Independence and Discrete Random Variables & Expectations

Two events are independent when knowing that one event has already occurred does not change the probability that the other event will occur.

Three ways to check for independence of 2 events

1. Is P(A ∩ B) = P(A)*P(B) ? [in light of the multiplication rule, which says P(A∩B)= P(A)*P(B|A) ]
2. Is P(B|A) = P(B) ?
3. Is P(A|B) = P(A) ?

If you answer “yes” to any one of these three questions then events A and B are independent.

X = Random Variable

Discrete x’s

X has values that are outcomes of an experiment

E(x) = “Expected value of x”

E(X) = “the long run average of the outcomes”

The variance of X, Var(X), is the long-run average squared distance from the xi’s to E(X).

The square root of Var(X), is the Standard Deviation of X, SD(X), which is the long-run average distance from the xi’s to E(X).

Step 1: Subtract each xi minus E(X) like

(1-0.81), (1-0.81),…,(1-0.81), (5-0.81), (5-0.81),…, (5-0.81), (10-0.81), (0-0.81), (0-0.81),…(0-0.81)

Step 2: Square each distance:

(1-0.81)2,(1-0.81)2,…,(1-0.81)2,(5-0.81)2,(5-0.81)2,…,(5-0.81)2,(10-0.81)2,(0-0.81)2,(0-0.81)2,…(0-0.81)2

Step 3: Average the squared distances

[(1-0.81)²+(1-0.81)²+…+(1-0.81)²+(5-0.81)²+(5-0.81)²+…+(5-0.81)²+(10-0.81)²+(0-0.81)²+(0-0.81)²+…+(0-0.81)²]/52=

[(1-0.81)²*12 + (5-0.81)²*4 + (10-0.81)²*1 + (0-0.81)²*35]/52=

Var(X)=$3.4246, SD(X)=sqrt(Var(X))=$1.85

So, on average, the winnings vary by $1.85 from the expected value of $0.81