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Continuous Random Variables Expectation Important Continuous Distributions Continuous Random Variables Expectation Important Continuous Distributions Random variables Recall, a Random variable X is a function X : Ω 7→ R on a sample space Ω.∗ TMS-062: Lecture 3. Part 1. Continuous Random Variables X is discrete R.V. if it takes values on a discrete (finite or countable) set and it is continuous if it takes values on a non-countable sets, e. g. height, weight, length, time, etc. We must find a way to describe how the unit of probability is distributed over this infinite set. Sergei Zuyev Formally, X is continuous, if its c.d.f. FX (t) = P{X ≤ t} is a continuous function. ∗ Sergei Zuyev TMS-062: Lecture 3. Part 1. Continuous Random Variables Continuous Random Variables Expectation Important Continuous Distributions There are subtleties for non-countable Ω, however! Sergei Zuyev TMS-062: Lecture 3. Part 1. Continuous Random Variables Continuous Random Variables Expectation Important Continuous Distributions Distribution of a continuous r.v. Discrete vs. Continuous R.V.’s Since now FX (t) is continuous, for any interval I of the form [a, b], [a, b), (a, b] or (a, b) we have that P{X ∈ I} = FX (b) − FX (a). If, in addition, it has a derivative fX (t) = FX0 (t) for almost all t such that Z When FX (t) is continuous, P(X = t) = FX (t) − FX (t−) = 0 for any t ∈ R. Thus no probability is associated with specific values of X when X is continuous! In contrast, in a discrete case all probabilities are associated with specific values of X . b FX (b) − FX (a) = fX (t) dt a then fX (t) is called the Probability Density Function (p.d.f). Such R.V.’s are called continuous (more exactly, absolutely continuous). We then have Z +∞ fX (t) ≥ 0 and fX (t) dt = P{X ∈ R} = 1. Since P(X ∈ (t − dt/2, t + dt/2)) ≈ fX (t)dt for small dt, p.d.f. in each point t shows how densely the probability is packed there, much the same as the particles making a stick weigh nothing, but not the stick itself which weight is the integral of the particles’ density. −∞ Sergei Zuyev TMS-062: Lecture 3. Part 1. Continuous Random Variables Sergei Zuyev TMS-062: Lecture 3. Part 1. Continuous Random Variables Continuous Random Variables Expectation Important Continuous Distributions Continuous Random Variables Expectation Important Continuous Distributions Density and Histogram Example: Discrete R.V. with Ω = {0, 1, 2, 3, 4}: P(X ≤ 2) = p0 + p1 + p2 , but P(X < 2) = p0 + p1 . For an (absolutely) continuous X with Ω = (0, 4) we have Z P(X ≤ 2) = P(X < 2) = FX (2) = 2 fX (t)dt 0 if a p.d.f. fX (t) exists. Sergei Zuyev TMS-062: Lecture 3. Part 1. Continuous Random Variables Continuous Random Variables Expectation Important Continuous Distributions Histogram of relative frequencies show how often the values from each bin occur. But fX (t) also shows the relative frequencies of values around each t to occur. So when bin size diminishes and the number of realisations of X grows in such a way, that the number of observations in each bin also grows, the histogram resembles the density function more more closely. Sergei Zuyev TMS-062: Lecture 3. Part 1. Continuous Random Variables Continuous Random Variables Expectation Important Continuous Distributions Expected value of a R.V. The expected value or expectation or the average or the mean value of a R.V. X is P k xk P(X = xk ) (Discrete) EX = R +∞ tf (t)dt (Abs. Continuous) −∞ X Sergei Zuyev TMS-062: Lecture 3. Part 1. Continuous Random Variables Sergei Zuyev TMS-062: Lecture 3. Part 1. Continuous Random Variables Continuous Random Variables Expectation Important Continuous Distributions Continuous Random Variables Expectation Important Continuous Distributions Variance The Variance of a continuous R.V. X is defined the same way: For any function g(X ) one has similarly P k g(xk )P(X = xk ) (Discrete) E g(X ) = R +∞ g(t)f (t)dt (Abs. Continuous) X −∞ σ 2 = var X = E(X − E X )2 = E X 2 − (E X )2 √ and the standard deviation is σ = var X . But for a r.v. X with p.d.f. fX (t), we have Z +∞ EX = E X2 = tfX (t)dt; −∞ Z +∞ t 2 fX (t)dt. −∞ Sergei Zuyev TMS-062: Lecture 3. Part 1. Continuous Random Variables Continuous Random Variables Expectation Important Continuous Distributions Z 10 0 10 = −e−t/100 0 TMS-062: Lecture 3. Part 1. Continuous Random Variables Continuous Random Variables Expectation Important Continuous Distributions Example: A machine is overhauled. Suppose time until breakdown T (in hours) has p.d.f. 1 −t/100 e , t >0 fT (t) = 100 Find the probabilities that the machine (a) breaks down within 10 hours; (b) runs for at least 50 hours. What is the c.d.f. and E T , σ? (a) P(T ≤ 10) = Sergei Zuyev 1 −t/100 e dt 100 Uniform distribution X has a Uniform Unif(a, b) distribution on [a, b] if X falls with the same probability in any subset (c, d) ⊂ (a, b) provided its length d − c is constant. This implies that p.d.f. is constant ( 1/(b − a) t ∈ (a, b) fX (t) = 0 otherwise = 1 − e−10/100 ≈ 0.095 (b) P(T > 50) = 1 − P(T ≤ 50) . . . = e−50/100 ≈ 0.393 Z ∞ t −t/100 ET = e dt = int. by parts = 100; 100 0 Z ∞ 2 t ET2 = e−t/100 dt = int. by parts = 20 000; 100 0 p σ = 20 000 − 1002 = 100 Sergei Zuyev TMS-062: Lecture 3. Part 1. Continuous Random Variables thus P(X ∈ (c, d)) = (d − c)/(b − a) E X = (a + b)/2 Sergei Zuyev var X = (b − a)/12 TMS-062: Lecture 3. Part 1. Continuous Random Variables Continuous Random Variables Expectation Important Continuous Distributions Continuous Random Variables Expectation Important Continuous Distributions Exponential distribution F X (t ) 1 A random variable T has Exponential Exp(λ) distribution with parameter (or rate) λ if 0 FT (t) = 1 − e−λt , a fT (t) = λe−λt b for t ≥ 0 and 0 otherwise. 0 FX (t) = (t − a)/(b − a) 1 Sergei Zuyev t <a t ∈ [a, b] t >b E T = 1/λ TMS-062: Lecture 3. Part 1. Continuous Random Variables Continuous Random Variables Expectation Important Continuous Distributions and Sergei Zuyev var T = 1/λ2 TMS-062: Lecture 3. Part 1. Continuous Random Variables Continuous Random Variables Expectation Important Continuous Distributions Normal distribution One of the most important distribution is Normal: N (m, σ 2 ) has bell-shaped density n (t − m)2 o exp − , t ∈ (−∞, +∞) σ2 2πσ EX = m and var X = σ 2 . fX (t) = √ 1 A Normal r.v. Z with parameters m = 0 and σ = 1 is called 2 standard normal and it has density fZ (t) = √1 e−t /2 . 2π Sergei Zuyev TMS-062: Lecture 3. Part 1. Continuous Random Variables Sergei Zuyev TMS-062: Lecture 3. Part 1. Continuous Random Variables Continuous Random Variables Expectation Important Continuous Distributions Continuous Random Variables Expectation Important Continuous Distributions Important property: Normal distribution is stable and preserved by linear transformations: if X1 ∼ N (m1 , σ12 ), X2 ∼ N (m2 , σ22 ) are independent, then X1 + X2 ∼ N (m1 + m2 , σ12 + σ22 ) aX + b ∼ N (am + b, a2 σ 2 ) for any const. a, b There is no explicit expression for integrals of fX (t), so computer codes are used. The c.d.f. Φ(t) = P(Z < t) for the standard normal distribution Z ∼ N (0, 1) is one of special functions called the Laplace function. Sergei Zuyev TMS-062: Lecture 3. Part 1. Continuous Random Variables Continuous Random Variables Expectation Important Continuous Distributions Sergei Zuyev TMS-062: Lecture 3. Part 1. Continuous Random Variables Continuous Random Variables Expectation Important Continuous Distributions Standartisation If X ∼ N (m, σ 2 ) then, by above, Z = (X − m)/σ has N (0, 1) distribution. Therefore Example: Assuming that the weight of screws is normally distr. with mean 2.10 gm and st.dev. 0.15 gm, find the proportion of screws weighing more than 2.55 gm. P(X < t) = P((X − m)/σ < (t − m)/σ) wt = X ∼ N (2.10, 0.152 ); = P(Z < (t − m)/σ) = Φ((t − m)/σ), denote Z = (X − 2.10)/0.15 P(X > 2.55) = P(Z > (2.55 − 2.10)/0.15) the latter can be computed in a software, e.g. Matlab. = P(Z > 3) = 1 − P(Z ≤ 3) ≈ 1 − 0.9987 = 0.0013 2 Note that the symmetry of the p.d.f. fZ (t) = Φ(t) = 1 − Φ(−t). Sergei Zuyev −t /2 e√ 2π implies that TMS-062: Lecture 3. Part 1. Continuous Random Variables Sergei Zuyev TMS-062: Lecture 3. Part 1. Continuous Random Variables Continuous Random Variables Expectation Important Continuous Distributions The rule of sigmas Let us compute the probability that X ∼ N (m, σ 2 ) deviates from its mean m by more than twice its standard deviation σ. By the standartisation procedure and the symmetry of the normal p.d.f., P(|X − m| > 2σ) = P(X < m − 2σ) + P(X > m + 2σ) = P(Z < −2) + Φ(Z > 2) = 2Φ(2) ≈ 0.04550 i.e. less than 5%. For 3 sigmas, we already have P(|X − m| > 3σ) = 2Φ(3) ≈ 0.0027. Thus, although a normal r.v. may take any real value, with probability at least 95% it does not deviate from its mean by more than 2 sigmas and with at least 99.7% probability it is within 3 sigmas. Sergei Zuyev TMS-062: Lecture 3. Part 1. Continuous Random Variables