Task 1. For a random variable $X$, given by the distribution function $F(x)=0$ for $x<1$, $F(x)=a(x-1)^2$ for $1 \leq x \leq 4$, and $F(x)=1$ for $x>4$, determine: the value of the parameter $a$, the density function $f(x)$, the initial and central moments of the first four orders, as well as the skewness and kurtosis. Plot the graphs of $f(x)$ and $F(x)$.
Solution. We have $1=F(4)=a(x-2)^{2}=4 a$. From this, $a=0.25$.

Next, $f(x)=F^{\prime}(x)=0.5(x-2), x \in(2,4)$. Thus (see Fig. 7.12 and 7.13)

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-159.jpg?height=288&width=448&top_left_y=941&top_left_x=129)

Fig. 7.12

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-159.jpg?height=288&width=503&top_left_y=941&top_left_x=647)

Fig. 7.13

$$
\begin{aligned}
& f(x)=\left[\begin{array}{ll}
0, & x \leqslant 2, x \geqslant 4 ; \\
0.5(x-2), & 24
\end{array}\right.
\end{aligned}
$$

Calculation of initial moments:

$$
\begin{aligned}
M_{1}(X)=0.5 \int_{2}^{4} x(x-2) d x=0.5\left(\frac{x^{3}}{3}-x^{2}\right) & \left.\right|_{2} ^{4}= \\
& =\frac{1}{2}\left(\frac{64}{3}-16-\frac{8}{3}+4\right)=\frac{10}{3}
\end{aligned}
$$

$$
\begin{aligned}
& M_{2}(X)=0.5 \int_{2}^{4} x^{2}(x-2) d x=0.5\left(\frac{x^{4}}{4}-\frac{2 x^{3}}{3}\right) & \left.\right|_{2} ^{4}= \\
& =\frac{1}{2}\left(64-\frac{128}{3}-4+\frac{16}{3}\right)=\frac{34}{3} \\
& M_{3}(X)=0.5 \int_{2}^{4} x^{3}(x-2) d x=0.5\left(\frac{x^{5}}{5}-\frac{x^{4}}{2}\right) & \left.\right|_{2} ^{4}= \\
& =\frac{1}{2}\left(\frac{1024}{5}-128-\frac{32}{5}+8\right)=\frac{196}{5}=39.2 ; \\
& M_{4}(X)=0.5 \int_{2}^{4} x^{4}(x-2) d x=0.5\left(\frac{x^{6}}{6}-\frac{2 x^{5}}{5}\right) & \left.\right|_{2} ^{4}= \\
& =\frac{1}{2}\left(\frac{4096}{6}-\frac{2048}{5}-\frac{64}{6}+\frac{64}{5}\right)=\frac{1376}{10}=137.6 .
\end{aligned}
$$

We will calculate the central moments using integrals (omitting some details):

$$
\begin{aligned}
m_{1}=0.5 \int_{2}^{4}\left(x-\frac{10}{3}\right)(x-2) d x= & 0.5 \int_{2}^{4}\left(x-\frac{10}{3}\right)\left(x-\frac{10}{3}+\frac{4}{3}\right) d x= \\
& =0.5 \int_{2}^{4}\left[\left(x-\frac{10}{3}\right)^{2}+\frac{4}{3}\left(x-\frac{10}{3}\right)\right] d x=0
\end{aligned}
$$

$$
\begin{aligned}
& m_{2}=0.5 \int_{2}^{4}\left(x-\frac{10}{3}\right)^{2}(x-2) d x=0.5 \int_{2}^{4}\left(x-\frac{10}{3}\right)^{3}+ \\
& +0.5 \int_{2}^{4} \frac{4}{3}\left(x-\frac{10}{3}\right)^{2} d x=\left.\frac{1}{8}\left(x-\frac{10}{3}\right)^{4}\right|_{2} ^{4}+\left.\left(x-\frac{10}{3}\right)^{3}\right|_{2} ^{4}= \\
& =\frac{1}{8} \cdot \frac{16-256}{81}+\frac{2}{9} \cdot \frac{8+64}{27}=-\frac{10}{27}+\frac{16}{27}=\frac{2}{9} \\
& D(X)=M_{2}-M_{1}^{2}=\frac{34}{3}-\frac{100}{9}=\frac{2}{9}, \quad \sigma(X)=\frac{\sqrt{2}}{3} \approx 0.47 \\
& m_{3}=0.5 \int_{2}^{4}\left(x-\frac{10}{3}\right)^{3}(x-2) d x=0.5 \int_{2}^{4}\left(x-\frac{10}{3}\right)^{4} d x+\frac{2}{3} \int_{2}^{4}\left(x-\frac{10}{3}\right)^{3} d x= \\
& =\left.\frac{1}{10}\left(x-\frac{10}{3}\right)^{5}\right|_{2} ^{4}+\left.\frac{1}{6}\left(x-\frac{10}{3}\right)^{4}\right|_{2} ^{4}=\frac{1}{10} \cdot \frac{1056}{243}-\frac{240}{6 \cdot 81}=-\frac{8}{135}
\end{aligned}
$$

$$
\begin{gathered}
m_{4}=0.5 \int_{2}^{4}\left(x-\frac{10}{3}\right)^{4}(x-2) d x=0.5 \int_{2}^{4}\left(x-\frac{10}{3}\right)^{5} d x+\frac{2}{3} \int_{2}^{3}\left(x-\frac{10}{3}\right)^{4} d x= \\
=\left.\frac{1}{12}\left(x-\frac{10}{3}\right)\right|_{2} ^{4}+\left.\frac{2}{15}\left(x-\frac{10}{3}\right)^{5}\right|_{2} ^{4}=-\frac{336}{3^{6}}+\frac{2112}{5 \cdot 3^{6}}=\frac{16}{135} \\
A s=\frac{m_{3}}{\sigma^{3}}=-\frac{8}{135}:\left(\frac{\sqrt{2}}{3}\right)^{3}=-\frac{8}{10 \sqrt{2}}=-\frac{2 \sqrt{2}}{5} \approx-0.56 \\
E k=\frac{m_{4}}{\sigma^{4}}-3=\frac{16}{135} \cdot \frac{81}{4}-3=-\frac{3}{5}=-0.6
\end{gathered}
$$

Separately, we highlight $M(X)=M_{1}(x)=\frac{10}{3}, D(X)=m_{2}(x)=\frac{2}{9}, \sigma(x)=$ $=\sqrt{D(x)}=\frac{\sqrt{2}}{3} \approx 0.47, A s(x) \approx-0.56$ and $E k(x)=-0.6$. Thus, the mean value of $X$ - the central value - is $\frac{10}{3}$, the standard deviation - the spread - is 0.47; the distribution is skewed to the left and is below the normal distribution.