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: Understanding the Uniform Distribution

The random variable \(X\) is uniformly distributed over the set \(S = \{1, 2, 3, \ldots, n\}\). This means each integer from 1 to \(n\) is equally likely to be selected. The probability for each outcome is given as \(P(X=k) = \frac{1}{n}\), where \(k\) is any integer within the set \(S\). The task requires us to find the expected value \(E(X)\) and the variance \(\operatorname{var}(X)\) of this distribution.

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

The expected value for a discrete random variable is calculated by the formula \(E(X) = \sum_{k=1}^{n} k \cdot P(X=k)\). Since \(P(X=k)=\frac{1}{n}\), we substitute and simplify: \[ E(X) = \sum_{k=1}^{n} k \cdot \frac{1}{n} = \frac{1}{n} \sum_{k=1}^{n} k \] Using the hint, \(\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\), substitute to find:\[ E(X) = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2} \]

Step 3: Calculate the Variance \(\operatorname{var}(X)\)

Variance is calculated by the formula \(\operatorname{var}(X) = E(X^2) - [E(X)]^2\). First, find \(E(X^2)\): \[ E(X^2) = \sum_{k=1}^{n} k^2 \cdot P(X=k) = \frac{1}{n} \sum_{k=1}^{n} k^2 \] Using the hint, \(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\), substitute to find:\[ E(X^2) = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6} \]Now calculate \(\operatorname{var}(X)\): \[ \operatorname{var}(X) = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2 \]Simplify this expression to obtain:\[ \operatorname{var}(X) = \frac{n^2 - 1}{12} \]

Step 4: Conclusion

Using the formula for the sums of integers and squares, the expected value \(E(X)\) is \(\frac{n+1}{2}\) and the variance \(\operatorname{var}(X)\) is \(\frac{n^2-1}{12}\). These results help to understand the properties of the uniform distribution on the given set.

Key Concepts

Expected Value

The expected value, often referred to as the mean, represents the average outcome we would anticipate if we repeated an experiment many times or if we took many samples of a random variable. For a discrete random variable, like our variable \(X\) which is uniformly distributed, the expected value is calculated by multiplying each possible outcome with its probability and then summing these products.

In our example, the random variable \(X\) can take any integer value from the set \( \{1, 2, 3, \ldots, n\} \). Since it is uniformly distributed, each outcome is equally likely with a probability of \( \frac{1}{n} \).

To find \(E(X)\), we use the formula:

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

Substituting in the sum of the first \(n\) integers, \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \), we simplify the formula to get:

• \( E(X) = \frac{n(n+1)}{2n} = \frac{n+1}{2} \)

This result tells us that the expected, or average, value for \(X\) in a uniform distribution is the midpoint of our set \(S\). If \(n\) is 10, for instance, the expected value is exactly 5.5.

Variance

Variance measures how much the values of a random variable deviate from the expected value. It provides an indication of the spread or dispersion of the distribution. For a discrete random variable, variance is calculated with the formula:

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

In our scenario, we need to first calculate \(E(X^2)\). This is done by averaging the squares of the outcomes, using the probability for each outcome just like we did with \(E(X)\):

• \( E(X^2) = \sum_{k=1}^{n} k^2 \cdot \frac{1}{n} \)

Using the hint, \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \), we substitute and simplify to find \( E(X^2) \):

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

So the variance is:

• \( \operatorname{var}(X) = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2 \)
• \( \operatorname{var}(X) = \frac{n^2 - 1}{12} \)

This formula shows how variance depends on \(n\), the number of elements in the set. More elements mean a larger spread, hence higher variance.

Discrete Random Variable

A discrete random variable is a variable that can take on a finite or countably infinite set of values. These are distinct and separate values - think of them as steps on a staircase where you can't land between steps. Examples of discrete random variables include the roll of a die, where outcomes are 1 through 6, or the result of flipping a coin, yielding heads or tails.

Our variable \(X\) represents the uniform distribution over the set \( \{1, 2, 3, \ldots, n\} \). This means that every number within this range is equally likely, which is a defining feature of uniform distributions. Each number is discrete, making our \(X\) a discrete random variable. In uniform distributions, every element has an identical probability, further distinguishing our variable as discrete.

This concept is pivotal in probability and statistics because it helps to simplify calculations and allows for easier predictions about a system's behavior based purely on probabilities. Advanced understanding of discrete random variables can lead to deeper insights into various real-world mechanisms and enhance statistical modeling and data science techniques.