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) \(\operatorname{var}(X) = 1.4726\), \(\operatorname{var}(Y) = 1.2975\); (d) \(\operatorname{var}(X+Y) = 2.7701\).

Step 1: Calculate Expected Value for X (E(X))

To find the expected value of random variable \(X\), we multiply each possible value \(k\) by its probability \(P(X=k)\) and sum them up:\[ E(X) = \sum (k \cdot P(X=k)) = (-3)(0.1) + (-1)(0.1) + (0)(0.2) + (0.5)(0.3) + (2)(0.15) + (2.5)(0.15) \]\[ E(X) = -0.3 - 0.1 + 0 + 0.15 + 0.3 + 0.375 = 0.425 \]

Step 2: Calculate Expected Value for Y (E(Y))

Similarly, calculate the expected value of \(Y\):\[ E(Y) = \sum (k \cdot P(Y=k)) = (-3)(0.1) + (-1)(0.2) + (0)(0.1) + (0.5)(0.3) + (2)(0.1) + (2.5)(0.2) \]\[ E(Y) = -0.3 - 0.2 + 0 + 0.15 + 0.2 + 0.5 = 0.35 \]

Step 3: Calculate E(X+Y)

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

Step 4: Calculate Variance of X (Var(X))

Variance is calculated by taking the expectation of squared deviations from the mean. First, calculate this for \(X\):\[ \operatorname{var}(X) = \sum ((k - E(X))^2 \cdot P(X=k)) \]Using previous \(E(X) = 0.425\), we substitute and calculate:\[ = (-3 - 0.425)^2(0.1) + (-1 - 0.425)^2(0.1) + (0 - 0.425)^2(0.2) + \]\[ (0.5 - 0.425)^2(0.3) + (2 - 0.425)^2(0.15) + (2.5 - 0.425)^2(0.15) \]\[ \operatorname{var}(X) = 1.4726 \]

Step 5: Calculate Variance of Y (Var(Y))

Compute variance for \(Y\) using a similar method:\[ \operatorname{var}(Y) = \sum ((k - E(Y))^2 \cdot P(Y=k)) \]Using \(E(Y) = 0.35\), substitute and calculate:\[ = (-3 - 0.35)^2(0.1) + (-1 - 0.35)^2(0.2) + (0 - 0.35)^2(0.1) + \]\[ (0.5 - 0.35)^2(0.3) + (2 - 0.35)^2(0.1) + (2.5 - 0.35)^2(0.2) \]\[ \operatorname{var}(Y) = 1.2975 \]

Step 6: Calculate Var(X+Y)

For independent variables, the variance of their sum is the sum of their variances:\[ \operatorname{var}(X+Y) = \operatorname{var}(X) + \operatorname{var}(Y) = 1.4726 + 1.2975 = 2.7701 \]

Key Concepts

Expected Value

The expected value in probability and statistics is a crucial concept when dealing with random variables. It's often thought of as the "average" or "mean" of all possible outcomes, weighted by their probabilities. For a random variable like \(X\), the expected value \(E(X)\) is calculated by:

• Multiplying each outcome \(k\) by its probability \(P(X = k)\).
• Summing up these products.

For instance, in our case, \(X\) has possible values with specific probabilities, and the calculation goes as:\[ E(X) = (-3)(0.1) + (-1)(0.1) + (0)(0.2) + (0.5)(0.3) + (2)(0.15) + (2.5)(0.15) = 0.425 \]
The same approach applies to \(Y\), giving us:\[ E(Y) = 0.35 \]
Thus, when combining two independent variables, \(X+Y\), the expected values add directly:\[ E(X+Y) = E(X) + E(Y) = 0.775 \]Together these operations yield a clear, quantifiable measure of central tendency for each variable.

Variance

Variance is another fundamental concept, quantifying the "spread" of a set of random variables. It helps us understand how much the values differ from the expected value, or mean. For a random variable \(X\), variance \(\operatorname{var}(X)\) is determined by the square of the differences between each value and the expected value, weighted by their probabilities:

• Calculate \((k - E(X))^2\) for each value \(k\).
• Multiply each squared term by its probability \(P(X = k)\).
• Sum the results.

This results in the formula:\[\operatorname{var}(X) = \sum ((k - E(X))^2 \cdot P(X=k))\]
Applying this to our variables, \(X\) and \(Y\):\[\operatorname{var}(X) = 1.4726 \quad \text{and} \quad \operatorname{var}(Y) = 1.2975\]
For independent variables \(X\) and \(Y\), the variances add, thus:\[\operatorname{var}(X+Y) = \operatorname{var}(X) + \operatorname{var}(Y) = 2.7701\]Variance provides a numerical value about how data is dispersed around the mean.

Independence of Random Variables

Independence between random variables is pivotal in probability theory. Two random variables \(X\) and \(Y\) are deemed independent if the occurrence of one does not influence the probability of occurrence of the other. Mathematically, this relationship is defined as:

• \[ P(X = x \text{ and } Y = y) = P(X = x) \times P(Y = y) \]
• This means the joint probability of both \(X\) and \(Y\) equating to specific values equals the product of their individual probabilities.

In practical terms, when \(X\) and \(Y\) are independent:

• The expected value of their sum \(E(X+Y)\) is simply the sum of their individual expected values.
• The variance of their sum \(\operatorname{var}(X+Y)\) is also just the sum of their variances.

This simplification arises because each variable acts autonomously, without interference from the other, reaffirming the integral role of independence in facilitating statistical computation.