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Suppose that we have $n$ independent normal random variables, $X_1,...,X_n$. What is the probability that normal random variable $Y$ is less than all normal random variables $X_1,...,X_n$ ? These random variables are all independent but they are not identically distributed.

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    $\begingroup$ You probably want to assume that the random variables $X_1,\dots,X_n, Y$ are independent, else we can say nothing. I expect the mean, variance of all of these are specified. It would make things easier if the $X_i$ and $Y$ were identically distributed. $\endgroup$ – André Nicolas Jan 27 '13 at 2:03
  • $\begingroup$ I edited the question thank you. $\endgroup$ – May Jan 27 '13 at 2:06
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Since all the random variables are independent, we have that $$P\{X_1 > a, X_2 > a, \cdots , X_n > a \mid Y = a\} = \prod_{i=1}^n P\{X_i > a \mid Y = a\} = \prod_{i=1}^n Q\left(\frac{a - \mu_i}{\sigma_i}\right)$$ and so $$P\{X_1 > Y, X_2 > Y, \cdots , X_n > Y\} = \int_{-\infty}^\infty \prod_{i=1}^nQ\left(\frac{a - \mu_i}{\sigma_i}\right)f_Y(a) \,\mathrm da$$ where $Q(x) = 1 - \Phi(x)$ is the complementary standard normal distribution function. Further simplification is not possible though if the $X_i$ are iid $N(0,1)$ and $Y \sim N(\mu, 1)$, an attack analogous to that described in this answer on stats.SE might yield some results.

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  • $\begingroup$ does it also work for the case that these random variables are not independent? $\endgroup$ – May Jan 31 '13 at 21:44
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    $\begingroup$ No, it does not "work" for the case of dependent random variables because the conditional probability shown in the first displayed equation above does not factor into the product of the conditional probabilities. You would just get $$\int_{-\infty}^\infty P\{X_1>a,X_2>a,\cdots,X_n>a\mid Y=a\}f_Y(a)\,\mathrm da$$ as the final "answer". $\endgroup$ – Dilip Sarwate Jan 31 '13 at 21:59
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Simply the product of the probabilities: $$\prod_iP(Y<X_i)$$

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  • $\begingroup$ does it also work for the case that these random variables are not independent? $\endgroup$ – May Jan 27 '13 at 3:12
  • $\begingroup$ @May - No it doesn't. The ability to use the "product of the probabilities" is almost the definition independent random variables. $\endgroup$ – nbubis Jan 27 '13 at 3:39
  • $\begingroup$ I think it may not be the product. For simplicity assume variances $1$. If $Y\lt X_1$, where say has small mean, that would seem to increase the probability that $Y\lt X_2$, where $X_2$ has large mean. $\endgroup$ – André Nicolas Jan 27 '13 at 4:00
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    $\begingroup$ I too don't think this answer is correct. The events $(Y < X_1)$ and $(Y < X_2)$ are not independent, and the product rule does not apply. $\endgroup$ – Dilip Sarwate Jan 27 '13 at 4:24

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