P: 18 1. The problem statement, all variables and given/known data Dear All, I have this problem here. Part(1) Let Y be a stochastic variable with the distribution function $$F_{Y}$$ given by: $$P(Y \leq y) = F_{Y}(y) = \left\{ \begin{array}{ccc} \ 0& \ \ \mathrm{if} \ y < 0 \\ sin(y)& \ \ \mathrm{if} \ y \in [0,\pi/2] \\ 1& \ \ \mathrm{if} \ y > 0. \end{array}$$ Explain why Y is absolute continious and give the density $$f_{Y}.$$ 3. The attempt at a solution Proof Using the following theorem: Let X be a stochastic variable with the distribution function F, Assuming that F is continuous and that F' exists in all but finite many points $$x_{1} < x_{2} < x_\ldots < x_{n}$$. Then X is absolutely continuous with the desensity $$f(x) = \left\{ \begin{array}{ccc} F'(x) \ \ &\mathrm{if} \ \ x \notin \{x_{1}, x_{2}, \ldots, x_{n} \} \\ 0 \ \ &\mathrm{if} \ \ x \in \{x_{1}, x_{2}, \ldots, x_{n} \}. \end{array}$$ By the theorem above its clearly visable that $$F_Y$$ is continous everywhere by the definition of continouty, then F' exists and thusly $$f_{Y} = \frac{d}{dy}(sin(y)) = cos(y).$$ Therefore Y is absolute continious. Part two Let X be a absolute continous stochastic variable with the probability density $$f_{X}$$ given by $$f_{X}(x) = \left\{ \begin{array}{ccc} \frac{1}{9}|x|& \ \ \mathrm{if} \ \ x \in ]-3,3[ \\ 0& \ \ \mathrm{otherwise.} \end{array}$$ Show that $$P(|X| \leq 1) = \frac{1}{9}.$$ Proof Since we know that the density function is given according to the definition $$\int_{-\infty}^{\infty} f(x) dx = \int_{-3}^{3} \frac{1}{9}|x| dx = 1$$ Then to obtain where $$P(|X| \leq 1)$$ we analyse the interval $$x \in \{-1,1\}$$ from which we obtain $$P(|X| \leq 1) = \int_{-1}^{1} \frac{1}{9}|x| dx = \frac{1}{9} \cdot \int_{-1}^{1} |x| dx = [\frac{x \cdot |x|}{18}]_{x=-1}^{1} = \frac{1}{9}.$$ How does part one and two look? Do I need to add more text if yes what? The problem is correctly formulated from my textbook. Thanks in advance. BR Beowulf....