Warm Up

Andrew Dickinson

Published

April 3, 2024

I would like you all to group up. Talk to your peers around you. Discuss the following:



The following two question use the following definition:

Expected values

The expected value \(E[X]\) of a discrete random variable \(X\) is defined as: \[ E[X] = \sum_{i} x_i P(x_i) \] where \(x_i\) are the possible values of \(X\), and \(P(x_i)\) is the probability that \(X\) equals \(x_i\). This equation computes the average value of \(X\) by weighing each possible value by its probability of occurrence.

Q01

Suppose we have a discrete random variable named \(A\). The probability distribution for \(A\) is:

Value of \(A\) Probability
-4 0.20
0 0.10
2 0.40
4 0.30

What would the expected value \(E[A]\) be?

The expected value is calculated as follows:

\(E[A] = (-4 \cdot 0.20) + (0 \cdot 0.10) + (2 \cdot 0.40) + (4 \cdot 0.30) = 1.2\)

Therefore, the expected value E[A] is 1.2

Q02

Consider the function \(g(A) = A^2\), where \(A\) is the same random variable as in Q01. Compute the expected value of \(g(A)\). Show the work you use to arrive at your answer.

To find the expected value of g(A), we follow these steps:

  1. Calculate g(A) for each value of A:
    • \(g(-4) = (-4)^2 = 16\)
    • \(g(0) = 0^2 = 0\)
    • \(g(2) = 2^2 = 4\)
    • \(g(4) = 4^2 = 16\)
  2. Multiply each g(A) value by its corresponding probability:
    • \(16 \cdot 0.20 = 3.2\)
    • \(0 \cdot 0.10 = 0\)
    • \(4 \cdot 0.40 = 1.6\)
    • \(16 \cdot 0.30 = 4.8\)
  3. Sum the results:
    • \(3.2 + 0 + 1.6 + 4.8 = 9.6\)

Therefore, the expected value of g(A) is 9.6.