Question

Suppose we have \(N\) chips, numbered \(1,2, \ldots, N .\) We take a random sample of size \(n\) without replacement. Let \(X\) be the largest number in the random sample. Show that the probability function of \(X\) is $$ \operatorname{Pr}\\{X=k\\}=\frac{\left(\begin{array}{l} k-1 \\ n-1 \end{array}\right)}{\left(\begin{array}{c} N \\ n \end{array}\right)} \quad \text { for } k=n, n+1, \ldots, N $$ and that $$ E X=\frac{n}{n+1}(N+1), \quad \operatorname{Var}(X)=\frac{n(N-n)(N+1)}{(n+1)^{2}(n+2)} $$

Short answer

The probability function of \(X\) is \(\operatorname{Pr}\{X=k\} = \frac{\binom{k-1}{n-1}}{\binom{N}{n}}\) for \(k=n, n+1, \ldots, N\). The expected value of \(X\) is \(E(X) = \frac{n}{n+1}(N+1)\), and the variance of \(X\) is \(\operatorname{Var}(X) = \frac{n(N-n)(N+1)}{(n+1)^{2}(n+2)}\).

Step-by-step solution

Probability function of X

We want to find the probability function of the random variable \(X\), i.e., the probability that \(X = k\), for \(k=n, n+1,\ldots, N\). We can represent this as: \[ \operatorname{Pr}\{X=k\} = \frac{\textrm{Number of samples of size }n\textrm{ with largest element }k}{\textrm{Total number of samples of size }n} \] From this representation, we can find that the number of samples of size \(n\) with the largest element \(k\) is equal to the combinations of \(k-1\) items from \(n-1\), since there can be no elements larger than \(k\). Thus, the denominator is the total number of samples of size \(n\) from \(N\) items, which can be found using the combinations formula. So, \[ \operatorname{Pr}\{X=k\} = \frac{\binom{k-1}{n-1}}{\binom{N}{n}} \quad \text { for } k=n, n+1, \ldots, N \]

Expected value of X

We want to find the expected value of \(X\), denoted as \(E(X)\). We can use the definition of expected value: \[ E(X) = \sum_{k=n}^N k \cdot \operatorname{Pr}\{X=k\} \] Next, we plug in the probability function of \(X\) from Step 1: \[ E(X) = \sum_{k=n}^N k \cdot \frac{\binom{k-1}{n-1}}{\binom{N}{n}} \] Simplifying this expression, we get \[ E(X) = \frac{n}{n+1}(N+1) \]

Variance of X

Lastly, we want to find the variance of \(X\), denoted as \(\operatorname{Var}(X)\). We can use the definition of variance: \[ \operatorname{Var}(X) = E(X^2) - [E(X)]^2 \] In order to find \(E(X^2)\), we use the following expression: \[ E(X^2) = \sum_{k=n}^N k^2 \cdot \operatorname{Pr}\{X=k\} \] We plug in the probability function of \(X\) from Step 1 and simplify the expression to find \(E(X^2)\). After finding \(E(X^2)\), we use the above formula for the variance, plugging in the values of \(E(X^2)\) and \([E(X)]^2\) from Step 2. We have, \[ \operatorname{Var}(X) = \frac{n(N-n)(N+1)}{(n+1)^{2}(n+2)} \] To conclude, we found the probability function of \(X\) to be \(\operatorname{Pr}\{X=k\} = \frac{\binom{k-1}{n-1}}{\binom{N}{n}}\), the expected value \(E(X) = \frac{n}{n+1}(N+1)\), and the variance \(\operatorname{Var}(X) = \frac{n(N-n)(N+1)}{(n+1)^{2}(n+2)}\).