What is the median of the minimum or maximum of multiple samples?

Question

Suppose I have a variable with a known distribution, and suppose I sample that variable k times and record the minimum. If I repeat this many times, will the median of the minimum converge to a predictable value, and if so, what is that value?

My instinct tells me that the median will converge to the value for which the cumulative probability is 1/(k+1), but I have not been able to confirm that.

Note that I found a discussion of what the expected value of the maximum would be for samples from a normal distribution: Expected value of maximum of samples from normal distribution. However, that discussion doesn't say anything about the median, which I think should be simpler.

Answer 1

Let the distribution function of the variable $X$ be $F.$ Given any number $y,$ the chance that the minimum of $k$ iid copies of $X$ exceeds $y$ is the chance that all of those $k$ values exceed $y.$ Because they are independent, this is the product of the individual chances of exceedance, $1-F(y),$ whence

$$\Pr(\min(X_1,\ldots, X_k)\gt y) = (1-F(y))^k.$$

Thus, the distribution function of the minimum is

$$F_{(k)}(y) = \Pr(\min(X_1,\ldots, X_k)\le y) = 1-(1-F(y))^k.$$

To find a median, look for a $y$ for which the the graph of $F_{(k)}$ crosses the value $1/2:$ that is, look for a solution to

$$\frac{1}{2} = 1 - (1 - F(y))^k$$

(or, possibly, the right hand side could exceed $1/2$ but would be strictly less than $1/2$ for any smaller $y$).

Such a solution would satisfy

$$F(y) = 1 - \left(\frac{1}{2}\right)^{1/k}.$$

That is,

Any median of the minimum of $k$ iid random values is a $1 - 2^{-1/k}$ quantile.

Whenever $k \gt 1,$ $1-2^{-1/k} \ne 1/(k+1).$

Answer 2

The cumulative distribution function $F_{(1)}$ for the minimum of $k$ draws from a distribution is $1-(1-F(x))^k$, the complement of the probability that all $k$ draws are above $x$.

So the median for that minimum occurs when $$F_{(1)}(x) = 1-(1-F(x))^k=\frac12$$ $$F(x) = 1-\left(\frac12\right)^{1/k}$$

In other words, the median of the minimum is the value for which the cumulative probability is $1-\left(\frac12\right)^{1/k}$.

As to what you might have been thinking of:

There is no general formula for the mean of that minimum. But for bounded distributions we can write the mean as $(\max x) - \int F_{(1)}(x) dx$, and for a uniform distribution between $0$ and $1$ this gives $$1-\int_0^1 (1-(1-x)^k) dx$$ $$= \int_0^1 (1-x)^k dx$$ $$= 1/(k+1)$$

In particular, the cdf or cumulative probability is always distributed uniformly between $0$ and $1$. So the mean of the cdf of the minimum is always $1/(k+1)$.