1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Stochashic variable result(Please verify) Urgent

  1. Dec 12, 2007 #1
    1. The problem statement, all variables and given/known data

    Dear All,

    I have this problem here.


    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


    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]


    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.

    Last edited: Dec 12, 2007
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted