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

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.

• There is a simple formula for the p.d.f.s of those minima and maxima, which could in principle be used to find their medians -- see stats.stackexchange.com/questions/155187/… Commented Jul 16 at 2:02
• @jwimberley It's easier and more general to work with the CDF: that gives you an effective formula, not just one in principle.
– whuber
Commented Jul 16 at 2:03

## 2 Answers

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

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 $$1/(k+1)$$.