Section1 stochastic
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1. The document summarizes key concepts in probability and statistics including Venn diagrams, conditional probability, random variables, probability density functions, cumulative distribution functions, the normal distribution, and the standard normal distribution. 2. It provides examples of using these concepts to calculate probabilities for various scenarios involving insurance policyholders, discrete and continuous random variables, and reaction times. 3. Key formulas are presented for probability, density functions, cumulative distribution functions, and transforming a normal distribution to the standard normal distribution for finding probabilities using tables.
Section1 stochastic
- 1. Stochastic Section # 1 Revision Eslam Adel Notes are inspired by sections made by TA Eng.Tamim Rushdi March 8, 2018 1 Venn Diagram P(A ∪ B) = P(A) + P(B) − P(A ∩ B) (1) 1.1 Problem 1 You are given P(A ∪ B) = 0.7 and P(A ∪ B ) = 0.9 where B is the complement of B. Find P(A). Solution P(A ∪ B) = P(A) + P(B) − P(A ∩ B) 0.7 = P(A) + P(B) − P(A ∩ B) (1) 0.9 = P(A) + P(B ) − P(A ∩ B ) (2) And P(B ) = 1 − P(B) P(A ∩ B ) = P(A) − P(A ∩ B) Substitution in (2) 0.9 = P(A) + 1 − P(B) − P(A) + P(A ∩ B) = 1 − P(B) + P(A ∩ B) (3) Sum (1) + (3) 1.6 = P(A) + 1 So P(A) = 0.6 1
- 2. 2 Conditional Probability (Bayes’ Rule) P(A|B) means Probability of event A given that another event B occurred. P(A|B) = P(A ∩ B) P(B) = P(A, B) P(B) (2) Product Rule : P(A, B) = P(A|B)P(B) = P(B|A)P(A) (3) Summation Rule: P(A) = ∞ i=0 P(A, Bi) (4) Note : If A and B are independent so P(A, B) = P(A)P(B) and here the information given is useless as P(A|B) = P(A) and P(B|A) = P(B) 2.1 Problem 2 Ten percent of an insurance companys policyholders are smokers. The rest are non smokers. For each non- smoker, the probability of dying during the year is 0.01. For each smoker, the probability of dying during the year is 0.05. Given that a policyholder has died, what is the probability that the policyholder was a smoker? (hint: use Bayes’ rule) Solution: Let S denotes that person is smoker and D Denotes person is died. Givens P(S) = 0.1 and P(S ) = 0.9 P(D|S ) = 0.01 and P(D|S) = 0.05 Required : P(S|D) P(S|D) = P (S,D) P (D) = P (S,D) P (D,S)+P (D,S ) P(S|D) = P (D|S)P (S) P (D|S)P (S)+P (D|S )P (S ) = 0.05∗0.1 0.05∗0.1+0.01∗0.9 = 5 14 3 Random Variables Random variable x is a variable represents the possible outcomes of stochastic or random process. Each random variable has f(x) which is a probability density function PDF. 3.1 Probability density function PDF It is defined as f(x) = P(X = x) (5) Properties x f(x)dx = 1 (6) or for discrete random variables x f(x) = 1 (7) 3.2 Cumulative distribution function (CDF) CDF is defined as F(x) = P(X < x) (8) Properties (i) CDF is always continuous for discrete and continous RV. 2
- 3. (ii) CDF is increasing function (iii) Maximum Value of CDF is 1 3.3 Problem 3 Suppose that X has a discrete distribution with the following PMF: f(x) = cx, for x = 1, 2, 3, 4, 5 0, otherwise Determine the Value of the constant c Solution: x f(x) = 1 c + 2c + 3c + 4c + 5c = 1 c = 1 15 3.4 Problem 4 A Discrete random variable x has a probability distribution f(x) = c(8 − x), for x = 0, 1, 2, 3, 4, 5 0, otherwise (i) Find constant c (ii) Find the CDF F(X) (iii) Find P(X > 2) Solution: (i) x f(x) = 1 → c = 1 33 (ii) PDF x 0 1 2 3 4 5 f(x) 8 33 7 33 6 33 5 33 4 33 3 33 Figure 1: PDF of x CDF 3
- 4. Figure 2: CDF of x x x < 0 0 ≤ x < 1 1 ≤ x < 2 2 ≤ x < 3 3 ≤ x < 4 4 ≤ x < 5 x ≥ 5 F(x) 0 8 33 15 33 21 33 26 33 30 33 1 (iii) P(X > 2) P(X > 2) = 1 − P(X ≤ 2) = 1 − F(2) = 1 − 21 33 = 12 33 3.5 Problem 5 Suppose cumulative distribution function (CDF) F(X)of random variable X is defined by F(X) = 0, x < 0 x4 , 0 ≤ x < 1 1, x ≥ 1 Find probability density function of X. Solution : f(x) = dF (x) dx f(x) = 0, x < 0 4x3 , 0 ≤ x < 1 0, x ≥ 1 4 Normal Distribution Normal distribution or Gaussian distribution is a continuous probability distribution of a random variable. Central limit theory stated that summation of many random variables always tends to normal distribution even original variables are not normally distributed. It has the form f(x) = 1 √ 2πσ2 e −(x−µ)2 2σ2 = 1 √ 2πσ e −(x−µ)2 2σ2 (9) 4
- 5. where µ is the mean and σ2 is the variance. So it is described as N(µ, σ2 ) 4.1 Standard Normal Distribution (SND) (i) It is a normal distribution N(0, 1) (ii) Symmetric around zero (iii) Area under the curve = 1 (iv) each have has area 0.5 (v) All normal distribution can be transformed to SND using this transformation Z = X − µ σ (10) where µ is the mean for X and σ is the standard deviation of X 4.2 Problem 6 Let Z N(0, 1) (i.e. Z is a standard normal random variable). Find following probabilities using table: (i) P(Z = 1) (ii) P(Z < 0.8) (iii) P(Z ≥ −1.43) (iv) P(−0.09 ≤ Z ≤ 1.91) (v) P(−2.2 ≤ Z ≤ −1.4) Solution : (i) P(Z = 1) = 0 (ii) P(Z < 0.8) = 0.7881 (iii) P(Z ≥ −1.43) = P(Z ≤ 1.43) = 0.9236 (iv) P(−0.09 ≤ Z ≤ 1.91) = (P(Z ≤ 0.09) − 0.5) + (P(Z ≤ 1.91) − 0.5) 0.5358 − 0.5 + 0.97193 − 0.5 = 0.50773 (v) P(−2.2 ≤ Z ≤ −1.4) = P(Z ≤ 2.2) − P(Z ≤ 1.4) = 0.9873 − 0.9192 = 0.0681 4.3 Problem 7 The reaction time of a driver to visual stimulus is normally distributed with a mean of 0.4 seconds and a standard deviation of 0.05 seconds (i) What is the probability that a reaction requires more than 0.5 seconds (ii) What is the probability that a reaction requires between 0.4 and 0.5 seconds? Solution : (i) P(X > 0.5) = P(X−µ σ > 0.5−0.4 0.05 ) P(Z > 2) = 1 − P(Z < 2) = 1 − 0.977 = 0.023 (ii) P(0.4 < X < 0.5) = P(0 < X−µ σ < 2) = P(Z ≤ 2) − 0.5 = 0.477 5
- 6. 5 Appendix A SND Table 6