Will Landau. Feb 26, 2013

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1 ,, and,, and Iowa State University Feb 26, 213 Iowa State University Feb 26, / 27

2 Outline,, and Iowa State University Feb 26, / 27

3 of continuous distributions The p-quantile of a random variable, X, is the number, Q(p), such that: P(X Q(p)) = p,, and In terms of the cumulative distribution function (cdf): F (Q(p)) = p Q(p) = F 1 (p) Iowa State University Feb 26, / 27

4 Example Let Y be the time delay (s) between a 6 Hz AC circuit and the movement of a motor on a different circuit. Q(.95) : f (y) = { 6 < y < 1 6 otherwise Q(.95).95 = P(Y Q(.95)) = f (y)dy Q(.95) = dx + 6dy = + (6 Q(.95) = 6Q(.95) Q(.95) =.95 6 = ,, and Interpretation: on average, 95% of the time delays will be below.158 seconds. Iowa State University Feb 26, / 27

5 Example You can also calculate quantiles directly from the cdf: y F (y) = 6y < y y > 1 6 Q(.25):.25 = P(Y Q(.25)) = F (Q(.25)) = 6 Q(.25),, and Hence: Q(.25) =.25 6 = Interpretation: on average, 25% of the time delays will be below.417 seconds. Iowa State University Feb 26, / 27

6 Your turn: calculating quantiles T Exp(α = 1/2):,, and f (t) = { t 2e 2t t F (t) { t < 1 e 2t t Find: 1. Q(.5) 2. Q(.5) 3. Q(p) for some p with p 1 Iowa State University Feb 26, / 27

7 Answers: calculating quantiles 1. Q(.5):.5 = P(T Q(.5)) = F (Q(.5)) = 1 e 2Q(.5).95 = e 2Q(.5) log(.95) = 2Q(.5) Q(.5) = log(.95) 2.256,, and 2. Q(.5):.5 = P(T Q(.5)) = F (Q(.5)) = 1 e 2Q(.5).5 = e 2Q(.5) log(.5) = 2Q(.5) Q(.5) = log(.5) Iowa State University Feb 26, / 27

8 Answers: calculating quantiles,, and 3. Q(p) p = P(T Q(p)) = F (Q(p)) = 1 e 2Q(p) 1 p = e 2Q(p) log(1 p) = 2Q(p) Q(p) = log(1 p) 2 Iowa State University Feb 26, / 27

9 Outline,, and Iowa State University Feb 26, / 27

10 Expected value,, and The expected value of a continuous random variable is: E(X ) = x f (x)dx As with continuous, E(X ) (often denoted by µ) is the mean of X, a measure of center. Iowa State University Feb 26, / 27

11 Example: time delay, Y E(Y ) = = f (y) = y f (y)dy y dy + ( y 2 = = 1 ( { 6 y 1 6 otherwise 1/6 ) 1/6 + ) 2 6 = 1 12 y 6dy + y dy 1/6,, and Iowa State University Feb 26, / 27

12 E(X) is the center of mass of a distribution,, and Iowa State University Feb 26, / 27

13 Your turn: calculate E(X ),, and f (x) = { x < 1 α e x/α x 1. X Exp(3) 2. X Exp(α) Iowa State University Feb 26, / 27

14 Answers: Calculate E(X ) 1. X Exp(3): integration by parts: E(X ) = x f (x)dx = x dx + x 1 3 e x/3 dx,, and ( ) = + x( e x/3 ) ( = e /3 + e /3) + = + ( 3e x/3) ( 3e /3 + 3e /3) = = 3 ( e x/3 )dx e x/3 dx 2. Similarly, E(X) = α when X Exp(α). Iowa State University Feb 26, / 27

15 Example: waiting time for the next student to arrive at the library From 12: to 12:1 PM, about 12.5 students per minute enter on average. 1 Hence, the average waiting time for the next student is 12.5 =.8 minutes for the next student. Let T Exp(.8) be the time until the next student arrives. P(wait is more than 1 seconds) = ( P (T > 1/6) = 1 F (1/6) = 1 1 e (.8 1/6)) =.12,, and Iowa State University Feb 26, / 27

16 Outline,, and Iowa State University Feb 26, / 27

17 The variance of a continuous random variable X is: Var(X ) = Shortcut formulas: Var(X ) = (x E(X )) 2 f (x)dx x 2 f (x)dx E 2 (X ) = E(X 2 ) E 2 (X ),, and The standard deviation is SD(X ) = Var(X ) Iowa State University Feb 26, / 27

18 Your turn: checkout time,, and Calculate: 1. E(X ) 2. Var(X ) Iowa State University Feb 26, / 27

19 Answers: checkout time 1. E(X ) = = x f (x)dx = ( x x 2 3 dx = 6 2 ) 2 x 1 2 xdx = ,, and 2. E(X 2 ) = = 2 x 2 f (x)dx = 2 Var(X ) = E(X 2 ) E 2 (X ) = 2 = 2 9 x xdx = 1 2 ( ) ( x x 3 4 dx = 8 ) 2 Iowa State University Feb 26, / 27

20 Your turn: ecology An ecologist wishes to mark off a circular sampling region having radius 1 m. However, the radius of the resulting region is actually a random variable R with pdf:,, and f (r) = { 3 2 (1 r)2 9 r 11 otherwise Calculate: 1. E(R) 2. SD(R) Iowa State University Feb 26, / 27

21 Answers: ecology 1. E(R) = r f (r)dr 11 = r (1 r)2 dr 11 ( ) 3 = 9 2 r 3 3r r dr ( ) 3 11 = 8 r 3 1r r 2 9 ( 3 = 8 (11)3 1(11) (11) 2 = 1 ) ( ) (9) (9) 2,, and Iowa State University Feb 26, / 27

22 Answers: ecology 2. E(R 2 ) = r 2 f (r)dr 11 = r (1 r)2 dr 11 ( ) 3 = 9 2 r 4 3r r 2 dr ( 3 = 1 r 5 15 ) 11 2 r 4 + 5r 3 9 ( 3 = 1 (11) (11)4 + 5(11) 3 = 53 5 = 1.6 Var(R) = E(R 2 ) E 2 (R) = = 3 5 =.6 SD(R) = Var(R) = ) ( 3 1 (9)5 15 ) 2 (9)4 + 5(9) 3,, and Iowa State University Feb 26, / 27

23 Outline,, and Iowa State University Feb 26, / 27

24 Expectation of a function of a random variable Why does E(X 2 ) = x 2 f (x)dx? It turns out that for any function g of a random variable: Hence: E(g(X )) = E(X 2 ) = g(x) f (x)dx x 2 f (x)dx if we take g(x ) = X 2. In the ecology example, the expected area of the circular sampling region is: E(πR 2 ) = πr 2 f (r)dr where πr 2 = g(r) is the sampling area.,, and Iowa State University Feb 26, / 27

25 Expectation of a linear function of X For constants a and b: E(aX + b) = = a (ax + b) f (x)dx x f (x)dx } {{ } E(X ) = ae(x ) + b +b f (x)dx } {{ } 1,, and Example: the expected diameter of the ecologist s sampling region is: E(2 R + ) = 2 E(R) + = 2 1 = 2 Iowa State University Feb 26, / 27

26 of a linear function of X For constants a and b: Var(aX + b) = E((aX + b) 2 ) E 2 (ax + b) = E(a 2 X 2 + abx + b 2 ) (ae(x ) + b) 2 = (a 2 E(X 2 ) + abe(x ) + b 2 ) (a 2 E 2 (X ) + abe(x ) + b 2 ) = a 2 (E(X 2 ) E 2 (X )) = a 2 Var(X ),, and Example: the variance of the diameter of the ecologist s sampling region is: Var(2 R + ) = 4Var(R) = = Iowa State University Feb 26, / 27

27 Standardization Standardization: converting a random variable X into another random variable Z by subtracting the mean and dividing by the standard deviation:,, and Z has mean : ( ) X E(X ) E(Z) = E SD(X ) = Z = X E(X ) SD(X ) 1 E(X ) E(X ) SD(X ) SD(X ) = ( 1 = E SD(X ) X E(X ) ) SD(X ) Z has variance (and standard deviation) 1: ( ) ( X E(X ) 1 Var(Z) = Var = Var SD(X ) SD(X ) X E(X ) ) SD(X ) 1 = SD 2 (X ) Var(X ) = Var(X ) 1 Var(X ) = 1 Iowa State University Feb 26, / 27

Definition: Suppose that two random variables, either continuous or discrete, X and Y have joint density

HW MATH 461/561 Lecture Notes 15 1 Definition: Suppose that two random variables, either continuous or discrete, X and Y have joint density and marginal densities f(x, y), (x, y) Λ X,Y f X (x), x Λ X,