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This is somewhat similar to a number of existing questions, e.g. this one, but I couldn't find a way to use them to solve this specific case.

Let $X_1..X_n$ be normally distributed variables, with means $\mu_1..\mu_n$ and standard deviations $\sigma_1..\sigma_n$. Let $z_1..z_n$ be numbers drawn from each of the respective distributions, and let $z_{max}=max\{z_i\}$.

What can be said about the probability that $z_{max}$ was drawn from each of the distributions $\{X_i\}$? In other words (and I'm not sure about the notation here), can $P[number\ drawn\ from\ X_i\ is\ maximal]$ be calculated?

At first glance this doesn't seem very tough, but I've tried various approaches here and none seemed to yield the correct answers for obvious cases, e.g. three normal distributions with the same mean and standard deviation.

Any ideas or hints will be appreciated.

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  • $\begingroup$ For notation you might say $Z=\max \{X_i\}$ and ask for $\Pr(Z=X_i)$ for each $i$. $\endgroup$ – Henry Dec 26 '12 at 9:40
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Denote the p.d.f. and c.d.f. of a normal distribution with mean $\mu$ and variance $\sigma^2$ by $\phi(x; \mu, \sigma^2)$ and $\Phi(x; \mu, \sigma^2)$ respectively. Then the required probability is just $$ \operatorname{Pr}\left(X_i=\max\{X_1,\ldots,X_n\}\right) =\int_{-\infty}^\infty \phi(x; \mu_i, \sigma_i^2) \prod_{j\not=i}\Phi(x; \mu_j, \sigma_j^2) dx. $$ When $n=3,\ \mu_1=\mu_2=\mu_3=\mu$ and $\sigma_1=\sigma_2=\sigma_3=\sigma$, we have \begin{align} \operatorname{Pr}\left(X_i=\max\{X_1,X_2,X_3\}\right) &=\int_{-\infty}^\infty \phi(x; \mu, \sigma^2)\, \Phi^2(x; \mu, \sigma^2) dx\\ &=\int_{-\infty}^\infty \Phi^2(x; \mu, \sigma^2)\ d\Phi(x; \mu, \sigma^2)\\ &=\frac13\left.\Phi^3(x; \mu, \sigma^2)\right|_{-\infty}^\infty =\frac13. \end{align}

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  • $\begingroup$ Thanks, I was close, but not quite there. One edit: The variable indexes under the phi sign should be j, not i. This is too short for an edit :) $\endgroup$ – etov Dec 26 '12 at 10:52
  • $\begingroup$ Corrected. Thanks for pointing out the typo. $\endgroup$ – user1551 Dec 26 '12 at 11:05

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