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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!

vonjd
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1 Answers1

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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 $1$.

tortue
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