Stochashic variable result(Please verify) Urgent

1. Dec 12, 2007

Beowulf2007

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.

BR
Beowulf....

Last edited: Dec 12, 2007