PDB_Stat_100_Lecture_15_Printable

PDB_Stat_100_Lecture_15_Printable - STA 100 Lecture 15 Paul...

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STA 100 Lecture 15 Paul Baines Department of Statistics University of California, Davis February 7th, 2011

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Admin for the Day I Please pick up Midterm+old homeworks I Important: Homework 3 Solutions posted – please read! I Homework 4 posted, due Wednesday by 5pm! I Hwk 4, Q2(i) – compare to the correct answer for Hwk3 Q2(k) I Hwk 4, Q3 – Weekly data (not monthly) I Project details coming soon. . . References for Today: Rosner, Ch 6.5-6.8, Ch 6.5 (7th Ed.) References for Wednesday: Rosner, Ch 6.5-6.8 (7th Ed.)
Recap: Linear Combinations We just covered linear transformations , where you add and multiply a single random variable by ﬁxed numbers ( a and b ). Now we consider linear combinations of random variables. This is where you add multiple random variables together. . . These are diﬀerent things! Linear transformations deal with transformations of a single variable (usually to change units), whereas linear combinations deal with what happens when you combine lots of random variables (usually to take a sum or average of them).

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Recap: Linear Combinations Let X 1 ,..., X n be any random variables, c 1 c n constants (just numbers). I The mean of a linear combination of any random variables: E [ c 1 X 1 + c 2 X 2 + ··· + c n X n ] = c 1 E [ X 1 ] + c 2 E [ X 2 ] + + c n E [ X n ] . I The variance of a linear combination of independent random variables: Var ( c 1 X 1 + c 2 X 2 + + c n X n ) = c 2 1 ( X 1 ) + c 2 2 ( X 2 ) + + c 2 n ( X n ) . I The SD of a linear combination of independent random variables: compute the variance and take a square root. Note: you cannot just add the standard deviations! I Linear combinations of normally distributed random variables are still normally distributed!
Recap: Linear Combinations Let X 1 ,..., X n be independent normally distributed random variables, c 1 c n constants (just numbers). I If we create a new random variable using the X ’s, say: Y = c 1 X 1 + c 2 X 2 + ··· + c n X n , then, using the properties on the previous slide we get: Y N ( μ Y 2 Y ) , where: μ Y = c 1 E [ X 1 ] + c 2 E [ X 2 ] + + c n E [ X n ] , σ 2 Y = c 2 1 Var ( X 1 ) + c 2 2 ( X 2 ) + + c 2 n ( X n ) .

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Example Two independent random variables: X 1 N (0 , 1), X 2 N (0 , 1). I Lets consider Y = X 1 - X 2 . I So, compare to the general formula: Y = 1 · X 1 + ( - 1) · X 2 Y = c 1 · X 1 + c 2 · X 2 I So, c 1 = 1 and c 2 = - 1 . Plugging in: E [ Y ] = 1 · E [ X 1 ] + ( - 1) · E [ X 2 ] = 0 , Var ( Y ) = 1 2 · ( X 1 ) + ( - 1) 2 · ( X 2 ) ( Y ) = ( Y ) + ( Y ) = 1 + 1 = 2 , i.e., Y N (0 , 2) . I Even though we subtract X 1 and X 2 we add their variances!
Subtracting Normals Lots of X 1 ’s and X 2 ’s, Y ’s: x1 x2 x1-x2 1 -0.733730691 0.77965838 -1.51338907 2 -0.077958086 0.02981548 -0.10777357 3 -1.364323284 1.28076141 -2.64508470 4 -1.434933965 0.33073996 -1.76567393 5 1.179683727 1.27726560 -0.09758187 6 -0.008452536 -0.57704471 0.56859218 ...

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X 1 and X 2 -6 -4 -2 0 2 4 6 0.0 0.1 0.2 0.3 0.4 Histogram of X_1 x1 -6 -4 -2 0 2 4 6 Histogram of X_2 x2
Y = X 1 - X 2 -4 -2 0 2 4 6 0.0 0.1 0.2 0.3 0.4 Histogram of Y = (X_1 - X_2) x1 - x2 Note: The spread is wider than for X 1 and X 2 !

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Example Example:
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PDB_Stat_100_Lecture_15_Printable - STA 100 Lecture 15 Paul...

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