Find the probability that the larger of two iid samples exceeds the population median.

Understanding
We assume that \(X_1\) and \(X_2\) are two continuous, independent, identically distributed (iid) random samples and that \(m\) is the population median. The probability of a continuous random sample exceeding the median is \(\frac{1}{2}\).

Method

 * 1) Note that the probability, that the maximum value of \(X1\) and \(X2\) is greater than \(m\), is equal to: \(1\) minus the probability that the maximum value of \(X1\) and \(X2\) is less than or equal to \(m\), because the sum of the two probabilities would give \(1\), as shown below: \[\begin{align*}P(\max(X_1,X_2)>m)&=1-P(\max(X_1,X_2)\le m)\end{align*}\]
 * 2) Note that this \(P(\max(X_1,X_2)\le m)=P(X_1\le m,X_2\le m)\) is true according to "well-order".
 * 3) Substitute the new expression, \(P(X_1\le m,X_2\le m)\), into the previous expression, as shown below: \[\begin{align*}P(\max(X_1,X_2)>m)&=1-P(X_1\le m,X_2\le m)\\&=1-P(X_1\le m)P(X_2\le m)&\hbox{by independence}\\&=1-\frac{1}{2}\cdot\frac{1}{2}\\&=\frac{3}{4}\end{align*}\]

Everything else
The last part of the method section is incomplete. It's written as a mathematical statement rather than a step-by-step sequence. This will be changed to fit with the Substepr format (as described on the Main Page).

The above solution can be generalized as follows. \[\begin{align*} P(\max(X_1,\ldots,X_n)>m)&=1-P(X_1\le m)\cdots P(X_n\le m)\\ &=1-\frac{1}{2^n} \end{align*}\]

The above solutions only work for continuous random variables. For discrete random variables, it is possible that the sample is exactly the same as the median. In this case the probability of exceeding the median is less that \(\frac{1}{2}\).

Reference
Casella and Berger, Statistical Inference, Second Edition, p. 259.

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