Uniform Distribution (Probability)

Click For Summary

Homework Help Overview

The discussion revolves around a discrete random variable X with a given probability mass function (p.m.f) and the task of generating a random value from this distribution using a uniform random variable u drawn from a uniform(0,1) distribution. Participants are exploring how to utilize the cumulative distribution function (C.D.F) derived from the p.m.f to determine the corresponding value of X for a specific u value.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the construction of the C.D.F from the provided p.m.f and the implications of using a uniform random variable to find a corresponding value of X. There is uncertainty about how to apply the uniform random variable to the C.D.F, with some questioning the validity of their approaches and assumptions.

Discussion Status

Some participants have provided insights into the relationship between the uniform distribution and the C.D.F of X, suggesting the need to identify the correct bin corresponding to the value of u. There is acknowledgment of the need to follow a logical mapping rather than simply finding the closest bin, indicating a productive direction in the discussion.

Contextual Notes

Participants are working under the constraints of the problem as posed, with specific values and a defined p.m.f. There is an ongoing exploration of how to interpret the C.D.F and its application to the uniform random variable.

Kites
Messages
39
Reaction score
0

Homework Statement



Let X be a discrete random variable with the p.m.f given in the following table:

x 10 20 30 40

p(x) .25 .2 .4 .15

Suppose you can generate a random value, u, from a uniform(0,1) distribution.


If u = 0.576, then what is the value of your random value from the distribution of X?


Homework Equations



F(x) = \int f(x)dx


Uniform:

f(x) = (a+b)/2

The Attempt at a Solution




I've attempted to find the C.D.F of the table:

x 10 20 30 40

p(x) .25 .45 .85 1

From here I am at a complete loss at how to use my value of U to get at a random value of X. I did attempt the following but am unsure of its validity:

u = \int(1/30)dx from 10 to x, finding approximately 27 as the value. This seems very wrong to me as that makes me assume that the table is a uniform distribution which it is not.

I could truly use some help. Thanks a lot.
 
Physics news on Phys.org
The CDF for a uniform number is a straight line with values going from 0 to 1. In particular, the CDF for U(0,1) is the line F(x)=x. You function does not have a linear CDF. This is not a problem.

Suppose you have a CDF F(x) for some random distribution. The distribution does not have to be uniform. Now suppose this CDF F(x) is an invertible function. In other words, given some value F, the inverse function F-1(F) finds x such that F(x)=F. If you have such a magic bullet, you can use numbers drawn from the standard uniform distribution to generate random numbers drawn from the distribution in question. How? Simple apply the inverse function F-1(F). This is essentially what you are to do with this problem. What you need to do is find which bin (10, 20, 30, or 40) corresponds to a CDF value of 0.576.
 
Okay, I've taken a look at the CDF and the values are:


x 10 20 30 40

p(x) .25 .45 .85 1


for the CDF of X.

By your logic I have to find what BIN that .567 goes into...



I would say that the closest bin is the P(x) = .45 or, x=20.


Is this correct?
 
Any value drawn from U(0,1) that falls between 0 and 0.25 obviously must be assigned to x=10. (What else?) Suppose you use a higher cutoff value than 0.25. This will make the probability of drawing a "10" will exceed 0.25. So anything between 0 and 0.25 maps to the first bin, x=10. Keep following this logic forward and you will find you do not want the closest bin.
 
You're exactly right thank you DH It's a true life saver for me.
 
Your'e welcome.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
9
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
13K
Replies
7
Views
1K
Replies
1
Views
1K