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

[EngWiPy]

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

 

[mathman]

It depends on the probability (Prob(X=L)). If all the probabilities are the same then it is 1/L.

 

[EngWiPy]

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?

 

[chiro]

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.

 

[EngWiPy]

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

 

[mathman]

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)=u^M.

 

[EngWiPy]

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)=u^M.

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

 

[chiro]

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.

 

[EngWiPy]

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?

 

[chiro]

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.

 

[mathman]

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].

 

[EngWiPy]

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.

 

[chiro]

S_David are you familiar with order statistics?

 

[EngWiPy]

S_David are you familiar with order statistics?

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

Thanks

 

[chiro]

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.

 

[EngWiPy]

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