Example 7. A random variable $X$ is given by the distribution function

$$
F(x)= \begin{cases}0 & \text { if } x \leq 0 \\ x^{2} & \text { if } 0 < x \leq 1 \\ 1 & \text { if } x > 1\end{cases}
$$

Find the initial and central moments of the first three orders of the random variable $X$.
Solution. First, we find the probability density function of the given random variable $X$. Since $p(x) = F'(x)$, the probability density function of this random variable is defined by the function

$$
p(x)= \begin{cases}0 & \text { if } x \leq 0 \\ 2 x & \text { if } 0 < x \leq 1 \\ 0 & \text { if } x > 1\end{cases}
$$

According to formulas (2.6.6), we find the initial moments:

$$
v_{1}=M(X)=\int_{-\infty}^{+\infty} x p(x) d x=\int_{0}^{1} x 2 x d x=2 \int_{0}^{1} x^{2} d x=\left.2 \cdot \frac{x^{3}}{3}\right|_{0} ^{1}=\frac{2}{3}
$$

$$
\begin{aligned}
& v_{2}=M\left(X^{2}\right)=\int_{-\infty}^{+\infty} x^{2} p(x) d x=\int_{0}^{1} x^{2} 2 x d x=2 \int_{0}^{1} x^{3} d x=\left.2 \cdot \frac{x^{4}}{4}\right|_{0} ^{1}=\frac{1}{2} \\
& v_{3}=M\left(X^{3}\right)=\int_{-\infty}^{+\infty} x^{3} p(x) d x=\int_{0}^{1} x^{3} 2 x d x=2 \int_{0}^{1} x^{4} d x=\left.2 \cdot \frac{x^{5}}{5}\right|_{0} ^{1}=\frac{2}{5}
\end{aligned}
$$

The first central moment $\mu_{1}=M(X-a)=0$ (the expected value of the deviation is zero).

According to formulas (2.6.11), we calculate the central moments of the second and third orders:

$$
\begin{aligned}
& \mu_{2}=M(X-a)^{2}=\int_{-\infty}^{+\infty}(x-a)^{2} p(x) d x=\int_{0}^{1}\left(x-\frac{2}{3}\right)^{2} \cdot 2 x d x= \\
& =\int_{0}^{1}\left(x^{2}-\frac{4}{3} x+\frac{4}{9}\right) 2 x d x=2 \int_{0}^{1} x^{3} d x-\frac{8}{3} \int_{0}^{1} x^{2} d x+\frac{8}{9} \int_{0}^{1} x d x= \\
& =\left.2 \cdot \frac{x^{4}}{4}\right|_{0} ^{1}-\left.\frac{8}{3} \cdot \frac{x^{3}}{3}\right|_{0} ^{1}+\left.\frac{8}{9} \cdot \frac{x^{2}}{2}\right|_{0} ^{1}=\frac{1}{2}-\frac{8}{9}+\frac{4}{9}=\frac{1}{18}, \\
& \mu_{3}=M(X-a)^{3}=\int_{-\infty}^{+\infty}(x-a)^{3} p(x) d x=\int_{0}^{1}\left(x-\frac{2}{3}\right)^{3} \cdot 2 x d x= \\
& =\int_{0}^{1}\left(x^{3}-3 x^{2} \cdot \frac{2}{3}+3 \cdot \frac{4}{9} x-\frac{8}{27}\right) 2 x d x=\int_{0}^{1}\left(2 x^{4}-4 x^{3}+\frac{8}{3} x^{2}-\frac{16}{27} x\right) d x= \\
& =\left.\left(2 \cdot \frac{x^{5}}{5}-x^{4}+\frac{8}{9} x^{3}-\frac{8}{27} x^{2}\right)\right|_{0} ^{1}=\frac{2}{5}-1+\frac{8}{9}-\frac{8}{27}=-\frac{1}{135} .
\end{aligned}
$$

## Problems

1. A discrete random variable $X$ is given by the distribution law

| $X$ | 2 | 3 |
| :---: | :---: | :---: |
| $P$ | 0.4 | 0.6 |

Find the initial and central moments of the first, second, and third orders of the random variable $X$.

2. A discrete random variable $X$ is given by the distribution law:

![](https://cdn.mathpix.com/cropped/2024_05_22_720b3cbdc89c57e2b629g-150.jpg?height=137&width=1110&top_left_y=11&top_left_x=11)

Find the initial and central moments of the first, second, and third orders of the random variable $X$.

3. A random variable $X$ is given by the density function

$$
p(x)= \begin{cases}0 & \text { if } x \leq 0 \\ 3 e^{-3 x} & \text { if } x > 0\end{cases}
$$

Find the initial and central moments of the first, second, third, and fourth orders of the random variable $X$.

4. A random variable $X$ is given by the density function

$$
p(x)= \begin{cases}0 & \text { if } x \leq 1 \\ \frac{6}{x^{7}} & \text { if } x > 1\end{cases}
$$

Find the initial and central moments of the random variable $X$.

5. A random variable $X$ is given by the distribution function

$$
F(x)= \begin{cases}0 & \text { if } x \leq 0 \\ x^{3} & \text { if } 0 < x \leq 1\end