Verifying Stochastic Variable Result: Urgent Homework

[Beowulf2007]

Homework Statement

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 $$f_{Y}.$$

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 continuous 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 continuous 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...

 

[mighty2000]

Dear Beowulf,

Part one looks good. You have correctly applied the theorem to show that Y is absolutely continuous and have found the density function f_Y. However, it would be helpful to explain why the density function is cos(y) and not just state it. You could mention that since sin(y) is the derivative of cos(y), it follows that cos(y) is the density function.

As for part two, your proof is correct but it would be helpful to explain why the interval [-1,1] is used and how you obtained it. You could also mention that the interval [-3,3] is chosen because it is the support of the density function f_X.

Overall, your solutions are well-written and clear. However, it would be good to add more explanation and justification to make them more understandable to someone who is not familiar with the problem. Also, it would be helpful to cite your sources for the theorems and definitions used.

Hope this helps!




Scientist