CSE 312: Foundations of Computing II Continuous Random Variables 8 Solutions Review of Main Concepts (a) Cumulative Distribution Function (cdf): For any random variable (discrete or continuous) X, the cumulative distribution function is defined as FX (x) = P (X ≤ x). Notice that this function must be monotonically nondecreasing: if x < y then FX (x) ≤ FX (y), because P(X ≤ x) ≤ P(X ≤ y). Also notice that since probabilities are between 0 and 1, that 0 ≤ FX (x) ≤ 1 for all x, with limx→−∞ FX (x) = 0 and limx!+1 FX (x) = 1. (b) Continuous Random Variable: A continuous random variable X is one for which its cumulative distribu- tion function FX (x): R ! R is continuous everywhere. A continuous random variable has an uncountably infinite number of values. (c) Probability Density Function (pdf or density): Let X be a continuous random variable. Then the d probability density function fX (x): R ! R of X is defined as fX (x) = dx FX (x). Turning this around, R x it means that FX (x) = P (X ≤ x) = −∞ fX (t) dt. From this, it follows that P(a ≤ X ≤ b) = R b R 1 FX (b) − FX (a) = a fX (x)dx and that −∞ fX (x)dx = 1. From the fact that FX (x) is monotonically nondecreasing it follows that fX (x) ≥ 0 for every real number x. If X is a continuous random variable, note that in general fX (a) 6= P(X = a), since P (X = a) = FX (a) − FX (a) = 0 for all a. However, the probability that X is close to a is proportional to fX (a): for δ δ small δ, P a − 2 < X < a + 2 ≈ δfX (a). (d) i.i.d. (independent and identically distributed): Random variables X1;:::;Xn are i.i.d. (or iid) if they are independent and have the same probability mass function or probability density function. (e) Univariate: Discrete to Continuous: Discrete Continuous PMF/PDF pX (x) = P(X = x) fX (x) 6= P(X = x) = 0 P R x CDF FX (x) = t≤x pX (t) FX (x) = −∞ fX (t) dt P R 1 Normalization x pX (x) = 1 −∞ fX (x) dx = 1 P R 1 Expectation E[g(X)] = x g(x)pX (x) E[g(X)] = −∞ g(x)fX (x) dx (f) Standardizing: Let X be any random variable (discrete or continuous, not necessarily normal), with 2 X−µ E[X] = µ and V ar(X) = σ . If we let Y = σ , then E[Y ] = 0 and V ar(Y ) = 1. (g) Closure of the Normal Distribution: Let X ∼ N (µ, σ2). Then, aX + b ∼ N (aµ + b; a2σ2). That is, linear transformations of normal random variables are still normal. (h) “Reproductive” Property of Normals: Let X1;:::;Xn be independent normal random variables with 2 E[Xi] = µi and V ar(Xi) = σi . Let a1; : : : ; an2 R and b2 R. Then, n n n ! X X X 2 2 X = (aiXi + b) ∼ N (aiµi + b); ai σi i=1 i=1 i=1 There’s nothing special about the parameters – the important result here is that the resulting random variable is still normally distributed. (i) Multivariate: Discrete to Continuous: 1 Discrete Continuous Joint PMF/PDF pX;Y (x; y) = P(X = x; Y = y) fX;Y (x; y) 6= P(X = x; Y = y) P R x R y Joint CDF FX;Y (x; y) = t≤x;s≤y pX;Y (t; s) FX;Y (x; y) = −∞ −∞ fX;Y (t; s) dsdt P R 1 R 1 Normalization x;y pX;Y (x; y) = 1 −∞ −∞ fX;Y (x; y) dxdy = 1 P R 1 Marginal PMF/PDF pX (x) = y pX;Y (x; y) fX (x) = −∞ fX;Y (x; y)dy P R 1 R 1 Expectation E[g(X; Y )] = x;y g(x; y)pX;Y (x; y) E[g(X; Y )] = −∞ −∞ g(x; y)fX;Y (x; y)dxdy Conditional PMF/PDF p (xjy) = pX;Y (x;y) f (xjy) = fX;Y (x;y) XjY pY (y) XjY fY (y) P R 1 Conditional Expectation E[X j Y = y] = x xpXjY (xjy) E[X j Y = y] = −∞ xfXjY (xjy)dx Independence 8x; y; pX;Y (x; y) = pX (x)pY (y) 8x; y; fX;Y (x; y) = fX (x)fY (y) (j) Law of Total Probability (Continuous): A is an event, and X is a continuous random variable with density function fX (x). Z 1 P(A) = P(AjX = x)fX (x)dx −∞ (k) Law of Total Expectation (Continuous): Y is a random variable, and X is a continuous random variable with density function fX (x). Z 1 E[Y ] = E[Y j X = x] fX (x)dx −∞ Zoo of Continuous Random Variables (a) Uniform: X ∼ Uniform(a; b) iff X has the following probability density function: 1 if x 2 [a; b] f (x) = b−a X 0 otherwise a+b (b−a)2 E[X] = 2 and V ar(X) = 12 . This represents each real number from [a; b] to be equally likely. (b) Exponential: X ∼ Exponential(λ) iff X has the following probability density function: λe−λx if x ≥ 0 f (x) = X 0 otherwise 1 1 −λx E[X] = λ and V ar(X) = λ2 . FX (x) = 1 − e for x ≥ 0. The exponential random variable is the continuous analog of the geometric random variable: it represents the waiting time to the next event, where λ > 0 is the average number of events per unit time. Note that the exponential measures how much time passes until the next event (any real number, continuous), whereas the Poisson measures how many events occur in a unit of time (nonnegative integer, discrete). The exponential random variable X is memoryless: for any s; t ≥ 0; P (X > s + t j X > s) = P(X > t) The geometric random variable also has this property. (c) Normal (Gaussian, “bell curve”): X ∼ N (µ, σ2) iff X has the following probability density function: 2 1 − 1 (x−µ) fX (x) = p e 2 σ2 ; x 2 R σ 2π 2 E[X] = µ and V ar(X) = σ . The “standard normal” random variable is typically denoted Z and has 2 X−µ mean 0 and variance 1: if X ∼ N (µ, σ ), then Z = σ ∼ N (0; 1). The CDF has no closed form, but we denote the CDF of the standard normal as Φ(z) = FZ (z) = P(Z ≤ z). Note from symmetry of the probability density function about z = 0 that: Φ(−z) = 1 − Φ(z). 2 Memorylessness of Exponential Let s; t > 0 be positive real numbers, and X ∼ Exponential(λ). Prove the memoryless property of the exponential −λx distribution: P(X > s + tjX > s) = P(X > t). (Hint: Use the fact that P(X > x) = 1 − FX (x) = e ). Interpret what this statement is saying. Note that this is also true of the Geometric distribution. Solution: −λ(s+t) P(X > s + t \ X > s) P(X > s + t) e −λt P(X > s + tjX > s) = = = = e = P(X > t) P(X > s) P(X > s) e−λs This means that, after waiting s units of time, the probability you wait t more units of time is the same as waiting t units of time from the beginning. New PDF? Alex came up with a function that he thinks could represent a probability density function. He defined the 1 potential pdf for X as f(x) = 1+x2 defined on [0; 1). Is this a valid pdf? If not, find a constant c such that c d −1 1 π the pdf fX (x) = 1+x2 is valid. Then find E[X]. (Hints: dx (tan x) = 1+x2 , tan 2 = 1, and tan 0 = 0.) Solution: Z 1 c −1 1 π 2 dx = c tan x j0 = c − 0 = 1 0 1 + x 2 so c = 2/π. Z 1 Z 1 cx 2 x 1 2 1 E[X] = 2 dx = 2 dx = ln(1 + x ) j0 = 1 0 1 + x π 0 1 + x π Transformations 1 Suppose X ∼ Uniform(0; 1) has the continuous uniform distribution on (0; 1). Let Y = − λ log X for some λ > 0. (a) What is ΩY ? Solution: ΩY = (0; 1) because log(x) 2 (−∞; 0) for x 2 (0; 1). (b) First write down FX (x) for x 2 (0; 1). Then, find FY (y) on ΩY . Solution: FX (x) = x for x 2 (0; 1). Let y 2 ΩY . 1 F (y) = (Y ≤ y) = (− log X ≤ y) = (log X ≥ −λy) = (X ≥ e−λy) = 1 − (X < e−λy) Y P P λ P P P Then, because e−λy 2 (0; 1) −λy −λy = 1 − FX (e ) = 1 − e (c) Now find fY (y) on ΩY . What distribution does Y have? Solution: 0 −λy fY (y) = FY (y) = λe Hence, Y ∼ Exponential(λ). 3 Uniform Distribution on the Circle Consider the closed unit circle of radius r, S = f(x; y): x2 + y2 ≤ rg. Suppose we throw a dart onto this circle and are guaranteed to hit it. However, we would like the probability of hitting any point equally likely. Let (X; Y ) be the coordinates of the circle that the dart hits. Find their joint density fX;Y (x; y) and be sure 2 to specify the values for all (x; y) 2 R . Are X and Y independent? Are the marginal distributions fX (x) and fY (y) uniform on [−r; r]? Solution: 1 if (x; y) 2 S f (x; y) = πr2 X;Y 0 otherwise X and Y cannot be independent since their range fails to be a rectangle (ΩX;Y = S 6= [−1; 1] × [−1; 1] = ΩX × ΩY ).