-v t}\right|_{0} ^{+\infty}=\frac{1}{v} .
\end{gathered}
$$

Knowing that

$$
\begin{aligned}
D T & =E T^{2}-(E T)^{2}, \text { and } E T^{2}=\int_{-\infty}^{+\infty} t^{2} p(t) d t=\int_{0}^{+\infty} t^{2} v e^{-v t} d t= \\
& =-\int_{0}^{+\infty} t^{2} d e^{-v t}=-\left.t^{2} e^{-v t}\right|_{0} ^{+\infty}+\int_{0}^{+\infty} e^{-v t} \cdot 2 t d t= \\
& =2 \cdot \frac{1}{v} \int_{0}^{+\infty} t v e^{-v t} d t=2 \cdot \frac{1}{v} \cdot \frac{1}{v}=\frac{2}{v^{2}}
\end{aligned}
$$

we find

$$
D T=\frac{2}{v^{2}}-\frac{1}{v^{2}}=\frac{1}{v^{2}}
$$