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Let $ \begin{matrix} X_i & \sim & N(\mu_X, \sigma_X^2) \\ Y_i & \sim & N(\mu_Y, \sigma_Y^2) \\ X & := & \max(X_i, i \in \{ 1,2, ..., n_X\}) \\ Y & := & \max(Y_i, i \in \{ 1,2, ..., n_Y\}) \end{matrix} $

What is $\mathbb{P}[X>Y]$, as a function of $( \mu_X, \sigma_X, n_X, \mu_Y, \sigma_Y, n_Y )$?

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    $\begingroup$ You presumably wish to assume that $X$ and $Y$ are independent $\endgroup$
    – wolfies
    Jul 25, 2021 at 2:44
  • $\begingroup$ One easy case is $$P[X>Y]=\Phi\Big(\frac{\mu_X-\mu_Y}{(\sigma_X^2+\sigma_Y^2)^{1/2}}\Big) \text { if } n_X=n_Y=1$$ One harder case is $\mu_X=\sigma_X=\sigma_Y$ and $\mu_Y=0$, for which there is a relevant answer at stats.stackexchange.com/questions/550015 $\endgroup$
    – Matt F.
    Jan 30 at 18:46

1 Answer 1

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$Pr(X>Y)=\sum_{k=1}^{n_x}Pr(X_k>max\{Y_i\}, X_k=max\{X_i\})$

$Pr(X_k>max\{Y_i\}, X_k=max\{X_i\})=\int_{x_k}Pr(x_k>max\{Y_i\}, x_k=max\{X_i\})f_X(x_k)dx_k=\int_{x_k}Pr(Y_1<x_k)^{n_y}Pr(X_1\leq x_k)^{n_x-1}f_X(x_k)dx_k=\int_{x_k}\Phi(\frac{x_k-\mu_y}{\sigma_y})^{n_y}\Phi(\frac{x_k-\mu_x}{\sigma_x})^{n_x-1}f_X(x_k)dx_k$

Thus $Pr(X>Y)=\sum_{k=1}^{n_x}\int_{x_k}\Phi(\frac{x_k-\mu_y}{\sigma_y})^{n_y}\Phi(\frac{x_k-\mu_x}{\sigma_x})^{n_x-1}f_X(x_k)dx_k$, where $f_X()$ is the density of $X_i$ and will depend on $\mu_x,\sigma_x$.

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