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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?

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  • $\begingroup$ 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. $\endgroup$
    – whuber
    Apr 16, 2015 at 16:23

1 Answer 1

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$$P\{Z_n > \alpha\} = P\{X_1 > \alpha, X_2 > \alpha, \cdots, X_n > \alpha\} = \prod_{i=1}^n P\{X_i > \alpha\}$$ 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 $1$.

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  • $\begingroup$ 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! $\endgroup$
    – mklingen
    Nov 18, 2014 at 23:38

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