How Do You Compute p(Z > l2) in Exponential Distributions?

[oscaralive]

Homework Statement

Let <d1,d2,d3...dN> be an odered set of samples from an exponential
random variable with parameter lambda.
Let <l1,l2,l3,...,lM> the same.

Let Z = min<d1,d2,...,dN> --> Z is exp with parameter lambda*N
Let U = min<l1,l2,...,lM> --> U is exp with parameter lambda*M

Homework Equations

Since we can write that:
p(Z > U) = M/(N+M), which is in fact, p(d1 > l1)= p(Z > l1)

How can I compute the following probability:

p(Z > l2), p(Z > l3),...,p(Z > lN).

The Attempt at a Solution

Until now I am "approximating" this probability by assuming that:

p(Z > l2) = (M-1)/((M-1)+N) and succesively, but I know it is wrong...

Thanks a lot in advance!

 

[KataKoniK]

Thank you for your question. To compute the probabilities p(Z > l2), p(Z > l3),...,p(Z > lN), we can use the fact that the minimum of exponential random variables follows a distribution known as the exponential order statistic. This means that the probability of Z being greater than any particular value, such as l2, can be calculated using the following formula:

p(Z > l2) = (N+M-2)!/(N-1)!(M-1)! * (1-e^(-lambda*l2))^(N-1) * (1-e^(-lambda*l2))^(M-1)

In this formula, N and M represent the number of samples from the exponential random variable, and lambda is the parameter of the exponential distribution. This formula can be derived using the properties of order statistics and the fact that the minimum of exponential random variables follows a gamma distribution.

I hope this helps answer your question. If you have any further questions, please don't hesitate to ask. Good luck with your research!

 

[Mene]

I would first clarify the context and assumptions of the problem. It seems that the samples are from two different exponential distributions, with different parameter values (lambda). It is also assumed that the samples are independent and identically distributed (i.i.d.).

Given this information, I would approach the problem by using the properties of the exponential distribution and the concept of order statistics.

Firstly, since Z is the minimum of N exponential samples, it follows that Z is also exponentially distributed with parameter lambda*N. This means that we can write p(Z > l2) as 1 - p(Z < l2), which is the same as 1 - F(l2), where F is the cumulative distribution function (CDF) of Z.

We can use the CDF of an exponential distribution to calculate this probability. The CDF of an exponential distribution with parameter lambda is given by F(x) = 1 - e^(-lambda*x). Therefore, p(Z > l2) = 1 - (1 - e^(-lambda*N*l2)).

Similarly, we can calculate p(Z > l3) = 1 - (1 - e^(-lambda*N*l3)), and so on for the remaining probabilities.

In general, for an ordered set of samples from an exponential distribution with parameter lambda, the probability of the minimum being greater than a certain value x is given by 1 - (1 - e^(-lambda*N*x)).

It is important to note that this approach assumes that the samples are i.i.d. and that the minimum is being taken from a set of N samples. If these assumptions do not hold, then the approach may not be applicable.