Expectation and Variance

Transcription

1 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/ / /36 P(=13) = 0 E = (2)P = 2 + (3)P = 3 + (4)P = P( = 12) = = = = 42 6 = 7

2 Expecttion nd Vrince : sum of two die rolls P(= = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, /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 = (12) 2 P = 12 P(= 8 5/36 9 1/9 10 1/ / /36 = (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) = = = Vr = E 2 E 2 = SD = = = = =

3 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 b 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 b K E K = c + b 1 μ 1 + b 2 μ 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 b K 2 Vr K = b 2 1 σ b 2 2 σ b 2 K σ 2 2 K = σ Y The stndrd devition of Y is just the squre root of σ Y 2.

4 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 = 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 = E 1 + E 2 = = 7 Vr Y = Vr = Vr 1 + Vr 2 = = = 35 6 = This is exctly the sme s we found by computed the expected vlue nd vrince of the sum of two dice directly!

5 Consider rndom vrible with vlues from 0 to 1 Discrete P(= Continuous x = 0.1? x = 0.01? x = 0.001? P(= P(= P(= Wht increment do you use???

6 Consider rndom vrible with vlues from 0 to 1 Discrete P(= Continuous x = 0.1? x = 0.01? x = 0.001? P(= P(= P(= Wht increment do you use???

7 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

8 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

9 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

10 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

11 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

12 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 )

13 Complement rule P = 1 P > ( > nd re complements) P( ) P = 1 P( > ) =

14 Intervl P b = P P( b) P b P( ) P( b) = b b