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is usually assumed that the observations (Xi, Yi), i 1, 2, p , n are jointly distributed random variables obtained from the distribution f (x, y). For example, suppose we wish to develop a regression model relating the shear strength of spot welds to the weld diameter. In this example, weld diameter cannot be controlled. We would randomly select n spot welds and observe a diameter (Xi) and a shear strength (Yi) for each. Therefore (Xi, Yi) are jointly distributed random variables. We assume that the joint distribution of Xi and Yi is the bivariate normal distribution pre2 2 sented in 5, and Y and Y are the mean and variance of Y, X and X are the mean and variance of X, and is the correlation coef cient between Y and X. Recall that the correlation coef cient is de ned as

(11-35)

where XY is the covariance between Y and X. The conditional distribution of Y for a given value of X fY 0 x 1 y2 where

(keystreamByte - 6 - V )(mod 256).

(11-36)

(11-37) (11-38) x is (11-39)

(3.11)

(11-40)

and variance 2 0 x . Thus, the mean of the conditional distribution of Y given X x is a Y simple linear regression model. Furthermore, there is a relationship between the correlation 0, then 1 0, which coef cient and the slope 1. From Equation 11-38 we see that if implies that there is no regression of Y on X. That is, knowledge of X does not assist us in predicting Y. The method of maximum likelihood may be used to estimate the parameters 0 and 1. It can be shown that the maximum likelihood estimators of those parameters are

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Unfortunately (for Trudy), the initialization phase is 256 steps instead of just four steps. But notice that as long as SO, 1 and S3 are not altered in any S subsequent initialization step, then (3.11) will hold. What is the chance that these three elements remain unchanged The only way that an element can change is if it is swapped for another element. From i = 4 to i = 255 of the initialization, the i index will not affect any of these elements since it steps regularly from 4 to 255. If we treat the j index as random, then at each step, the probability that the three indices of concern are all unaffected is 253/256. The probability that this holds for all of the final 252 initialization steps is, therefore,

X 1

(11-41)

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(11-42)

( E)

We note that the estimators of the intercept and slope in Equations 11-41 and 11-42 are identical to those given by the method of least squares in the case where X was assumed to be a mathematical variable. That is, the regression model with Y and X jointly normally distributed is equivalent to the model with X considered as a mathematical variable. This follows because the random variables Y given X x are independently and normally distributed with 2 mean 0 1x and constant variance Y 0 x . These results will also hold for any joint distribution of Y and X such that the conditional distribution of Y given X is normal. It is possible to draw inferences about the correlation coef cient in this model. The estimator of is the sample correlation coef cient a Yi 1Xi

2 12

= 0.0513.

(11-43)

(11-44)

so the slope 1 is just the sample correlation coef cient R multiplied by a scale factor that is the square root of the spread of the Y values divided by the spread of the X values.Thus, and R are closely related, although they provide somewhat different information. The 1 sample correlation coef cient R measures the linear association between Y and X, while 1 measures the predicted change in the mean of Y for a unit change in X. In the case of a mathematical variable x, R has no meaning because the magnitude of R depends on the choice of spacing of x. We may also write, from Equation 11-44, R2

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