Probability distribution and cumulative distribution function q's

[Evo8]

Homework Statement

$$P_x(x)=A(1- \frac{|x|}{2}) \ \ \ |x|≤2$$
$$P_x(x)=0 \ \ |x|>2$$

Find A

Homework Equations

The Attempt at a Solution

This one shouldn't be too bad but I wanted to verify that I am on the right track.

I basically have $P_x(x)=A(1- \frac{x}{2})$ when x≤2. 0 otherwise.

So I write the integral $\int_{-\infty}^{2} A(1-\frac{|x|}{2})dx$

I end up with $A (x(\frac{-x^2}{4})dx$ evaluated from 2 to $-\infty$ which will equate to $\infty$

The section in my text that talks about probability distribution function mentions this
$$\int_{-\infty}^{\infty} p_x(x)dx=1$$

So I could do $A\int_{-\infty}^{\infty} 1-\frac{|x|}{2} dx=1$? Solving this would yield A=$\infty$ which is not what I am looking for I don't think. Its something to do with the absolute value of x i think...

The form that the question writes the equation in always screws me up.

Thanks for any help

 

[SteamKing]

Your limits of integration do not match the domain specified for P(x). Look carefully. The ABSOLUTE VALUE of x is less than or equal to 2.

 

[Evo8]

Your limits of integration do not match the domain specified for P(x). Look carefully. The ABSOLUTE VALUE of x is less than or equal to 2.

How about something like this $\int_{-2}^{2}A(1-\frac{|x|}{2}=1$

$\frac{1}{A}=\int_{-2}{2}(1-\frac{|x|}{2}=4$

so $A=\frac{1}{4}$?

 

[SteamKing]

That looks much better.

 

[Evo8]

Thanks again for your help!

I think i actually made a mistake with the absolute value. A would equal $A=\frac{1}{2}$ not $\frac{1}{4}$

I have another question about this problem.

The second part is to find the CDF (cumulative distribution function) for x.

From what I understand the pdf is $p_x(x)$ as the problem states. The CDF is $P_x(x)$, and $P_x(x)= \int p_x(x)$?

Isnt that what I've already done to find the constant A?

So $P_x(x)=\frac{1}{2}\int_{-2}^2(1-\frac{|x|}{2})dx$

Isnt this what I've already done?

 

[Evo8]

Thanks again for your help!

I think i actually made a mistake with the absolute value. A would equal $A=\frac{1}{2}$ not $\frac{1}{4}$

I have another question about this problem.

The second part is to find the CDF (cumulative distribution function) for x.

From what I understand the pdf is $p_x(x)$ as the problem states. The CDF is $P_x(x)$, and $P_x(x)= \int p_x(x)$?

Isnt that what I've already done to find the constant A?

So $P_x(x)=\frac{1}{2}\int_{-2}^2(1-\frac{|x|}{2})dx$

Isnt this what I've already done?

Ok so I did a little more digging/reading and found a few video examples too. I found this one to quite helpful Cumulative Distribution Functions : Introduction

So to find $P_x(x)$ I did the following

$$P_x(x)=\frac{1}{2}\int_{-2}^{x}(1-\frac{|x|}{2}) \\
=\frac{1}{2}|_{-2}^{x} \ x-\frac{x-|x|}{4} \\
=\frac{1}{2}[x-\frac{x^2}{2}-1]\\
=\frac{-(x^2-4x+4)}{8}
$$
So my CDF is

$$\begin{cases}0 \ \ \ \ x<-2\\ \\
\frac{-(x^2-4x+4)}{8} \ \ \ \ -2≤x≤2\\ \\
1 \ \ \ \ x>2 \end{cases}$$

Does this look somewhat correct? I got a little confused when it came to values of x greater then 2 and the probability being 1. I guess I would have expected the probability to be 0 as the value is o above that upper limit...