by ee8291
Last Updated July 09, 2019 02:19 AM

I have no starting point on which formula to use:

One draws 5 samples from a cont. distribution. Now, what is the approx. probability p that the median of the Population falls within the smallest and largest samples drawn?**

Can anybody guide me on which formula to use or if any other approach is better suited?

I'm not sure about the general answer but if the continuous distribution is the standard normal then the probability that the median lies between the min and the max of a sample of size $n$ equals

$$ \frac{ 2^{n-1} - 1 }{2^{n-1}} $$

You can work this out based on the distribution of normal order statistics, which I will leave to you as an exercise since this appears to be a self study question.

Let $m$ denote the true median of the distribution. Since the distribution is continuous, by definition there is a probability of one-half that a given value will fall above/below the median. Hence, you have:

$$\mathbb{P}(X_{(1)} > m) = \prod_{i=1}^n \mathbb{P}(X_i > m) = \prod_{i=1}^n \frac{1}{2} = \frac{1}{2^n},$$

$$\mathbb{P}(X_{(n)} < m) = \prod_{i=1}^n \mathbb{P}(X_i < m) = \prod_{i=1}^n \frac{1}{2} = \frac{1}{2^n}.$$

Putting this together and using the inclusion-exclusion principle you get:

$$\begin{equation} \begin{aligned} \mathbb{P}(X_{(1)} \leqslant m \leqslant X_{(n)}) &= 1 - \mathbb{P}(X_{(1)} > m \ \text{ or } \ X_{(n)} < m) \\[6pt] &= 1 - \mathbb{P}(X_{(1)} > m) - \mathbb{P}(X_{(n)} < m) + \mathbb{P}(X_{(n)} < m < X_{(1)}) \\[6pt] &= 1 - \mathbb{P}(X_{(1)} > m) - \mathbb{P}(X_{(n)} < m) + 0 \\[6pt] &= 1 - \frac{1}{2^n} - \frac{1}{2^n} \\[6pt] &= 1 - \frac{1}{2^{n-1}} \\[6pt] &= \frac{2^{n-1} - 1}{2^{n-1}}. \\[6pt] \end{aligned} \end{equation}$$

Note that this result holds for any continuous distribution. Unsurprisingly, as $n \rightarrow \infty$ we have $\mathbb{P}(X_{(1)} \leqslant m \leqslant X_{(n)}) \rightarrow 1$, so the range of the sample will eventually come to encompass the true median, with probability approaching one.

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