Example. A random variable $\xi$ is distributed according to the normal law $N(0,4)$. Find the probability of the random event

$$
A=\left\{\frac{1}{\xi}>\frac{1}{1+\xi}\right\}
$$

*) Using unified notations and names of functions $\varphi(x)$ and $\Phi(x)$, different authors define them by different formulas.
## Solution.

1. We represent the event $A$ in the form (1). For this, we solve the inequality

$$
\frac{1}{x}>\frac{1}{x+1}
$$

We obtain

$$
x \in(-\infty,-1) \cup(0,+\infty) \text {. }
$$

Therefore, the random event $A=\{1 / \xi>1 /(1+\xi)\}$ can be written as

$$
A=\{\xi \in(-\infty,-1) \cup(0,+\infty)\} \text {. }
$$

By formula (2), we get that

$\mathrm{P}\left(\frac{1}{\xi}>\frac{1}{1+\xi}\right)=\mathrm{P}(\xi \in(-\infty,-1) \cup(0,+\infty))=$

$=\Phi\left(\frac{-1-0}{2}\right)-\Phi(-\infty)+\Phi(+\infty)-\Phi\left(\frac{0-0}{2}\right)=\Phi\left(-\frac{1}{2}\right)+1-\Phi(0)$.

Here, it is taken into account that $\Phi(-\infty)=0, \Phi(+\infty)=1$. From the tables, we find that

$$
\Phi\left(-\frac{1}{2}\right)=1-\Phi\left(\frac{1}{2}\right) \approx 1-0.6915=0.3085, \quad \Phi(0)=0.5
$$

Therefore, the required probability is approximately equal to

$$
0.3085+1-0.5=0.8085 \text {. }
$$

Answer. $\mathrm{P}\left(\frac{1}{\xi}>\frac{1}{1+\xi}\right) \approx 0.8085$.

Conditions of the Problem. The random variable $\xi$ is distributed according to the normal law $N(0,4)$. Find the probability of the event $A$.

$$
\text { 1. } A=\{\xi \in(0,2)\} . \quad \text { 2. } A=\{\xi>0.8\} . \quad \text { 3. } A=\{2 \xi-36\}$. 6. $A=\left\{\xi^{2}>2.25\right\}$. 7. $A=\left\{\xi^{2}-3 \xi+2-2\}$. 9. $A=\left\{\log _{4}^{2} \xi<0.25\right\}$. 10. $A=\left\{1 /\left(1+\xi^{2}\right)<0.5\right\}$.

Answers. 1. 0.3414. 2. 0.3446. 3. $0.7734 . \quad 4.0 .1747$. 5. 0.0064 . 6. 0.4533 . 7. 0.1499 . 8. 0.9181. 9. 0.2426. 10. 0.6171 .

### 6.25. Central Limit Theorem

Problem Statement. Let $\xi_{1}, \xi_{2}, \ldots$ be independent and identically distributed random variables, such that $\mathrm{M} \xi_{1}=m, \quad \mathrm{D} \xi_{1}=\sigma^{2}$, $0<\sigma^{2}<+\infty$. Let $S_{n}=\xi_{1}+\ldots+\xi_{n}, n=1,2, \ldots$ Find the probability of some random event related to the sum $S_{n}$ for large $n$ or to several sums $S_{n_{1}}, S_{n_{2}}, \ldots$ for large $n_{1}, n_{2}, \ldots$

Plan of Solution. The Central Limit Theorem states that regardless of the distribution of the individual summand, for large $n$ the sum $S_{n}$ has a distribution that is approximately normal $N\left(n m, n \sigma^{2}\right)$. A similar statement is true for the difference $S_{n_{2}}-S_{n_{1}}=\sum_{i=n_{1}+1}^{n_{2}} \xi_{i}$, if the quantity $n_{2}-n_{1}$ is large, and the random variables $S_{n_{1}}$ and $S_{n_{2}}-S_{n_{1}}$ are independent. The calculation of the required probability is done as described in problem 6.24. In this process, the following fact is used: if the random variables $\eta_{1}$ and $\eta_{2}$ are independent and distributed according to the normal laws $N\left(a_{1}, \sigma_{1}^{2}\right)$ and $N\left(a_{2}, \sigma_{2}^{2}\right)$ respectively, then the random variable $c_{1} \eta_{1}+c_{2} \eta_{2}$, where $c_{1}, c_{2}$ are constants, is distributed according to the normal law $N\left(c_{1} a_{1}+c_{2} a_{2}, c_{1}^{2} \sigma_{1}^{2}+c_{2}^{2} \sigma_{2}^{2}\right)$.