Problem 19 - Calculus for Biology and Medicine

Question

Suppose that the probability mass function of a discrete random variable \(X\) is given by the following table: $$ \begin{array}{cc} \hline \boldsymbol{x} & \boldsymbol{P}(\boldsymbol{X}=\boldsymbol{x}) \\ \hline-2 & 0.1 \\ -1 & 0.4 \\ 0 & 0.3 \\ 1 & 0.2 \\ \hline \end{array} $$ (a) Find \(E(X)\). (b) Find \(E\left(X^{2}\right)\). (c) Find \(E[X(X-1)]\).

Step-by-Step Solution

Answer

(a) \(E(X) = -0.4\); (b) \(E(X^2) = 1.0\); (c) \(E[X(X-1)] = 1.4\).

Step 1: Calculate Expectation E(X)

To find the expectation \(E(X)\), we need to use the formula for the expected value: \(E(X) = \sum{x \, P(X = x)}\). Here we calculate it as follows:\[E(X) = (-2) \cdot 0.1 + (-1) \cdot 0.4 + 0 \cdot 0.3 + 1 \cdot 0.2 = -0.2 - 0.4 + 0 + 0.2 = -0.4.\]

Step 2: Calculate E(X^2)

For \(E(X^2)\), use the formula \(E(X^2) = \sum{x^2 \, P(X = x)}\). We need to square each \(x\) and multiply by the probability:\[E(X^2) = (-2)^2 \cdot 0.1 + (-1)^2 \cdot 0.4 + 0^2 \cdot 0.3 + 1^2 \cdot 0.2 = 4 \cdot 0.1 + 1 \cdot 0.4 + 0 \cdot 0.3 + 1 \cdot 0.2 = 0.4 + 0.4 + 0 + 0.2 = 1.0.\]

Step 3: Calculate E[X(X-1)]

Use the formula \(E[X(X-1)] = \sum{x(x-1) \, P(X = x)}\). Here, compute for each value of \(x\):\[E[X(X-1)] = (-2)(-3) \cdot 0.1 + (-1)(-2) \cdot 0.4 + (0)(-1) \cdot 0.3 + (1)(0) \cdot 0.2 = 6 \cdot 0.1 + 2 \cdot 0.4 + 0 \cdot 0.3 + 0 \cdot 0.2 = 0.6 + 0.8 + 0 + 0 = 1.4.\]

Key Concepts

Expected Value

The expected value (denoted as \(E(X)\)) of a discrete random variable is a fundamental concept in probability theory. It represents the average or mean value that you would anticipate by conducting an experiment over an infinite number of trials. The concept can be thought of as the "weighted average" of all possible values that \(X\) can take, with the weights being their respective probabilities.
In general, the formula to calculate the expected value is as follows:

• First, multiply each possible value \(x\) of the random variable by the probability that it occurs, \(P(X = x)\).
• Then, sum up all these products.

Let's apply these steps to our original exercise:

• The value \(-2\) with probability \(0.1\) contributes \(-2 \times 0.1 = -0.2\).
• The value \(-1\) with probability \(0.4\) contributes \(-1 \times 0.4 = -0.4\).
• The value \(0\) with probability \(0.3\) contributes \(0 \times 0.3 = 0\).
• The value \(1\) with probability \(0.2\) contributes \(1 \times 0.2 = 0.2\).

By summing up these contributions, we find that the expected value \(E(X) = -0.4\). Understanding expected value helps in making predictions based on the average outcome of a random variable's distribution.

Discrete Random Variable

A discrete random variable, like the one in our exercise, is a variable that takes on a countable number of distinct values. The distinct values may be finite or countably infinite. Each value corresponds to a specific probability that represents the likelihood of that value occurring.
In probability theory, we often encounter discrete random variables when dealing with scenarios that involve counting events, like rolling a die or flipping a coin.
Each possible outcome of a discrete random variable comes with its own probability that follows certain rules:

• The sum of all individual probabilities must equal 1. This is because one of the outcomes must occur when you perform the experiment.
• The probability of any specific outcome is a non-negative number, meaning you will never encounter a negative probability.

In the context of our table, we have a discrete random variable \(X\) that can take on values \(-2\), \(-1\), \(0\), and \(1\) with their respective probabilities of \(0.1\), \(0.4\), \(0.3\), and \(0.2\). All probabilities add up to 1, meeting the requirements for a discrete random variable.

Probability Theory

Probability theory forms the backbone for understanding how random events operate and interact. It's a mathematical framework that allows us to calculate the likelihood of different outcomes. The major area where probability theory applies is in solving problems that involve randomness and uncertainty, such as the one presented in the exercise.
Key principles of probability theory include:

• Probability Mass Function (PMF): This is used for discrete random variables and assigns a probability to each possible value. It's essentially a map of probabilities associated with each discrete outcome.
• Laws of probability: Probability of an event must lie between 0 and 1. The probability of the sample space is always 1, representing the certainty that one of the possible outcomes must occur.

Within the PMF provided in our exercise, probability theory helps us compute expected values and variances. For each value that \(X\) could take, probability theory dictates that these values are multiplied by their associated probabilities to perform calculations such as expected value.
Through understanding probability theory, students can solve problems involving uncertainty and randomness, and apply this knowledge to more complex statistical analyses.