I've come across an interesting problem in my research that I don't quite know the answer to. Suppose I have a bunch of random variables:

$$ X_1, X_2, X_3, ... X_N $$

They are not identical but they are independently distributed. As a rule, they have different distributions. I am interested in the quantity:

$$ Z_N = \text{min}(X_1, ... X_N) $$

My first question: What can I say, in general, about the properties of $Z_N$? (Such as its expected value and variance) I understand that if the variables are IID, I can use Extreme Value Theory, but what if the variables have arbitrary distributions (say, different Gaussians?)

My second question: if all I have are samples of $X_i$, and not their distributions, and I get them one at a time online, is it possible to compute $E[Z_N]$ and $\text{Var}(Z_N)$ without storing all the samples?

  • The second question has a simple, satisfactory answer: from the online sequence of $(X_{1i}, \ldots, X_{Ni})$ you can construct the sequence of associated $Z_{i}$. Its sample moments can each be computed with $O(1)$ storage because all you have to do is accumulate the contributions from each observation. – whuber Apr 16 '15 at 16:23
up vote 5 down vote accepted

$$P\{Z_n > \alpha\} = P\{X_1 > \alpha, X_2 > \alpha, \cdots, X_n > \alpha\} = \prod_{i=1}^n P\{X_i > \alpha\}$$ and so $$F_{Z_n}(\alpha\} = 1-P\{Z_n > \alpha\} = 1 - \prod_{i=1}^n \left(1- F_{X_i}(\alpha)\right).$$ For the special case when the $X_i$ are exponential random variables with parameter $\lambda_i$ (and so $P\{X_i > \alpha\} = e^{-\lambda_i \alpha}$), we easily get that $Z_n$ is an exponential random variable with parameter $\sum_{i=1}^n \lambda_i$.

More generally, taking the derivative of $F_{Z_n}(\alpha\}$ with respect to $\alpha$ via the product rule, we get the density $f_{Z_n}(\alpha)$ as $$f_{Z_n}(\alpha) = \frac{\mathrm d}{\mathrm d\alpha}F_{Z_n}(\alpha) = \sum_{i=1}^n \left(\prod_{j=1, j\neq i}^n \left(1- F_{X_j}(\alpha)\right)\right)f_{X_i}(\alpha)$$ which might at first glance seem to be a mixture of the $X_i$ densities, but really isn't because the "weights" of the various densities are also functions of $\alpha$ instead of being constants not depending on $\alpha$ and summing to $1$.

  • If I'm following you, you're computing the CDF to infer the PDF, using an (elegant) property of the CDF and the independence of each X_i. That's very useful. I will have to think on this a bit to see if it answers my other questions. Thanks! – mklingen Nov 18 '14 at 23:38

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.