Probability distribution and cumulative distribution function q's

In summary, the conversation discusses finding the constant A in the probability distribution function P(x) for the given equation. The correct limits of integration are discussed, and it is determined that A=1/2. The conversation then moves on to finding the cumulative distribution function P(x), and the correct integral is shown. The resulting CDF is then calculated for values of x less than -2, between -2 and 2, and greater than 2. The final CDF is given as:P(x) = 0 for x < -2P(x) = -(x^2-4x+4)/8 for -2 ≤ x ≤ 2P(x) = 1 for x > 2.
  • #1
Evo8
169
0

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
 
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  • #2
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.
 
  • #3
SteamKing said:
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}##?
 
  • #4
That looks much better.
 
  • #5
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?
 
  • #6
Evo8 said:
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...
 

What is a probability distribution function (PDF)?

A probability distribution function (PDF) is a mathematical function that describes the likelihood of a random variable taking on a certain value. It shows the possible outcomes of an event and the probability of each outcome occurring.

What is a cumulative distribution function (CDF)?

A cumulative distribution function (CDF) is a mathematical function that shows the probability that a random variable is less than or equal to a certain value. It is the integral of the PDF and provides information about the entire range of possible outcomes.

What is the difference between a discrete and continuous probability distribution?

A discrete probability distribution is one where the random variable can only take on a finite or countably infinite number of values. A continuous probability distribution is one where the random variable can take on any value within a certain range.

How do you calculate the mean and standard deviation of a probability distribution?

The mean of a probability distribution is calculated by multiplying each possible outcome by its probability and then summing all the values. The standard deviation is calculated by taking the square root of the sum of the squared differences between each outcome and the mean, multiplied by its respective probability.

What are some real-world applications of probability distribution and cumulative distribution functions?

Probability distribution and cumulative distribution functions are used in various fields such as finance, physics, and engineering. They are used to model and predict outcomes in situations where there is uncertainty, such as in stock market trends, radioactive decay, and machine failure rates. They are also commonly used in statistical analysis and data science to understand and interpret data.

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