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Two points are realized on a $l\times w$ rectangle. A point has higher chance of being realized in the top-left quadrant $Q_{t,l}$. Of interest is the maximum x-coordinate among the two points. Given that one point was realized in $Q_{t,l}$ and the other not, does it suffice to find the expected value of the point that is not in the quadrant, i.e.

$$\frac{1}{3} \times \frac{l}{4} + \frac{2}{3} \times \frac{3l}{4} $$

(*) or do I explicitly need to take care of the case where the point in $Q_{t,l}$ happens to be realized at a higher x-coordinate than the point that is not in $Q_{t,l}$?

How do you solve for order statistics when there is no closed-form survival function or c.d.f. (in the variation where it is known that both points are NOT realized in $Q_{t,l}$). Opposed to the following case where there is no discontinuity: Expected value of maximum of two random variables from uniform distribution.

(*) Three quadrants remain that are not in $Q_{t,l}$. The probability that it is in the buttom-left is 1/3, 2/3 that it is one the right. If it is on the right, its expected value is halfway between l/2 and l. If it is on the bottom left, it is halfway in between 0 and l/2. Ignore the point in $Q_{t,l}$ because the expected value of the point not in $Q_{t,l}$ is father away than the expected value of the point in $Q_{t,l}$.

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Wuschelbeutel Kartoffelhuhn
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  • It might help answer the question usefully if you describe how you arrived at the idea that it might suffice to find the expected value of the point that's not in the quadrant. – joriki Oct 31 '12 at 04:06
  • @joriki: Just added it – Wuschelbeutel Kartoffelhuhn Oct 31 '12 at 04:13
  • Sorry, I don't understand how what you added might suggest that you could just ignore the other point. – joriki Oct 31 '12 at 04:15
  • @joriki: Youre right. Added more info. – Wuschelbeutel Kartoffelhuhn Oct 31 '12 at 04:17
  • Ah, that made it clearer, thanks :-). That one has higher expectation value than the other isn't enough to disregard one of them. Consider the extreme case where they're both nearly uniformly distributed, with a very small bias to the left in one case and to the right in the other. By your argument, you could ignore one of them, and then the expected value of the maximum would be the expected value of the other one, almost exactly in the middle, which is clearly not the right expected value for the maximum of two nearly uniformly distributed coordinates. – joriki Oct 31 '12 at 05:13
  • @joriki: "Consider the extreme case where they're both nearly uniformly distributed, with a very small bias to the left in one case and to the right in the other." If there is a .51 probability of a point being in $Q_{t,l}$ and a .49 probability of being in $Q_{t,l}^C$, then, knowing that exactly one of the two points is trapped in $Q_{t,l}$, its expected location has still a lower value than that of the point in $Q_{t,l}^C$. Danke für Ihre Hilfe übrigens. – Wuschelbeutel Kartoffelhuhn Oct 31 '12 at 05:30
  • What I'm really curious about is the second question. How to compute order statistics when there is no clean way (without dummy variables) to represent the c.d.f. due to discontinuity. Perhaps that warrants another thread. – Wuschelbeutel Kartoffelhuhn Oct 31 '12 at 05:32

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NOT AN ANSWER Difficult to write in the comments, perhaps this can get you started...

Here the y-dimension is irrelevant. Reducing it to one dimension and in the unit range

$$ E\{X\}=\int_0^1 P(X⩾x)dx = \int_0^1 (1-P(X \le x))dx \\ E\{\text{max}(X_1,X_2)\}=\int_0^1 (1-P(X_1 \le x)P(X_2 \le x))dx = \int_0^1 (1-P(X \le x)^2)dx $$

Assuming uniform distribution will give you $\int_0^1 (1-x^2)dx = \frac23$

scale with the actual dimension $l$.

karakfa
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