Computing the probability $P(\exists X\in\{X_1,\ldots, X_N\}:X>\text{max}\{Y_1,\ldots, Y_M\})$

I am trying to somewhat generalize a question, which has been asked in one way or another a several times here on StackExchange**. However, I have not managed to find an answer to the below problem.

Suppose, $$X_1,\ldots,X_N\sim\mathcal{N}(\mu_X, \sigma_X)$$ and $$Y_1,\ldots,Y_M\sim\mathcal{N}(\mu_Y, \sigma_Y)$$, which are independently drawn. We are interested in the following probability:

$$P(\text{there exists at least one }X_i\text{ such that } X_i \text{ greater than max}\{Y_1,\ldots, Y_M\})=P(\exists X\in\{X_1,\ldots, X_N\}:X>\text{max}\{Y_1,\ldots, Y_M\})\tag{1}$$

I have tried to decompose this problem into smaller problems by writing the "there exists" part as

$$P(\exists X\in\{X_1,\ldots, X_N\}:X>\text{max}\{Y_1,\ldots, Y_M\})=P(X_1>\text{max}\{Y_1,\ldots, Y_M\}\cup\ldots\cup X_N>\text{max}\{Y_1,\ldots, Y_M\}),$$

but the RHS does not contain any information about the partial order within the set of $$\{X_i\}$$. Trying to add this information to the RHS makes it overly complicated and I am not even sure if this would work out. Is there a known closed form expression or a numerical way to calculate the value in Eq.(1)?

• I assume the draws are independent? Dec 27 '20 at 13:26
• @Forgottenscience Good point! Yes, the draws are indeed independent. Dec 27 '20 at 13:34

First of all, you can simplify this condition to $$X_{(N)} > Y_{(M)}$$ using the notation for order statistics, which will simpify your question a fair bit. Assuming the values are all independent, these two random variables will also be independent, with respective distribution functions:$$^\dagger$$

$$F_{X_{(N)}}(x) = \Phi \Big( \frac{x-\mu_X}{\sigma_X} \Big)^N \quad \quad \quad \quad \quad F_{Y_{(M)}}(y) = \Phi \Big( \frac{y-\mu_Y}{\sigma_Y} \Big)^M,$$

and corresponding densities:$$^\dagger$$

\begin{align} f_{X_{(N)}}(x) = \frac{N}{\sigma_X} \cdot \phi \Big( \frac{x-\mu_X}{\sigma_X} \Big) \cdot \Phi \Big( \frac{x-\mu_X}{\sigma_X} \Big)^{N-1} \\[18pt] f_{Y_{(M)}}(y) = \frac{M}{\sigma_Y} \cdot \phi \Big( \frac{y-\mu_Y}{\sigma_Y} \Big) \cdot \Phi \Big( \frac{y-\mu_Y}{\sigma_Y} \Big)^{M-1}. \\[6pt] \end{align}

Thus, we can write the probability of interest as:

\begin{align} \mathbb{P}(X_{(N)} > Y_{(M)}) &= \int \limits_\mathbb{R} f_{X_{(N)}}(y) \cdot \mathbb{P}(Y_{(M)} \leqslant y) \ dy \\[6pt] &= \int \limits_\mathbb{R} f_{X_{(N)}}(y) \cdot F_{Y_{(M)}}(y) \ dy \\[6pt] &= \frac{N}{\sigma_X} \cdot \int \limits_\mathbb{R} \phi \Big( \frac{y-\mu_X}{\sigma_X} \Big) \cdot \Phi \Big( \frac{y-\mu_X}{\sigma_X} \Big)^{N-1} \cdot \Phi \Big( \frac{y-\mu_Y}{\sigma_Y} \Big)^M \ dy. \\[6pt] \end{align}

There is no closed form expression for this integral, but it can be evaluated numerically to obtain the probability of interest to you.

$$^\dagger$$ Here we use $$\Phi$$ and $$\phi$$ to denote the cumulative distribution function and density function for the standard normal distribution.

• This is a great explanation! Very useful indeed, thank you. Dec 27 '20 at 16:55