Expected Value and Variance

STAT 20: Introduction to Probability and Statistics

Adapted by Gaston Sanchez

Agenda

• Concept review
• Concept questions
• Worksheet

Recap

Summary

• Data
• Variables
  • Categorical
  • Numerical
• Histogram
• Mean
• Standard Deviation
• Variance = \(SD^2\)

Probability

• Chance processes
• Random Variables
  • Discrete
  • Continuous
• Probability Histogram
• Expected Value
• Standard Deviation
• Variance

Expected Value of a Discrete RV

\[ E(X) = \sum_x x \times P(X=x) = \mu \]

EV is the long-term average: the value that you would expected to get if you were to repeat the chance process a very large number of times.

Variance of a Discrete RV

\[ \text{Var}(X) = \sum_x (x - \mu)^2 \times P(X=x) \]

Var is the average squared deviation from the mean (measures the typical spread around EV)

Example: Expected Value

Let \(X\) be the spots when we roll a fair six-sided die.

\[ \Omega = \left \{ 1, 2, 3, 4, 5, 6 \right \} \]

\[ P(X = x) = 1/6, \quad \text{for all } x \]

\[ \begin{align*} EV(X) &= \sum_x x \times P(X=x) \\ &= 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6) \\ &= \frac{1 + 2 + 3 + 4 + 5 + 6}{6} \\ &= 3.5 \end{align*} \]

Example: Variance

Let \(X\) be the spots when we roll a fair six-sided die.

\[ \Omega = \left \{ 1, 2, 3, 4, 5, 6 \right \} \]

\[ \begin{align*} \text{Var}(X) &= \sum_x (x - \mu)^2 \times P(X=x) \\ &= (1 - 3.5)^2 (1/6) + (2 - 3.5)^2 (1/6) + \dots + (6 - 3.5)^2 (1/6) \\ &= \frac{(1 - 3.5)^2 + (2 - 3.5)^2 + (6 - 3.5)^2}{6} \\ & \approx 2.916 \end{align*} \]

Properties of EV

\(E(c) = c\)

\(E(cX) = c \cdot E(X)\)

\(E(X + Y) = E(X) + E(Y)\)

\(E(aX + bY) = a \cdot E(X) + b \cdot E(Y)\)

\(Y = g(X) \ \rightarrow \ E(Y) = E(g(X)) = \sum_x g(x) \times P(X=x)\)

Properties of Variance

\(\textit{Var}(c) = 0\)

\(\textit{Var}(cX) = c^2 \cdot \textit{Var}(X)\)

\(\textit{Var}(X + c) = \textit{Var}(X)\)

\(\textit{Var}(X + Y) = \textit{Var}(X) + \textit{Var}(Y) \quad \text{ for } X, Y \text{ indep}\)

\(\textit{Var}(X - Y) = \textit{Var}(X) + \textit{Var}(Y) \quad \text{ for } X, Y \text{ indep}\)

Concept Review

Let \(X\) be a random variable such that \[ X = \begin{cases} -1, & \text{ with probability } 1/3\\ 0, & \text{ with probability } 1/6\\ 1, & \text{ with probability } 4/15 \\ 2, & \text{ with probability } 7/30 \\ \end{cases} \]

1. Draw the graph of the CDF of \(X\)

write down pmf with denominator 30, and draw cdf on the board. P(X = -1)= 5/30, P(X = 0)= 10/30, P(X = 1)= 8/30, P(X = 2)= 7/30

Let \(X\) be a random variable such that \[ X = \begin{cases} -1, & \text{ with probability } 1/3\\ 0, & \text{ with probability } 1/6\\ 1, & \text{ with probability } 4/15 \\ 2, & \text{ with probability } 7/30 \\ \end{cases} \]

2. Compute the expected value and variance of \(X\)

write down pmf with denominator 30, and draw cdf on the board. P(X = -1)= 5/30, P(X = 0)= 10/30, P(X = 1)= 8/30, P(X = 2)= 7/30 E(X) = (-1)(5/30) + 0(10/30) + 1(8/30) + 2(7/30) (-10 + 0 + 8 + 14)/30 = 12/30 = 2/5 Var(X) = E(X^2) - 4/25 = (10 + 0 + 8 +28)/30 -4/25 = 23/15- 4/25 ~~ 1.37333 Graph the pmf and mark the expectation The numbers are not pretty but the point is to show them the computation and graph The rest of the ideas can be recapped as you go over the concept questions You may want to review linearity of expectation here, or save it for when going over the PS

Concept Questions

\(X\) is a random variable with the distribution shown below: \[ X = \begin{cases} 3, \; \text{ with prob } 1/3\\ 4, \; \text{ with prob } 1/4\\ 5, \; \text{ with prob } 5/12 \end{cases} \]

Consider the box with tickets: \(\fbox{3}\, \fbox{3}\, \fbox{3} \,\fbox{4} \,\fbox{4} \,\fbox{4} \,\fbox{4} \,\fbox{5} \,\fbox{5}\, \fbox{5} \,\fbox{5} \,\fbox{5}\)

Suppose we draw once from this box and let \(Y\) be the value of the ticket drawn.

Which random variable has a higher expectation?

This is a reading question. Note that the pmf implied by the box is not the same as the given pmf. In the box, 4 has a higher probability, so the average is higher. They do not need to actually do the computation. At this point, ask them what box would they use so that if \(X\) is the value of a randomly selected ticket, the distribution of \(X\) is exactly the distribution shown here.

Prof. Stoyanov’s Zoom office hours are not too crowded this spring. She observes that number of Stat 20 students coming to her Thursday office hours have a Poisson(2) distribution. There is one Data 88 student from a previous semester who is always there (they want a letter of recommendation).

What is the expected value (EV) and variance (V) of the number of students in her Zoom office hours?

This is also a reading question, with < 50% getting it right. Let \(X\) be the number of Stat 20 students who go to the office hours. Then \(E(X) = 2 = Var(X)\). But the number of students in the office is \(X+1\) since that one student is always there. \(E(X+1) = 2+1 = 3\) and \(Var(X+1) = Var(X) = 2\)

Let \(X\) be a discrete uniform random variable on the set \(\{-1, 0, 1\}\).

If \(Y=X^2\), what is \(E(Y)\)?

Let \(X\) be a discrete uniform random variable on the set \(\{-1, 0, 1\}\).

If \(W = \min(X, 0.5)\), what is \(E(W)\)?

begin by asking them what is What is \(E(X)\)? And pointing out that if we have a symmetric distribution then EX is in the middle. Emphasize the idea of it being a “typical” value and an average or a mean. This is so they practice computing the EV of a function of a rv. Have them make a table (or perhaps you draw a table) of \(X\), \(Y\), \(W\) and \(f(x)\)

Worksheet

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