Problem 30 - Calculus for Biology and Medicine

Question

Let \(X\) and \(Y\) be two independent random variables with probability mass function described by the following table: $$\begin{array}{rcc} \hline {\boldsymbol{k}} & \boldsymbol{P}(\boldsymbol{X}=\boldsymbol{k}) & \boldsymbol{P}(\boldsymbol{Y}=\boldsymbol{k}) \\ \hline-3 & 0.1 & 0.1 \\ -1 & 0.1 & 0.2 \\ 0 & 0.2 & 0.1 \\ 0.5 & 0.3 & 0.3 \\ 2 & 0.15 & 0.1 \\ 2.5 & 0.15 & 0.2 \\ \hline \end{array}$$ (a) Find \(E(X)\) and \(E(Y)\). (b) Find \(E(X+Y)\). (c) Find \(\operatorname{var}(X)\) and \(\operatorname{var}(Y)\). (d) Find \(\operatorname{var}(X+Y)\).

Step-by-Step Solution

Answer

(a) \( E(X) = 0.425, E(Y) = 0.35 \); (b) \( E(X+Y) = 0.775 \); (c) \( \operatorname{var}(X) = 2.431875, \operatorname{var}(Y) = 2.7025 \); (d) \( \operatorname{var}(X+Y) = 5.134375 \).

Step 1: Find E(X)

To find the expected value of a random variable, we use the formula:\[ E(X) = \sum_{k} k \cdot P(X=k) \]Using the values from the table:\[ E(X) = (-3 \cdot 0.1) + (-1 \cdot 0.1) + (0 \cdot 0.2) + (0.5 \cdot 0.3) + (2 \cdot 0.15) + (2.5 \cdot 0.15)\]\[ = -0.3 - 0.1 + 0 + 0.15 + 0.3 + 0.375 \]\[ = 0.425 \]

Step 2: Find E(Y)

Similarly, we find the expected value for the variable Y using the formula for expected value:\[ E(Y) = \sum_{k} k \cdot P(Y=k) \]Using the values from the table:\[ E(Y) = (-3 \cdot 0.1) + (-1 \cdot 0.2) + (0 \cdot 0.1) + (0.5 \cdot 0.3) + (2 \cdot 0.1) + (2.5 \cdot 0.2)\]\[ = -0.3 - 0.2 + 0 + 0.15 + 0.2 + 0.5 \]\[ = 0.35 \]

Step 3: Find E(X+Y)

Since X and Y are independent, the expected value of their sum is the sum of their expected values:\[ E(X+Y) = E(X) + E(Y) \]\[ E(X+Y) = 0.425 + 0.35 = 0.775 \]

Step 4: Find Var(X)

Variance is calculated using the formula:\[ \operatorname{var}(X) = E(X^2) - [E(X)]^2 \]First, find \( E(X^2) \):\[ E(X^2) = (-3)^2 \cdot 0.1 + (-1)^2 \cdot 0.1 + (0)^2 \cdot 0.2 + (0.5)^2 \cdot 0.3 + (2)^2 \cdot 0.15 + (2.5)^2 \cdot 0.15 \]\[ = 0.9 + 0.1 + 0 + 0.075 + 0.6 + 0.9375 \]\[ = 2.6125 \]Now calculate variance:\[ \operatorname{var}(X) = 2.6125 - (0.425)^2 \]\[ = 2.6125 - 0.180625 = 2.431875 \]

Step 5: Find Var(Y)

Similarly, find \( \operatorname{var}(Y) \):\[ E(Y^2) = (-3)^2 \cdot 0.1 + (-1)^2 \cdot 0.2 + (0)^2 \cdot 0.1 + (0.5)^2 \cdot 0.3 + (2)^2 \cdot 0.1 + (2.5)^2 \cdot 0.2 \]\[ = 0.9 + 0.2 + 0 + 0.075 + 0.4 + 1.25 \]\[ = 2.825 \]Now calculate variance:\[ \operatorname{var}(Y) = 2.825 - (0.35)^2 \]\[ = 2.825 - 0.1225 = 2.7025 \]

Step 6: Find Var(X+Y)

Since X and Y are independent, the variance of their sum is the sum of their variances:\[ \operatorname{var}(X+Y) = \operatorname{var}(X) + \operatorname{var}(Y) \]\[ \operatorname{var}(X+Y) = 2.431875 + 2.7025 = 5.134375 \]

Key Concepts

Variance

Variance is a measure of how much a set of numbers differ from their mean. In simpler terms, it tells us how spread out a data set is. For a random variable like \(X\) or \(Y\), variance is calculated using the formula:

• First, compute the expected value of the square of the random variable, \(E(X^2)\) or \(E(Y^2)\).
• Then subtract the square of the expected value, \( [E(X)]^2 \) or \( [E(Y)]^2 \), from it.

This calculation is done because the expected value of the squares helps measure the average of the squared deviations from the mean. A higher variance indicates a wider spread.
For example, in our exercise, we find \( \operatorname{var}(X) = 2.431875 \) and \( \operatorname{var}(Y) = 2.7025 \). These values suggest that \(Y\) has a slightly larger spread or variability in its distribution compared to \(X\).
It is important when dealing with real-world datasets because variance informs decisions about risk and uncertainty in fields like finance and engineering.

Independent Random Variables

When we say two random variables \(X\) and \(Y\) are independent, it implies that knowing the value of one does not affect the probability of the other. This is a crucial concept in probability and statistics because it simplifies calculations, particularly when finding the expected value or variance of sums of random variables.

• The independence of \(X\) and \(Y\) allows us to calculate the expected value of their sum as simply the sum of their expected values: \(E(X+Y) = E(X) + E(Y) \).
• For variance, the property of independence is also handy. It states that \( \operatorname{var}(X+Y) = \operatorname{var}(X) + \operatorname{var}(Y) \). There are no extra terms because there's no covariance between \(X\) and \(Y\).

Independence is assumed in many practical situations to streamline complex problems, though caution should be exercised, as real-world variables may not always be independent,

Probability Mass Function

A probability mass function (PMF) is a fundamental concept associated with discrete random variables. It presents the probability that a discrete random variable is exactly equal to some value. In our exercise, the PMF describes how probabilities are distributed over the values that a random variable can take.

• The PMF must satisfy two conditions: each probability must be between 0 and 1, and the sum of all probabilities must equal 1. This ensures that one of the possible outcomes occurs.
• For example, for the variable \(X\), the PMF is given by the values alongside the possible values of \(X\). As you can see in the table provided, \(P(X=-3) = 0.1\), \(P(X=-1) = 0.1\), and so forth.

Using the PMF, you can compute expected values and variances of the random variables, which are essential in gauging central tendency and variability.
Understanding and using PMFs is essential when analyzing discrete random variables in scenarios such as lotteries, dice games, and any scenario involving countable data.