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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)?

** $P(X_1 < \min(X_i,\ldots, X_n))$ across different normal random variables

** What is $P(X_1>X_2 , X_1>X_3,... , X_1>X_n)$?

** What is a method to calculate precisely $P(Y \geq X, Y\leq Z)$, given three independent random variables $X, Y$, and $Z$

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

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  • $\begingroup$ This is a great explanation! Very useful indeed, thank you. $\endgroup$ – quantB Dec 27 '20 at 16:55

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