Expecttion nd Vrince : sum of two die rolls P(= P(= = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 P(=2) = 1/36 P(=3) = 1/18 P(=4) = 1/12 P(=5) = 1/9 P(=7) = 1/6 P(=13) =? 2 1/36 3 1/18 4 1/12 5 1/9 6 5/36 7 1/6 8 5/36 9 1/9 10 1/12 11 1/18 12 1/36 P(=13) = 0 E = (2)P = 2 + (3)P = 3 + (4)P = 4 + + 12 P( = 12) = 2 1 36 + 3 1 18 + 4 1 12 + 5 1 9 + 6 5 36 + 7 1 6 + 8 5 36 + 9 1 9 + 10 1 12 + 11 1 18 + 12 1 36 = 2 36 + 6 36 + 12 36 + 20 36 + 30 36 + 42 36 + 40 36 + 36 36 + 30 36 + 22 36 + 12 36 = 252 36 = 42 6 = 7
Expecttion nd Vrince : sum of two die rolls P(= = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 2 1/36 P(=2) = 1/36 P(=3) = 1/18 3 1/18 P(=4) = 1/12 4 1/12 P(=5) = 1/9 5 1/9 P(=7) = 1/6 6 5/36 Vr = E 2 E 2 7 1/6 E 2 = (2) 2 P = 2 + (3) 2 P = 3 + (4) 2 P = 4 + + (12) 2 P = 12 P(= 8 5/36 9 1/9 10 1/12 11 1/18 12 1/36 = (2) 2 1 36 + (3)2 1 18 + (4)2 1 12 + (5)2 1 9 + (6)2 5 36 + (7)2 1 6 + (8)2 5 36 + (9)2 1 9 + (10)2 1 12 + (11)2 1 18 + (12)2 1 36 = 4 36 + 18 36 + 48 36 + 100 36 + 180 36 + 294 36 + 320 36 + 324 36 + 300 36 + 242 36 + 144 36 = 1974 36 = 329 6 54.8333 Vr = E 2 E 2 = 1974 42 36 6 SD = 5.8333 = 2.4152 2 = 1974 36 1764 36 = 210 36 = 35 6 5.8333
Expecttion nd Vrince Suppose we hve multiple Discrete Rndom Vribles 1, 2,, K with ll rndom vribles mutully independent. Suppose we crete new rndom vrible by tking weighted sum of the others: Expected vlue of Y: K Y = c + b k k = c + b 1 1 + b 2 2 + + b K K. k=1 E Y = E K c + b k k k=1 K K = E c + E(b k k ) = c + b k E( k ) k=1 k=1 = c + b 1 E 1 + b 2 E 2 + + b K E K = c + b 1 μ 1 + b 2 μ 2 + + b K μ K = μ Y Vrince of Y: Vr Y = Vr c + σk k=1 b k k = σk k=1 Vr b k k = σk k=1 b 2 k Vr( k ) = b 1 2 Vr 1 + b 2 2 Vr 2 + + b K 2 Vr K = b 2 1 σ 2 1 + b 2 2 σ 2 2 + + b 2 K σ 2 2 K = σ Y The stndrd devition of Y is just the squre root of σ Y 2.
Expecttion nd Vrince i : fce vlue of fir die roll on the ith roll. i = 1, 2, 3, 4, 5, 6 P i = 1 = 1/6 P i = 2 = 1/6 P i = 3 = 1/6 P i = 4 = 1/6 P i = 5 = 1/6 P i = 6 = 1/6 We know tht E i = 3.5, nd tht Vr i = 35 12. i P( i = x i ) 1 1/6 2 1/6 3 1/6 4 1/6 5 1/6 6 1/6 Now, suppose tht we wnt the expected vlue nd vrince of new rndom vrible Y, defined s the sum of two independently thrown dice. We find the following: E Y = E 1 + 2 = E 1 + E 2 = 3.5 + 3.5 = 7 Vr Y = Vr 1 + 2 = Vr 1 + Vr 2 = 35 12 + 35 12 = 70 12 = 35 6 = 5.8333 This is exctly the sme s we found by computed the expected vlue nd vrince of the sum of two dice directly!
Consider rndom vrible with vlues from 0 to 1 Discrete P(= 0 0.4 1 0.6 0 0.1 0.2 Continuous x = 0.1? x = 0.01? x = 0.001? 0.9 1.0 P(= 0 0.01 0.02 0.99 1.00 P(= 0 0.001 0.002 0.999 1.000 P(= Wht increment do you use???
Consider rndom vrible with vlues from 0 to 1 Discrete P(= 0 0.4 1 0.6 0 0.1 0.2 Continuous x = 0.1? x = 0.01? x = 0.001? 0.9 1.0 P(= 0 0.01 0.02 0.99 1.00 P(= 0 0.001 0.002 0.999 1.000 P(= Wht increment do you use???
Every continuous rndom vrible is defined by probbility density function (pdf) Probbility density function for continuous rndom vrible : f( In this exmple, vlues of rnge from to Probbility of less thn or equl to some vlue, P = F (), is defined s the re under the curve to the left of P = f x dx
Every continuous rndom vrible is defined by probbility density function (pdf) Probbility density function for continuous rndom vrible : f( In this exmple, vlues of rnge from to Probbility of less thn or equl to some vlue, P = F (), is defined s the re under the curve to the left of P = f x dx Probbility of greter thn or equl to some vlue, P, is defined s the re under the curve to the right of P = f x dx
Every continuous rndom vrible is defined by probbility density function (pdf) Probbility density function for continuous rndom vrible : f( In this exmple, vlues of rnge from to b Probbility of less thn or equl to some vlue, P = F (), is defined s the re under the curve to the left of P = f x dx Probbility of greter thn or equl to some vlue, P, is defined s the re under the curve to the right of P = f x dx Probbility of between nd b, P b, is defined s the re under the curve between nd b P b = b f x dx
Every continuous rndom vrible is defined by probbility density function (pdf) Probbility density function for continuous rndom vrible : f( In this exmple, vlues of rnge from to Probbility of the entire rnge of, P, is the re under the entire curve P = f x dx = 1 Probbility of ll possible vlues of
Every continuous rndom vrible is defined by probbility density function (pdf) Probbility density function for continuous rndom vrible : f( In this exmple, vlues of rnge from to Probbility of the entire rnge of, P, is the re under the entire curve P = f x dx = 1 Probbility of ll possible vlues of Probbility of between nd, P = P =, is the re under the curve between nd P = = f x dx = 0 Are doesn t exist! Are of line is 0. o A continuous rndom vrible theoreticlly hs n infinite number of possible vlues o The probbility of equl to n exct point vlue out of n infinite number of choices = 1 = 0
For continuous, since P = = 0, nd re interchngeble with < nd >, respectively P = P < + P = = P < + 0 = P < P = P > + P = = P > + 0 = P( > ) P b = P < b = P < b = P( < < b) o Cution: not true for discrete becuse P = is not 0 (given tht is possible vlue of )
Complement rule P = 1 P > ( > nd re complements) P( ) P = 1 P( > ) =
Intervl P b = P P( b) P b P( ) P( b) = b b