What is the Probability of the Maximum Value for Discrete Random Variables?

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Discussion Overview

The discussion revolves around the probability of the maximum value of discrete random variables, particularly focusing on uniform distributions and the derivation of probability mass functions (p.m.f). Participants explore various aspects of the topic, including definitions, assumptions, and mathematical formulations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose that the probability of the maximum value for uniformly distributed discrete random variables is 1/L when all probabilities are equal.
  • Others clarify that for M random variables, the probability mass function of the maximum can be derived from the cumulative distribution function (CDF).
  • A participant suggests that if the random variables are uniformly distributed between 0 and 1, the probability that the maximum is less than a certain value can be expressed as P(max(x_m) < u) = u^M.
  • There is a discussion about the transition from the CDF to the p.m.f, with some noting that the CDF for discrete variables is not differentiable.
  • One participant emphasizes the need to clarify whether the original random variables are discrete or continuous, as this affects the interpretation of uniform distribution.
  • Another participant mentions the concept of order statistics and its application to continuous random variables, questioning how it translates to discrete cases.
  • One participant concludes that they have solved the problem, indicating a resolution for their inquiry.

Areas of Agreement / Disagreement

Participants express various viewpoints on the derivation of the p.m.f and the nature of the distributions involved. There is no clear consensus on the application of order statistics to discrete random variables, and discussions remain unresolved regarding specific mathematical steps and interpretations.

Contextual Notes

Participants highlight limitations in understanding the distribution types and the mathematical treatment of discrete versus continuous random variables. The discussion includes unresolved questions about the application of CDFs and p.m.fs in discrete cases.

EngWiPy
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Hi,

Suppose we have a random variable X takes on disctere values xi, i=1,2,...,L. What is the probability of the maximum value?

Thanks in advance
 
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It depends on the probability (Prob(X=L)). If all the probabilities are the same then it is 1/L.
 
mathman said:
It depends on the probability (Prob(X=L)). If all the probabilities are the same then it is 1/L.

Yes, all RVs are equiprobable. But how you got the answer, please?
 
S_David said:
Yes, all RVs are equiprobable. But how you got the answer, please?

A Random Variable (discrete or continuous) as a distribution, which basically says how much "weight" each element gets that is related to its frequency of occurring.

If every element in the distribution has the same chance of occurring as every other element, then its known as a uniform distribution. So if you had say "L" blocks with equal probability, then as the above posted has stated, the probability will be 1/L.

Why is this? Well probability density functions must have the property that the sum of all probabilities is equal to 1 (if it is continuous then the sum will turn into an integral).

So we know that Sum of all probabilities = 1. There are L blocks with the same probability so basically we have L * a = 1 where a is the probability of one "block". Re-arranging we get a = 1/L.
 
let me clarify my question: suppose we have M RVs say x1,x2,...x_M. Let y=max{x_m}, m=1,2,...,M, then what is the p.m.f of y, given that the RVs are uniformly distributed. I hope this is clear now.

Thanks
 
For simplicity I'll assume the random variables are uniform between 0 and 1, so that P(x_m < u) = u for 0<u<1.
Your question is about P(max(x_m) < u) = P(x_1 < u and x_2 < u and...x_M <u).
Since the x_m are independent, P(max(x_m) < u) = P(x_1 < u)*...*P(x_M < u)=uM.
 
mathman said:
For simplicity I'll assume the random variables are uniform between 0 and 1, so that P(x_m < u) = u for 0<u<1.
Your question is about P(max(x_m) < u) = P(x_1 < u and x_2 < u and...x_M <u).
Since the x_m are independent, P(max(x_m) < u) = P(x_1 < u)*...*P(x_M < u)=uM.

Now we are going somewhere. You found the CDF, how to find the p.m.f?
 
S_David said:
Now we are going somewhere. You found the CDF, how to find the p.m.f?

To get the PDF from the CDF you differentiate.
 
chiro said:
To get the PDF from the CDF you differentiate.

But the CDF is not differentiable, because it is discrete. I know that p(X=xi)=F(xi)-F(xi-1), but how to apply that here?
 
  • #10
S_David said:
But the CDF is not differentiable, because it is discrete. I know that p(X=xi)=F(xi)-F(xi-1), but how to apply that here?

For the distribution that math-man presented, the PDF was continuous.
 
  • #11
Note to S. David. You need to clarify as plainly as possible what the distribution is for your original random variables. Is it discrete? When you said uniform, I assumed the standard meaning which has a density function which is constant over some finite interval, usually [0,1].
 
  • #12
mathman said:
Note to S. David. You need to clarify as plainly as possible what the distribution is for your original random variables. Is it discrete? When you said uniform, I assumed the standard meaning which has a density function which is constant over some finite interval, usually [0,1].

It is discrete not continuous.
 
  • #13
S_David are you familiar with order statistics?
 
  • #14
chiro said:
S_David are you familiar with order statistics?

I know some. I mean for continuous random variables, I have no problem. For example, let [tex]X_1,X_2,\ldots,X_M[/tex] be M i.i.d RVs. Let [tex]Y=\underset{l}{max}\{Xl\}[/tex]. Then:
[tex]F_Y(y)=\left[F_X(y)\right]^M[/tex]
and the PDF is just the differentiation of it. But what is the counterpart in discrete RVs, and to deal with them?

Thanks
 
  • #15
The difference is that instead of an integral you replace it with a sigma sign over the domain of the RV (ie the CDF of the RV). Get rid of the sigma and you basically have your pdf for a given index.
 
  • #16
chiro said:
The difference is that instead of an integral you replace it with a sigma sign over the domain of the RV (ie the CDF of the RV). Get rid of the sigma and you basically have your pdf for a given index.

Ok I solved the problem. Thank you all
 

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