# Reasoning error in “maximin” type problem

I have a problem that I have approached two different ways. The approaches give me two different answers. I have matched the answers against a simulation, so I think I know which one is right. But, I'm not certain. And, regardless, I don't understand why both approaches are not correct (and, since apparently at least one of them is wrong, maybe they both are). I would appreciate any insight: Is either method I used correct? If one or both is not, why not?

The problem: Assume there are 4 pairs of draws from a variable $X$ represented by $F(x) = (1-(1-x)^2)^2$ defined over $(0,1)$.

Now, take the minimum of each pair. Then take the maximum of the four minimums. Call that variable $X_m$. Now, take the other draw from the same pair as $X_m$. Call it $X_p$. Then, average $X_m$ and $X_p$. Call the random variable represented by that average $Y$. I am looking for $\mathbb{E}Y$.

Here is how I have approached it:

Method 1: $X_m$ is the max of 4 draws from the min of 2 draws from $X$. Thus, $F_m(x) =(1-(1-(1-(1-x)^2)^2)^2)^4$. Now, $X_p >X_m$ by definition. So, $F_p(x) = P[X_p < x | X_p> x_m] = \int_0^x \frac{F(x)-F(y)}{1-F(y)}f_m(y) \, dy$ Then, I average $f_m(x)$ and $f_p(x)$ to get $f_y(x)$. This way, $\mathbb{E}Y = \int_0^1yf_y(y) \, dy =0.61278$

Method 2: Given that there are 8 original draws from $X$, they can be labeled as 8 order statistics. Without changing up the pairings, label the smallest value as $X_1$, the next smallest as $X_2$, etc. Now, $X_m$ by definition can only be $X_4$ to $X_7$, and $X_p$ can only be $X_5$ to $X_8$. Working out all possible arrangements the order statistics could have been randomly paired in, we see that all values $X_4$ to $X_8$ are equally likely to occur in the set $\{X_m, X_p\}$. Thus, $Y$ is the average of the 4th -8th marginal order distributions of 8 draws from $X$.

To get $F_y(y)$ I use the standard equation for order statistics. $f_y(y) = \frac{1}{5}\sum_{k=4}^{8}\frac{8!}{(k-1)!(8-k)!}F(x)^{k-1}(1-F(x))^{8-k}f(x)$ Then, $\mathbb{E}Y = \int_0^1yf_y(y) \, dy= 0.590928$

I created a simulation in Excel, where the average $y$ over 500,000 trials is the same to four decimal places as one of the answers above. I do not know, however, if that means the method is correct (it depends partially on how good Excel's rnd() function is). Can anyone tell me why these methods do not produce the same results; if (and which) one of them is correct; and where my reasoning errors are?

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PS -- I know I've been asking a lot of questions lately. I hope this is not annoying. It's just that I only recently found out about this site, so I have a lot "built up." – OctaviaQ Aug 19 '11 at 4:01

What you are looking at is the conditional expectation. According to your question, you have 4 pairs $\left\{ x_i, y_i\right\}$, and each $x_i$ and $y_i$ are i.i.d. You seek to find

$$\mathbb{E}\left(\frac{x_1+y_1}{2} \Big| \land_{i=1}^4 (x_i<y_i) \land_{i=2}^4 x_1 > x_i \right)$$

This gives the same result as your Method 1. The method 2 is flawed because you do not know that $y_1$ is equally likely to be 5-th, 6-th, 7-th or 8-th order statistics.

Again, here is direct computation (N.B.: takes few minutes to run) of the above conditional expectation in Mathematica which confirm your first method:

In[23]:= Expectation[(x1 + y1)/
2 \[Conditioned] (x1 < y1 && x2 < y2 && x3 < y3 && x4 < y4 &&
x1 > x2 && x1 > x3 && x1 > x4)
, {x1, y1, x2, y2, x3, y3, x4, y4} \[Distributed]
ProductDistribution[{ProbabilityDistribution[{"CDF", (1 - (1 -
x)^2)^2}, {x, 0, 1}], 8}]]

Out[23]= 658390056901/1074431197125

In[24]:= N[%]

Out[24]= 0.61278

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you rock! Thank you for all your help this week. But, I still don't understand why Method 2 is wrong. It seems that the 4th -8th order stats do appear with equal frequency in the set of all possible order stats of {$x_m$, $x_p$}. – OctaviaQ Aug 19 '11 at 6:17
and yes, Method 1 is the one that matches the simulation, too, so at that makes sense too :) – OctaviaQ Aug 19 '11 at 6:19
as in, if $X_m = 4$, $X_p = \{5,6,7,8\}$ If $X_m = 5$, $X_p = \{6,7,8\}$ $X_m = 6$, $X_p = \{7,8\}$ , $X_m = 7$, $X_p = \{8\}$. Count them all up, an there are 4 instances of every number 4-8. Where am I going wrong? – OctaviaQ Aug 19 '11 at 6:27
Oh! I get it! The issue is that while there are an equal number of stats in combinations, the combinations are not equally probable. – OctaviaQ Aug 19 '11 at 6:39