Prob and stat expected value of x

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Homework Help Overview

The problem involves calculating the expected value of a random variable X given its distribution function F(x). The original poster expresses uncertainty about how to approach the calculations for E(X), E(X^{2} - 2|X|), and E(X|X|) based on the provided distribution function.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the definition of E(X) and the relationship between the distribution function F(x) and the probability density function f(x). There are hints about deriving f(x) from F(x) and calculating expected values using the distribution function.

Discussion Status

Some participants have provided hints and guidance on how to derive the probability density function from the distribution function. The original poster and others have attempted calculations and shared their results, indicating a collaborative exploration of the problem.

Contextual Notes

The original poster mentions confusion regarding the lack of an equation and the formatting of their question. There is also a reference to the need for clarity in calculations related to expected values.

Proggy99
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Homework Statement


The distribution function of a random variable X is given by

F(x) = {
0 if x < -3
3/8 if -3 [tex]\leq[/tex] x < 0
1/2 if 0 [tex]\leq[/tex] x < 3
3/4 if 3 [tex]\leq[/tex] x < 4
1 if x [tex]\geq[/tex] 4


Calculate E(X), E(X[tex]^{2}[/tex] - 2|X|), E(X|X|).


Homework Equations





The Attempt at a Solution


I actually have no idea how to start this. None of the chapter examples seem to give me a clue on what to do. I have calcuated E(X) for probability mass function equations, but not distributive functions like the above. Can someone give me a strong hint on how to approach this for E(X) and I can take it from there? Thank you for any help.

And sorry for the formatting, I could not get it to look right in latex, but I think it should be understandable.
 
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Well, what is the definition of E(X)? Further on, what is the relation between F(X) and f(x) (i.e. the probability density function f of the random variable X)?
 
Okay, another hint. Since you need to find f(x) to calculate E(X), is there a way to calculate f(x) from F(x)? [Hint 2: f(xi) = F(xi) - F(xi-), where F(xi-) is the left limit of F when x --> xi form the left.]
 
radou said:
Well, what is the definition of E(X)? Further on, what is the relation between F(X) and f(x) (i.e. the probability density function f of the random variable X)?

Well, E(X) is the expected value of X. E(X) = [tex]\sumxp(x)[/tex] where p(x) is the probability mass function. My main issue with the problem is that I am confused by the lack of an equation...

okay, so I started typing that and now I am wondering if it is as simple as this:
-3 * 3/8 + 0 * 1/8 + 3 * 2/8 + 4 * 2/8 = 5/8

I only looked at the numbers where it jumped on a graph and I looked at actual chances of each happening. In other words I subtracted f(x2) - f(x1) to get the chances of f(x2). I know the answer is 5/8, am I doing this correct or did I just get a coincidentally equal number?
 
Yes, my result is 5/8 too. So, the random variable is given with [tex]X = \[ \left( \begin{array}{cccc}-3 & 0 & 3 & 4 \\ 3/8 & 1/8 & 1/4 & 1/4 \end{array} \right)\][/tex]. If you "calculate" F(x) from X, you'll get your given function F.
 
radou said:
Okay, another hint. Since you need to find f(x) to calculate E(X), is there a way to calculate f(x) from F(x)? [Hint 2: f(xi) = F(xi) - F(xi-), where F(xi-) is the left limit of F when x --> xi form the left.]

so then what I really mean to say would be:
-3 * (3/8 - 0) + 0 * (1/2 - 3/8) + 3 * (3/4 - 1/2) + 4 * (1 - 3/4) = 5/8

technically the same thing, but clearer in showing what I was doing

and for E(X - 2|X|) would I then plug in -3 and replace the -3 above with the solution? such as (-3[tex]^{2}[/tex] - 2|-3|) = 3
so the equation above would start as 3 * (3/8 - 0) + ... ?
*edit*dropped the square, fixed now
 
answered my own question by working it out to find that I got the correct answer. Thanks for the hints radou, it is making sense now!
 
Proggy99 said:
answered my own question by working it out to find that I got the correct answer. Thanks for the hints radou, it is making sense now!

No problemo! It's nice to hear when someone works the answers out alone. :wink:
 

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