What is the distribution of maximum of a pair of iid draws, where the minimum is an order statistic of other minima?

Consider n\cdot m independent draws from cdf F(x), which is defined over 0-1, where n and m are integers. Arbitrarily group the draws into n groups with m values in each group. Look at the minimum value in each group. Take the group that has the greatest of these minima. Now, what is the distribution that defines the maximum value in that group?
More generally, what is the distribution for the j-th order statistic of m draws of F(x), where the kth order of those m draws is also the pth order of the n draws of that kth order statistic?

All of that is at the most abstract, so here is a more concrete example.
Consider 8 draws of F(x). Group them into 4 pairs of 2. Compare the minimum value in each pair. Select the pair with the highest of these 4 minima. Label that draw “a”. Label the other value in that same pair as “b”. What is the distribution F_b(b)? We know b>a. We know a is the maximum of 4 minimums of F(x), of F_a(a) = (1-(1-F(x))^2)^4. What is F_b(b)?

Answer

I answer this: “Arbitrarily group the draws into n groups with m values in each group. Look at the minimum value in each group. Take the group that has the greatest of these minima. Now, what is the distribution that defines the maximum value in that group?”
Let X_{i,j} the i-th random variable in group j and f(x_{i,j}) (F(x_{i,j})) its density (cdf) function.
Let X_{\max,j}, X_{\min,j} the maximum and minimum in group j. Let X_{final} the variable that results at the end of all process.
We want to calculate P(X_{final}<x) which is
P(X_{\max,j_0}<x \hbox{ and } X_{\min,j_0}=\max_j{X_{\min,j}} \hbox { and } 1\leq j_0\leq n)
=nP(X_{max,1}<x \hbox{ and } X_{\min,1}=\max_j{X_{\min,j}})
=nmP(X_{1,1}<x\hbox{ and } X_{1,1}=\max_i(X_{i,1})\hbox{ and } X_{\min,1}=\max_j{X_{\min,j}})
=nmP(X_{1,1}<x, X_{1,1}>X_{2,1}>\max_{j=2\ldots n} X_{min,j},\ldots,X_{1,1}>X_{m,1}>\max_{j=2\ldots n} X_{min,j})
Now, let Y=\max_{j=2\ldots n} X_{min,j} and W=X_{1,1}.

A reminder: if X_1,\ldots X_n are iid with pdf (cdf) h (H), then X_{\min} has pdf h_{\min}=nh(1-H)^{n-1} and X_{\max} has pdf h_{max}=nhH^{n-1}.
Using this, we get the pdf of Y is
g(y)=(n-1)mf(1-F)^{m-1}[\int_0^y mf(z)(1-F(z))^{m-1} dz]^{n-2},n\geq 2

Note that Y is a statistics that is independent of group 1 so its joint density with any variable in the group 1 is the product of densities.
Now the above probability becomes
nm\int_0^x f(w)[\int_0^w \int_y^w f(x_{2,1})dx_{2,1}\ldots\int_y^w f(x_{m,1})dx_{m,1}g(y)dy]dw
=nm\int_0^x f(w)[\int_0^w (F(w)-F(y))^{m-1}g(y)dy]dw
By taking derivative of this integral wrt x and using binomial formula we obtain the pdf of X_{final}.

Example: X is uniform, n=4, m=3. Then
g(y)=9(1-y)^2(3y+y^3-3y^2)^2,

P(X_{final}<x)=(1/55)x^{12}-(12/55)x^{11}
+ (6/5)x^{10}-(27/7)x^9+(54/7)x^8-(324/35)x^7+(27/5)x^6.

Mean of X_{final} is 374/455=0.822 and its s.d. is 0.145 .

Attribution
Source : Link , Question Author : OctaviaQ , Answer Author : mpiktas

Leave a Comment