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Suppose we have a normal distribution with mean $\mu$ and variance $\sigma^2$. Suppose we draw $n$ samples from this distribution. What is the expectation of the minimum of these samples?

user14717
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    If you want $E(V)$ for $V =\min(Z_1, \dots, Z_{10}),$ where $Z_i$ are independent standard normals, the following simulation in R gives $E(V) \approx -1.538,$ with 95% margin $0.001$ of simulation error. m = 10^6; v = replicate(m, min(rnorm(10))); mean(v) – BruceET Dec 29 '17 at 22:21

1 Answers1

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$X_1, ..., X_n$ are i.i.d $\mathcal{N}(\mu, \sigma^2)$.

$Y = \min\{X_i, 1\leq i \leq n\}$.

$F_Y(y) = \mathrm{Pr}\{Y \leq y\} = 1 - \mathrm{Pr}\{Y > y\} = 1 - \mathrm{Pr}\{\min\{X_i\} > y\} = 1 - \mathrm{Pr}\{\cap_i X_i > y\}$

If $X_i$ are independent,

$F_Y(y) = 1 - (\mathrm{Pr}\{X > y\})^n = 1 - (1 - F_X(y))^n$

$f_Y(y) = \frac{\partial}{\partial y} F_Y(y) = n \left(1 - F_X(y)\right)^{n-1} \frac{\partial}{\partial y} F_X(y) = n \left(1 - F_X(y)\right)^{n-1} f_X(y)$

Finally,

$\mathbb{E}[Y] = \int y f_Y(y)\mathrm{d}y = n\int y\left(1 - F_X(y)\right)^{n-1} f_X(y) \mathrm{d}y$

where

$$ f_X(y) = \frac{1}{\sqrt{2 \pi \sigma^2}} \exp \left(-\frac{1}{2\sigma^2}(y-\mu)^2 \right) $$

$$ F_X(y) = \int_{-\infty}^y f_X(x)\mathrm{d}x = \int_{-\infty}^y \frac{1}{\sqrt{2 \pi \sigma^2}} \exp\left(-\frac{1}{2\sigma^2}(x-\mu)^2\right) \mathrm{d}x $$

You better use numerical evaluation.

AlexTP
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