Problem 29 - Calculus for Biology and Medicine

Question

Let \(X\) and \(Y\) be two independent random variables with probability mass function described by the following table: $$\begin{array}{ccc} \hline \multicolumn{1}{c} {\boldsymbol{k}} & \boldsymbol{P}(\boldsymbol{X}=\boldsymbol{k}) & \boldsymbol{P}(\boldsymbol{Y}=\boldsymbol{k}) \\ \hline-2 & 0.1 & 0.2 \\ -1 & 0 & 0.2 \\ 0 & 0.3 & 0.1 \\ 1 & 0.4 & 0.3 \\ 2 & 0.05 & 0 \\ 3 & 0.15 & 0.2 \\ \hline \end{array}$$ (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) \(E(X) = 0.75\), \(E(Y) = 0.3\). (b) \(E(X+Y) = 1.05\). (c) \(\operatorname{var}(X) = 1.7875\), \(\operatorname{var}(Y) = 3.01\). (d) \(\operatorname{var}(X+Y) = 4.7975\).

Step 1: Calculate Expected Value of X

To find the expected value of \(X\), we use the formula \(E(X) = \sum_{k} k \cdot P(X=k)\). Calculating:\[ E(X) = (-2)(0.1) + (-1)(0) + (0)(0.3) + (1)(0.4) + (2)(0.05) + (3)(0.15) = -0.2 + 0 + 0 + 0.4 + 0.1 + 0.45 = 0.75 \]

Step 2: Calculate Expected Value of Y

To find the expected value of \(Y\), use the formula \(E(Y) = \sum_{k} k \cdot P(Y=k)\). Calculating:\[ E(Y) = (-2)(0.2) + (-1)(0.2) + (0)(0.1) + (1)(0.3) + (2)(0) + (3)(0.2) = -0.4 -0.2 + 0 + 0.3 + 0 + 0.6 = 0.3 \]

Step 3: Calculate Expectation of Sum of X and Y

For independent random variables, the expected value of their sum is the sum of their expected values: \(E(X+Y) = E(X) + E(Y)\). Thus:\[ E(X+Y) = 0.75 + 0.3 = 1.05 \]

Step 4: Calculate Variance of X

Variance is calculated using \( \operatorname{var}(X) = E(X^2) - [E(X)]^2 \). First, find \(E(X^2)\):\[ E(X^2) = (-2)^2 (0.1) + (-1)^2(0) + 0^2(0.3) + 1^2(0.4) + 2^2(0.05) + 3^2(0.15) \]\[ = 0.4 + 0 + 0 + 0.4 + 0.2 + 1.35 = 2.35 \]Then, \( \operatorname{var}(X) = 2.35 - (0.75)^2 = 2.35 - 0.5625 = 1.7875 \).

Step 5: Calculate Variance of Y

Using the same approach, compute \( \operatorname{var}(Y) = E(Y^2) - [E(Y)]^2 \). First, find \(E(Y^2)\):\[ E(Y^2) = (-2)^2 (0.2) + (-1)^2(0.2) + 0^2(0.1) + 1^2(0.3) + 2^2(0) + 3^2(0.2) \]\[ = 0.8 + 0.2 + 0 + 0.3 + 0 + 1.8 = 3.1 \]Then, \( \operatorname{var}(Y) = 3.1 - (0.3)^2 = 3.1 - 0.09 = 3.01 \).

Step 6: Calculate Variance of Sum of X and Y

If X and Y are independent, the variance of their sum is the sum of their variances: \( \operatorname{var}(X+Y) = \operatorname{var}(X) + \operatorname{var}(Y) \). Thus:\[ \operatorname{var}(X+Y) = 1.7875 + 3.01 = 4.7975 \]

Key Concepts

Expected Value

When dealing with random variables, the expected value is a crucial concept that helps us determine the average outcome. For any discrete random variable like \(X\), the expected value, denoted by \(E(X)\), is a weighted average of all possible values that \(X\) can take. Each value is weighted by its probability of occurrence. The formula used to calculate the expected value is:

• \(E(X) = \sum_{k} k \cdot P(X=k)\)

For example, in our problem, the expected value for \(X\) is calculated as follows:

• \(E(X) = (-2)(0.1) + (-1)(0) + (0)(0.3) + (1)(0.4) + (2)(0.05) + (3)(0.15) = 0.75\)

Similarly, the expected value for \(Y\) is also calculated using the same approach, giving us \(E(Y) = 0.3\). The expected value of a sum of two independent random variables \(X\) and \(Y\) is the sum of their expected values, \(E(X + Y) = E(X) + E(Y) = 1.05\). By knowing this, we can predict the average result when both variables are combined.

Variance

Variance provides us with the measure of how spread out a set of random values are. It's a solid indicator of variability, letting us know how much the outcomes differ from the expected value. The formula for variance for a random variable \(X\) is:

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

To calculate this, you need to find the expected value of the squares of the random variable's values, \(E(X^2)\). Once you have this, subtract the square of the expected value \(E(X)\) from it. In our scenario, we calculated:

• \(E(X^2) = 2.35\), leading to \(\text{var}(X) = 1.7875\)
• \(E(Y^2) = 3.1\), leading to \(\text{var}(Y) = 3.01\)

For independent random variables, the variance of their sum is the sum of their variances:

• \(\text{var}(X + Y) = \text{var}(X) + \text{var}(Y) = 4.7975\)

This symmetry simplifies calculations significantly when analyzing aggregated random outcomes.

Independent Random Variables

Independence between random variables is a fundamental concept in probability theory and statistics. Two random variables \(X\) and \(Y\) are independent if the occurrence of one does not affect the probability of occurrence of the other. Mathematically, this means:

• \(P(X=k, Y=m) = P(X=k) \cdot P(Y=m)\) for all values \(k\) and \(m\).

In our context, \(X\) and \(Y\) are independent as given, which influences how we calculate combined properties.

Importance of Independence

Independence makes it straightforward to calculate the expected value and variance of the sum of variables. For expected value:

• \(E(X+Y) = E(X) + E(Y)\)

And for variance:

• \(\text{var}(X + Y) = \text{var}(X) + \text{var}(Y)\)

Understanding independence is key to simplifying problems involving multiple random variables.

Probability Mass Function

A Probability Mass Function (PMF) provides the probabilities of the possible outcomes of a discrete random variable. For each outcome \(k\), the PMF is denoted by \(P(X=k)\), giving you a function that describes the likelihood of each outcome:

• If \(X\) is our variable, then \(P(X=k)\) is the probability \(X\) equals \(k\).

The PMF must satisfy two conditions:

• \(0 \leq P(X=k) \leq 1\) for every \(k\).
• The sum of all probabilities equals 1: \(\sum_{k} P(X=k) = 1\).

In the exercise, tables for \(X\) and \(Y\) show their probabilities, helping compute quantities like expected value or variance. By understanding the PMF, you can predict and compute the likelihood of outcomes, assisting in decision-making based on the probabilistic models.