probability Y> max of independent random variables with same distribution, different means

I am extremely enthused at finding this website. Thanks to Dilip Sarwate for some comments. I still don't have a final solution, however. Below is edited to better focus the problem.

Here is my problem. The situation is N random variables X(i), and one more, Y. The N+1 variables are independently distributed, with the same type of distribution, but they are not IID; each variable has a different mean. (For example, all N+1 variables might be exponentially distributed). These distribution means are drawn from a known distribution with a small standard deviation. Ultimately, I want to be able to take observed probabilities (e.g., I observe that Y exceeds the max of the X(i), for 1.1% of the 20,000 draws), and infer the mean of Y.

This is for an economics application, ultimately to infer "quality" from market share data. I am seeking functional forms such that these questions can be answered.

Since they are independent, then as Dilip pointed out, it is not difficult to figure out P(X(i)<x) or P(Y<x) but I am having difficulty figuring out P(Y>Max(X(i)). Apologies if this is basic. If substantive theory is involved, I am not averse to including someone as a coauthor.

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The hard part is not the different distributions but the lack of independence if when you say "They are not IID", you are denying both I's in the acronym IID. For $N$ independent random variables, $$P\{\max_i X_i \leq a\} = \prod_k P\{X_k \leq a\}.$$ – Dilip Sarwate Apr 3 '13 at 3:07
Hi Dilip, thanks for your comment --and yes, independent. I did get your expression, but what I need to be able to do is figure out P[Y>Max(X(i)] for some known distribution. For example, with exponentials with various means, it is easy to figure out P{Y>x,Max(X(i))<x} but I don't know how to move forward from there to P[Y>Max(X(i)}. Apologies if this is incredibly easy. – Randy V Apr 4 '13 at 13:58
What is $Y$? For exponential random variables, $P\{X_k \leq a\} = 1-\exp(-a\lambda_k)$ and so $$P\{\max_i X_i > a\} = 1-\prod_k P\{X_k \leq a\}=1-\prod_k (1-\exp(-a\lambda_k)).$$ Warning: do not multiply out the product on the right; evaluate each $1-\exp()$ to get a numerical value and multiply those $n$ numbers together and subtract from $1$. Multiplying those $1-\exp()$'s out will give you the answer as obtained by the principle of inclusion-exclusion which requires a lot more calculation. – Dilip Sarwate Apr 4 '13 at 14:23
Apologies for not being clear. <i>Y</i> is the variable of interest, say another exponential (with parameter lambda). I want to figure out P[Y>Max{X(i)}]. I know how to compute P[Y>x,Max{X(i)}<x] and for that matter P[Y>x,Max{X(i)}>x] but I don't know how to compute P[Y>Max{X(i)}. – Randy V Apr 4 '13 at 21:34