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Dear All,

I have this problem here.

Part(1)

Let Y be a stochastic variable with the distribution function [tex]F_{Y}[/tex] given by:

[tex]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}[/tex]

Explain why Y is absolute continious and give the density [tex]f_{Y}.[/tex]

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 [tex]x_{1} < x_{2} < x_\ldots < x_{n}[/tex]. Then X is absolutely continuous with the desensity

[tex]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}[/tex]

By the theorem above its clearly visable that [tex]F_Y[/tex] is continous everywhere by the definition of continouty, then F' exists and thusly

[tex]f_{Y} = \frac{d}{dy}(sin(y)) = cos(y).[/tex]

Therefore Y is absolute continious.

Part two

Let X be a absolute continous stochastic variable with the probability density [tex]f_{X}[/tex] given by

[tex]f_{X}(x) = \left\{ \begin{array}{ccc} \frac{1}{9}|x|& \ \ \mathrm{if} \ \ x \in ]-3,3[ \\ 0& \ \ \mathrm{otherwise.} \end{array}[/tex]

Show that [tex]P(|X| \leq 1) = \frac{1}{9}.[/tex]

Proof

Since we know that the density function is given according to the definition

[tex]\int_{-\infty}^{\infty} f(x) dx = \int_{-3}^{3} \frac{1}{9}|x| dx = 1[/tex] Then to obtain where [tex]P(|X| \leq 1)[/tex] we analyse the interval

[tex]x \in \{-1,1\}[/tex] from which we obtain

[tex]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}.[/tex]

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....

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