The Probability of the Minimum of X Being Greater than the Maximum of Y

Statistics
Probability Distributions
Derivation
Change of Variables
Transformations
Stochastic Orders
Identical and Independently Distributed
Random Variables
Statistical Independence
Smooth Functions
Minimum
Maximum
Probability
Order Statistics
Convolution
Convolution Theorem
Fourier Transform
Author

Galen Seilis

Published

March 2, 2024

Suppose we have a collection of IID random variables \(\{ X_1, \ldots, X_n \}\), and we also have a second collection of IID random variables \(\{ Y_1, \ldots, Y_m \}\). Each \(X_i \sim F_X\) and \(Y_i \sim F_Y\) and we will assume that all these variables are statistically independent. Let us also assume that that \(F_X\) and \(F_Y\) are in the \(\mathcal{C}^1\) smoothness class.

Suppose we would like to find \(Pr \left[ \min (X_1, \ldots, X_n) > \max (Y_1, \ldots, Y_n) \right]\), which is equal to \(Pr \left[ \min (X_1, \ldots, X_n) - \max (Y_1, \ldots, Y_n) > 0 \right]\). The relevance of this observation is that \(\min (X_1, \ldots, X_n) - \max (Y_1, \ldots, Y_n)\) is an expression for which we can derive the distribution

For the minimum of the collection of \(X\) variables we can use order statistics to obtain:

\[F_{X_{(1)}}(x) = Pr \left[ \min \{X_1, \ldots, X_n \} \leq x \right] = 1 - \left[1 - F_X(x) \right]^n.\]

Likewise, the maximum of the \(Y\) variables comes from order statistics:

\[\max (Y_1, \ldots, Y_m) \sim \left[ F_Y \right]^m\]

We would like to put our problem into the form of adding two independent random variables \(U + V\) because then we can convolve them to obtain the distribution of the sum. Taking \(U = X_{(1)}\) as our minimum of the \(X\) variables, and \(V = - Y_{(m)}\) of the \(Y\) variables, we can next consider the distribution of \(V\) to be a reflection of \(Y_{(m)}\). The smooth change in variables works out to be

\[f_V(v) = m f_Y(-y) \left[ F_Y(-y)\right]^{m-1}.\]

To compute the convolution of the densities \(f_U \star f_V\) we need the density \(f_U\):

\[f_U(u) = \frac{d}{dx} F_X(x) = n \left[ 1 - F_X(x) \right]^{n-1}f_X(x)\]

We can use the convolution theorem to obtain the result via the Fourier transform \(\mathcal{F}\) and its inverse \(\mathcal{F}^{-1}\).

\[f_{X_{(1)} - Y_{(m)}} = \mathcal{F}^{-1} \left\{ \mathcal{F} \left\{ n \left[ 1 - F_X(x) \right]^{n-1} f_X(x) \right\} \mathcal{F} \left\{ m f_Y(-y) \left[ F_Y(-y) \right]^{m-1} \right\} \right\}\]

Finally, we can obtain the cumulative distribution by integrating:

\[F_{X_{(1)} - Y_{(m)}}(x,y) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{X_{(1)} - Y_{(m)}}(x,y) dx dy\]