Question

Найти функцию распределения F(х)

Answer 1

$F(x)= \int\limits^{x}_{-\infty } f(t)\, dt\\\\f(x)= \left\{\begin{array}{l}0,\; esli\; \; \; x \leq 1\\x-\frac{1}{2}\; ,\; esli\; 1\ \textless \ x \leq 2\\0\; ,\; esli\; \; \; x\ \textgreater \ 2\end{array}\right\\\\a)\; \; x \leq 1:\; \; F(x)=\int\limits^{x}_{-\infty }\, f(t)\, dt=\int \limits _{-\infty }^{x}\, 0\, dt=0\\\\b)\; \; 1\ \textless \ x \leq 2:\; \; F(x)=\int\limits^{x}_{-\infty }\, f(t)dt= \int\limits^1_{-\infty }\, 0\, dt+\\\\+\int\limits^{x}_1\, (t-\frac{1}{2})\, dt=0+(\frac{t^2}{2}-\frac{1}{2}\, t)\Big |_1^{x}=\frac{x^2}{2}-\frac{x}{2}=\frac{1}{2}\cdot (x^2-x)$

$c)\; \; x\ \textgreater \ 2:\; \; F(x)=\int \limits _{-\infty }^1\, 0\, dt+\int\limits^2_1\, (t-\frac{1}{2})\, dt+\int\limits^{x}_3\, 0\, dt=\\\\=(\frac{t^2}{2}-\frac{t}{2})\Big |_1^2=\frac{4}{2}-1-(\frac{1}{2}-\frac{1}{2})=2-1-0=1\\\\F(x)= \left\{\begin{array}{l}0\; ,\; esli\; x\leq 1\\\frac{1}{2}\cdot (x^2-x)\; ,\; esli\; 1<x\leq 2\\1\; ,\; esli\; x>2\end{array}\right$