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: $$ \begin{array}{ccc} \hline \boldsymbol{k} & \boldsymbol{P}(\boldsymbol{X}=\boldsymbol{k}) & \boldsymbol{P}(\boldsymbol{Y}=\boldsymbol{k}) \\ \hline-2 & 0.1 & 0.2 \\ -1 & 0 & 0.2 \\ 0 & 0.3 & 0.1 \\ 1 & 0.4 & 0.3 \\ 2 & 0.05 & 0 \\ 3 & 0.15 & 0.2 \\ \hline \end{array} $$ (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

E(X) = 0.75, E(Y) = 0.3, E(X+Y) = 1.05, Var(X) = 1.7875, Var(Y) = 3.01, Var(X+Y) = 4.7975.

Step 1: Calculate Expected Value of X

To find the expected value of \(X\), multiply each value \(k\) by its corresponding probability \(P(X=k)\), and sum them up: \[E(X) = \sum_{k} k \cdot P(X=k) = (-2) \cdot 0.1 + 0 \cdot 0 + 0 \cdot 0.3 + 1 \cdot 0.4 + 2 \cdot 0.05 + 3 \cdot 0.15 = -0.2 + 0 + 0 + 0.4 + 0.1 + 0.45 = 0.75.\]

Step 2: Calculate Expected Value of Y

Similarly, for \(Y\):\[E(Y) = \sum_{k} k \cdot P(Y=k) = (-2) \cdot 0.2 + (-1) \cdot 0.2 + 0 \cdot 0.1 + 1 \cdot 0.3 + 0 \cdot 0 + 3 \cdot 0.2 = -0.4 - 0.2 + 0 + 0.3 + 0 + 0.6 = 0.3.\]

Step 3: Find Expected Value of X+Y

Since \(X\) and \(Y\) are independent, the expected value of their sum is the sum of their expected values: \[E(X+Y) = E(X) + E(Y) = 0.75 + 0.3 = 1.05.\]

Step 4: Calculate Variance of X

The variance of \(X\) is calculated by finding the expected value of \(X^2\) and using the formula \(\operatorname{var}(X) = E(X^2) - (E(X))^2\).First calculate \(E(X^2)\):\[E(X^2) = \sum_{k} k^2 \cdot P(X=k) = (-2)^2 \cdot 0.1 + 0^2 \cdot 0 + 0^2 \cdot 0.3 + 1^2 \cdot 0.4 + 2^2 \cdot 0.05 + 3^2 \cdot 0.15 = 0.4 + 0 + 0 + 0.4 + 0.2 + 1.35 = 2.35.\]Now find the variance:\[\operatorname{var}(X) = E(X^2) - (E(X))^2 = 2.35 - (0.75)^2 = 2.35 - 0.5625 = 1.7875.\]

Step 5: Calculate Variance of Y

Similarly, calculate the variance for \(Y\) using the same method as for \(X\):\[E(Y^2) = \sum_{k} k^2 \cdot P(Y=k) = (-2)^2 \cdot 0.2 + (-1)^2 \cdot 0.2 + 0^2 \cdot 0.1 + 1^2 \cdot 0.3 + 0^2 \cdot 0 + 3^2 \cdot 0.2 = 0.8 + 0.2 + 0 + 0.3 + 0 + 1.8 = 3.1.\]The variance of \(Y\):\[\operatorname{var}(Y) = E(Y^2) - (E(Y))^2 = 3.1 - (0.3)^2 = 3.1 - 0.09 = 3.01.\]

Step 6: Find Variance of X+Y

For independent random variables, the variance of the sum is the sum of the variances. Therefore, \[\operatorname{var}(X+Y) = \operatorname{var}(X) + \operatorname{var}(Y) = 1.7875 + 3.01 = 4.7975.\]

Key Concepts

Expected Value

The expected value is a fundamental concept in probability theory. It allows us to find the average or mean outcome of a random variable over a long period. Essentially, it is the long-term average value that you would expect if you could repeat a random process an infinite number of times.
To calculate the expected value of a random variable, like in the case of the variables \(X\) and \(Y\), you multiply each possible value \(k\) by its corresponding probability \(P(X=k)\) or \(P(Y=k)\), and then sum all these products together. This method provides a clear quantitative expression of what should happen on average.
For example:

• The expected value of \(X\) is calculated as \(E(X) = (-2) \times 0.1 + 0 \times 0 + 1 \times 0.4 + 2 \times 0.05 + 3 \times 0.15 = 0.75\).
• Similarly, \(E(Y) = (-2) \times 0.2 + (-1) \times 0.2 + 1 \times 0.3 + 3 \times 0.2 = 0.3\).

This method reflects how much we "expect" certain values to show up, providing an average snapshot of the overall distribution.

Variance

Variance is another crucial concept in probability theory. While expected value gives us a measure of the central tendency, variance gives us insight into the variability or spread of the random variables around the mean. It essentially quantifies how much the outcomes deviate from the expected value.
To compute the variance of a random variable \(X\), you need to find the expected value of the squared difference from the mean (\(E((X - E(X))^2)\)). However, it's more convenient to use the formula \(\operatorname{var}(X) = E(X^2) - (E(X))^2\).
Here's how:

• Calculate \(E(X^2)\), which is the expected value considering squared outcomes of \(X\), then subtract the square of \(E(X)\).
• For \(X\), \(E(X^2) = (-2)^2 \times 0.1 + 0^2 \times 0.3 + 1^2 \times 0.4 + 2^2 \times 0.05 + 3^2 \times 0.15 = 2.35\). Thus, \(\operatorname{var}(X) = 2.35 - (0.75)^2 = 1.7875\).

The process is similar for \(Y\):

• \(E(Y^2) = (-2)^2 \times 0.2 + (-1)^2 \times 0.2 + 1^2 \times 0.3 + 3^2 \times 0.2 = 3.1\).
• Thus, \(\operatorname{var}(Y) = 3.1 - (0.3)^2 = 3.01\).

Variance gives us a way to understand how concentrated or spread out our random variable's outcomes are.

Independent Random Variables

Understanding independent random variables is essential in probability, especially when dealing with sums of random variables. Two random variables, \(X\) and \(Y\), are independent if the occurrence of events for one variable does not influence the occurrence of events for the other. This means the joint probability distribution is simply the product of their individual distributions.
Independence simplifies calculations:

• For expected values, the expected value of the sum of independent random variables is the sum of their expected values: \(E(X+Y) = E(X) + E(Y)\).
• The variance of the sum of independent random variables is the sum of their variances: \(\operatorname{var}(X+Y) = \operatorname{var}(X) + \operatorname{var}(Y)\).

This is particularly useful because it allows complex problems to be broken down into simpler individual components. For the given exercise, since \(X\) and \(Y\) are independent, we find:

• \(E(X+Y) = 1.05\).
• \(\operatorname{var}(X+Y) = 1.7875 + 3.01 = 4.7975\).

Independence helps in handling multiple random variables efficiently and predicting outcomes more accurately.