University:
McMaster University
Course:
STATS 1LL3 | Introduction to Probability and Statistics
Academic year:

2023
Views:

242

Pages:

4
Author:

Sabrina S.

CH 4: Random Variables Pg. 91 In order to manipulate the events defined on a sample space mathematically, it is necessary to attach a numerical value to each elementary event. When events are quantitative (i.e., hours, count of items), there is an obvious and natural way to assign numbers to them. However, if elementary events are expressed qualitatively, we have to assign a number to each event. Once all elementary events in a sample space have numerical values assigned to them, we follow the classic algebraic tradition and let X stand for the numerical values of the elementary events. We then call X a "random variable." X is a variable because it can take on different values. X is a random variable because the particular value it takes on depends on the outcome of a random experiment. Discrete random variables take on isolated values along the real line, usually integer values. PROBABILITY MASS FUNCTIONS Continuous random variables can be measured to any degree of accuracy. PROBABILITY DENSITY FUNCTIONS Probabilities for discrete random variables are found by calculating the values of the probability mass function p(x) at the points of interest and summing them. However, for continuous random variables, the probability density function f(x) is constructed in such a way that probabilities of events are found by integration. The area under the graph between the numbers c and d represents the probability of the event. Video 9: Long Run Average The expected value of X can be viewed as a weighted sum of the possible values of X. E(X) = ∑(x * p(x)) where X is the number of dots on an upturned face of a randomly thrown die E(X) = (1/6) * 1 + (1/6) * 2 + (1/6) * 3 + (1/6) * 4 + (1/6) * 5 + (1/6) * 6 = 3.5 The expected value of X is also referred to as the mean of X. How does this apply to the theory of continuous random variables? E(X) = ∫ x * f(x) dx Example: Let x be a continuous random variable with pdf given as: f(x) = 1/10 (1 - x²) for -1 ≤ x ≤ 1 f(x) = 0 elsewhere E(X) = ∫₋₁¹ x * (1/10)(1 - x²) dx E(X) = (1/10) * ∫₋₁¹ (x - x³) dx E(X) = (1/10) * [(x²/2) - (x⁴/4)] |₋₁¹ E(X) = (1/10) * [(1/2 - 1/4) - (-1/2 + 1/4)] E(X) = 0 Var(X) = E[X²] - (E[X])² How do we calculate? E(A + B) E(A - B) Var(A + B) Var(A - B) We use these rules: ① E(A + B) = E(A) + E(B) ② E(A - B) = E(A) - E(B) ③ E(cA) = c * E(A) in the case where A and B are independent variables Var(A + B) = Var(A) + Var(B) Var(A - B) = Var(A) + Var(B) Var(cA) = c² * Var(A) Excel example mean = AVERAGE(A1:A1000) variance = VAR(A1:A1000) Table: p(x) x x*p(x) x²*p(x) 0.167 1 0.167 0.167 0.167 2 0.334 0.668 0.167 3 0.501 1.503 0.167 4 0.668 2.672 0.167 5 0.835 4.175 0.167 6 1.002 6.012 E[x] = SUM(x * p(x)) Var[x] = E[x²] - (E[x])² SD[x] = sqrt(VAR[x]) Pr(0.8) E[x] = ∑ x * p(x) = 0( ) + 1( ) + 2( ) + 3( ) 0 = 0.2 * 3 = 0.008 1 = 0.8 * 0.2 = 0.032 2 = 0.9 * 0.2 = 0.128 3 = 0.3 * 0.2 = 0.512 ∫₀¹ x * f(x) dx = [x * x]₀¹ = 1 ∫₁² x * g(x) dx = [x * (x/10)]₁² = [x²/10]₁² = (4/10) - (1/10) = 3/10 Var[x] = ∫₀¹ (x - 1)² * f(x) dx = ∫₀¹ (x² - 2x + 1) * (x/10) dx = (1/10) * ∫₀¹ (x³ - 2x² + x) dx = (1/10) * [(x⁴/4) - (2x³/3) + (x²/2)]₀¹ = (1/10) * [(1/4 - 2/3 + 1/2) - (0)] = -24/30