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This is a follow-up question on this one: Expected value for maximum of n normal random variable

@RobertIsrael states the following:

Presumably the $X_i$ are independent. If $\Phi$ is the standard normal cdf, $$P(\max_i X_i < \mu + t \sigma) = \prod_i P(X_i < \mu + t \sigma) = \Phi(t)^n$$ so $$ E[\max_i X_i] = \mu + \sigma \int_{-\infty}^\infty t \dfrac{d}{dt} \Phi(t)^n\ dt $$

I can see some of the ideas that lead to this formula (e.g. the multiplication because of independence) but I don't see all of the details. It would be very helpful if somebody could explain the derivation in detail... Thank you!

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It is basically using the definition of expected value and one simple substitution. The first equation can be rewritten as $$ \mathbb{P}(\max_i X_i < t) = \Phi\left(\frac{t - \mu}{\sigma}\right)^n. $$ Then, by definition $$ \mathbb{E} \max_i X_i = \int_{-\infty}^{+\infty} t \cdot \frac{d}{dt} \Phi\left(\frac{t - \mu}{\sigma}\right)^n dt = \int_{-\infty}^{+\infty} \sigma(t + \mu) \cdot \sigma \frac{d}{dt} \Phi(t)^n \frac{dt}{\sigma} = \mu + \sigma \int_{-\infty}^{+\infty} t\cdot \frac{d}{dt} \Phi(t)^n dt, $$ as desired.

So, in the first equation we used the substitution $t \to \sigma t + \mu$, in the second one we simplified the expression and used that integral of the density function over the whole domain is $1$.

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  • $\begingroup$ Thank you, which definition are you using for the $d/dt$? $\endgroup$ – vonjd Jan 20 at 9:54
  • $\begingroup$ $\frac{d}{dt}$ is the derivative operator, so $\frac{d}{dt} \Phi(t)^n$ becomes the density function. is it clear now? $\endgroup$ – pointguard0 Jan 20 at 9:56
  • $\begingroup$ because the probability density function (which is used in the definition of expected value) is the derivative of cumulative distribution function $\endgroup$ – pointguard0 Jan 20 at 10:15
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    $\begingroup$ Yes, got it - Thank you! $\endgroup$ – vonjd Jan 20 at 11:02

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