Problem 24 - Calculus for Biology and Medicine

Question

Let \(X\) be uniformly distributed on the set $$ S=\\{1,2,3, \ldots, n\\} $$ where \(n\) is a positive integer; that is, $$ P(X=k)=\frac{1}{n}, \quad k \in S $$ (a) Find \(E(X)\). (b) Find \(\operatorname{var}(X)\). Hint: Recall that $$ \sum_{k=1}^{n} k=\frac{n(n+1)}{2} $$ and $$ \sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6} $$

Step-by-Step Solution

Answer

(a) \(E(X) = \frac{n+1}{2}\). (b) \(\operatorname{var}(X) = \frac{n^2 - 1}{12}\).

Step 1: Understand the Problem Setup

We are dealing with a uniform distribution over the set \(S = \{1, 2, 3, \ldots, n\}\), which means all outcomes are equally likely. Each outcome \(k \in S\) has a probability \(P(X=k) = \frac{1}{n}\). We need to find the expected value \(E(X)\) and the variance \(\operatorname{var}(X)\).

Step 2: Calculate the Expected Value \(E(X)\)

The expected value of a discrete uniform distribution \(X\) on the set \(S = \{1, 2, 3, \ldots, n\}\) is calculated as the average of the outcomes: \[ E(X) = \sum_{k=1}^{n} k \cdot P(X=k) = \sum_{k=1}^{n} k \cdot \frac{1}{n} = \frac{1}{n}\sum_{k=1}^{n} k. \] Using the provided formula for the sum of the first \(n\) integers, \(\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\), we find: \[ E(X) = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2}. \]

Step 3: Calculate the Expected Value of \(X^2\)

To find the variance, we need \(E(X^2)\). Use the property: \[ E(X^2) = \sum_{k=1}^{n} k^2 \cdot P(X=k) = \frac{1}{n}\sum_{k=1}^{n} k^2. \] The given formula for \(\sum_{k=1}^{n} k^2\) is \(\frac{n(n+1)(2n+1)}{6}\): \[ E(X^2) = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6}. \]

Step 4: Calculate the Variance \(\operatorname{var}(X)\)

The variance is calculated using \(\operatorname{var}(X) = E(X^2) - (E(X))^2\). We established \(E(X) = \frac{n+1}{2}\) and \(E(X^2) = \frac{(n+1)(2n+1)}{6}\). Let's calculate \((E(X))^2\): \[ (E(X))^2 = \left(\frac{n+1}{2}\right)^2 = \frac{(n+1)^2}{4}. \] Substituting these into the variance formula gives: \[ \operatorname{var}(X) = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}. \] Simplify to find the variance: \[ \operatorname{var}(X) = \frac{(n+1)(2n+1) \cdot 2}{12} - \frac{3(n+1)^2}{12} = \frac{(n+1)((2n+1) - 3(n+1))}{12} = \frac{n^2 - 1}{12}. \]

Step 5: Conclusion

Thus, the expected value \(E(X)\) is \(\frac{n+1}{2}\), and the variance \(\operatorname{var}(X)\) is \(\frac{n^2 - 1}{12}\).

Key Concepts

Expected Value

The expected value, often denoted as \(E(X)\), is a measure of the "central" or "typical" value that a random variable takes. It's essentially a weighted average of all possible outcomes, where the weights are their probabilities. In the context of a discrete uniform distribution, where every outcome is equally likely, computing the expected value becomes straightforward.

For a discrete uniform distribution over a set \(S=\{1,2,3, \ldots, n\}\), each number has the same chance of occurring, specifically \(\frac{1}{n}\). Therefore, the expected value formula is:

• \[ E(X) = \sum_{k=1}^{n} k \cdot P(X=k) = \frac{1}{n} \sum_{k=1}^{n} k \]

We use the arithmetic series sum formula to get \(\frac{n(n+1)}{2}\). This gives us the elegant result:

• \[ E(X) = \frac{n+1}{2} \]

This formula tells us the expected value of a uniformly distributed set is simply the middle point or "average" of the numbers in the set.

Variance

Variance, denoted \(\operatorname{var}(X)\), quantifies how much the values of a random variable differ from the expected value. In other words, it measures the spread or "dispersion" of a set of values. For a discrete uniform distribution, calculating variance involves determining both \(E(X^2)\) and \((E(X))^2\).

The formula to calculate variance is:

• \[ \operatorname{var}(X) = E(X^2) - (E(X))^2 \]

To get \(E(X^2)\), we use:

• \[ E(X^2) = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6} \]

Now, \((E(X))^2\) can be calculated as:

• \[ (E(X))^2 = \left(\frac{n+1}{2}\right)^2 = \frac{(n+1)^2}{4} \]

By plugging these values back into the variance formula, we simplify and arrive at:

• \[ \operatorname{var}(X) = \frac{n^2 - 1}{12} \]

This result provides a measure of how tightly or loosely the numbers in the set \(S\) are clustered around their average.

Discrete Probability

Discrete probability applies to scenarios where outcomes fall into distinct, separate categories. Here, it refers to the likelihood of each specific event occurring. In our problem, \(X\) is defined by a uniform distribution on set \(S = \{1, 2, 3, \ldots, n\}\), meaning each number has the same probability of occurring.

In a uniform distribution, particularly discrete, every element shares the probability evenly distributed across all possible outcomes. Thus, the probability of drawing any specific number \(k\) from \(S\) is:

• \[ P(X=k) = \frac{1}{n} \]

Discrete probability is incredibly useful for creating models where outcomes are countable and equally probable, like rolling a fair die. Each side has a \(\frac{1}{6}\) probability, mirroring our scenario where selecting any number from a set of \(n\) choices holds a \(\frac{1}{n}\) probability.

Using discrete probability, we can effectively compute deeper statistical properties like expected values and variance. It is a fundamental aspect that allows us to perform reliable calculations in various discrete scenarios.