stat indep repairs REVD

Statistical Independence Repair PhL 8.1.13 I had to go through all of FT and clean away this wrong idea, it got done OK. I made the common mistake of equating these to statements X and Y are statistically independent X and Y are uncorrelated. The web gives a nice example of how uncorrelated is possible for X and Y dependent. (I stole it! ) In this example, X and X2 obviously are not independent, but for the symmetric pdf p(x) and zero mean X he gets E(X3) = ∫dx x3p(x) = 0 So then cov(X,X2 ) = E(X3) - μxμx2 = E(X3) = 0. ********************* try to edit in a repair ****************** (c) Basic Probability Theory In the above discussion, if we have random variables A,B,C.... , we can write a joint probability density function in this manner pdf(A=a, B=b, C=c.....) (G.2) where now we don't distinguish whether the various sample spaces are continuous or discrete, we just write anything as a pdf (with the understanding of (G.1) above). The meaning here is that pdf(...) is the probability that random variable A has value a, while at the same time (the same experiment) random variable B has value b, and so on. For the continuous case, pdf(...) da db dc... is the same probability but for the range da of a and db of b, etc. If all these random variables are statistically independent (such as A = number of dust particles on your pillow and B = temperature at some location on Pluto), this joint probability distribution factors, pdf(A=a,B=b,C=c.....) = pdf(A=a) pdf(B=b) pdf(C=c) .... (G.3a) For two random variables A and B, one would have pdf(X=x,Y=y) = pdf(X=x) pdf(Y=y) . (G.4a) We now adopt some shorthand notations: p(a,b,c....) ≡ pdf(A=a,B=b,C=c.....) = pdfABC...(a,b,c...) = pABC..(a,b,c..) (G.5) When one sees p(2, -4, 0. 4...) one must remember that the arguments correspond to values of specific random variables, and if things become unclear, one must revert to the fuller notation. One trick is to use a parameter name that reminds the reader of the random variable name, such as a for A: p(a) = P(A=a). In the shorthand notation we have p(a,b,c...) = p(a)p(b)p(c)..... // N statistically independent random variables (G.3b) p(x,y) = p(x)p(y) // 2 statistically independent random variables (G.4b) Digression: One way to understand the concept of statistical independence is by the use of conditional probabilities. Consider: pX( x | y ) = probability just for X where X = x given that Y = y which we compare to pXY(x,y) = probability for X and Y where X = x and Y = y . The connection is given by, pXY(x,y) = pX( x | y ) pY(y). If X and Y are independent, then pX( x | y ) has no dependence on y, X knows nothing about Y, and in this case we have pX( x | y ) = pX(x) . This then yields the factored form pXY(x,y) = pX(x) pY(y) . For three variables we can define pX( x | y, z ) = probability just for X where X = x given that Y = y and Z = z and then (start on the right end when reading this) pXYZ(x,y,z) = pX( x | y, z ) pY(y | z) pZ(z) . If X,Y,Z are statistically independent, then certainly X knows nothing about Y and Z, and Y knows nothing about Z, so then pX( x | y, z ) = pX(x) and pY(y | z) = pY(y) and then pXYZ(x,y,z) = pX(x) pY(y) pZ(z) giving the factored form. Often this discussion appears with the symbol replace our commas, and one can verify the various claims with Venn diagrams. We shall not digress more on this subject and shall take the factored form as our definition of statistical independence. We return now to pXY(x,y) = pX(x)pY(y) // 2 statistically independent random variables (G.4b) If one regards pXY(x,y) as a function fy(x) for various fixed values of y, (G.4b) says that the shape of this function is not influenced by the values of y, only the overall scale of fy(x) is affected by y. After we define the correlation measure corr(X,Y) below, we will see that (G.4b) being true implies that corr(X,Y) = 0 which means X and Y are uncorrelated. This arrangement works only one way: X,Y statistically independent X,Y uncorrelated. It is easy to find examples where X,Y are uncorrelated but are not independent, we shall look at an example later. Any pdf is normalized to 1 since the probability of all possible outcomes (mapped from sample spaces to the random variables) is 1. Thus, ∫∫.... p(x,y....) dx dy... = 1 or Σx,y.... p(x,y....) = 1 . (G.6) ********************** repair has been installed ****************** But I need to continue through G and make other repairs, I think all is OK at least through the start of the covariance definition. We get finally to G.26 and there are two things I want to add. I added (1) the range -1 to 1 with a simple argument, (2) the counterexample of dependent but uncorrelated. I am now repaired up to (G.31) and there more is needed. I could just cut it down to Fact: If X and Y are statistically independent random variables, then (G.31) (a) p(x,y) = p(x)p(y) (b) E(XY) = E(X)E(Y) (c) cov(X,Y) = 0 (d) corr(X,Y) = 0 (e) X and Y are uncorrelated (a) follows from our definition of statistical independence (G.3a) or (G.3b). (b) E(XY) = ∫∫ x y p(x,y) dx dy = ∫∫ x y p(x)p(y) dx dy = [∫x p(x) dx] [∫y p(y) dy] = E(X)E(Y) . (c) cov(X,Y) ≡ E(XY) - μxμy = E(X)E(Y) - μxμy = μxμy - μxμy = 0 (d) corr(X,Y) ≡ cov(X,Y) / [σ(X) σ(Y) ] = 0 / [σ(X) σ(Y) ] = 0 (e) This is the same as (d), just stated in words QED ******************* the above has been installed **************8 Now, let's scan through the rest of the doc and see what other repairs are needed. Pause to worry about independent versus correlation. near (35.5) needs work near D.3 near G.3a and G.4a G.31 Lets try to fix G.31. Repair of (G.31) independent OK, after a lot of effort I think I completely purged this error from FT. It may exist somewhere in Scrambler, but that is for later.