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µ µ σ2 V(X) = E(X )2 = E(X 2)− 2 = x In the discrete case, this is equivalent to = =∑ − All X V(X) σ2 (x µ)2 P(x) E Standard deviation of X The standard deviation is the positive square root of the variance, ie SD(X) =σ= σ2 Expectations Page 1 F Examples 1 Hayes (p 96) gives the probability distribution for the number of spots appearing on two fair dice Find theG H Z M % !ø d v H á , N Y ³ 5 Ö 0 5 0 j g @ é H á , @ ¦ Ü Y n µ 9 õ ) Û ð ½ H á , ç P P Z v R E ÿ 1 \ R Þ û µ 9 1 j g N v ¸ Ô 1 I « \ ñ 1 ) Î Z ¾ Û 1 5 À s Æ ¯ 6 ' ¬ Û 1 2 e b l I H á , ¸ J 5 H á , æ µ _ 5 µ I

µ = E X = ∞ J −∞ xf (x)dx σ 2 g(x) = λ α x e α−1 −λx, x > 0, Γ(α) So U is Gamma(α, λ) with α = 1/2 and λ = 1/2 Moment generating function M U (t) = E e tU = 1 − t/λ −α = (1 − 2t) −1/2 MIT Distributions Derived 4From the Normal Distribution Distributions Derived from Normal Random Variables χ 2 , t, and F Distributions Statistics74 SG1RGQ/308 للجن Ø© ا€X ¯Ø±Ø§Ø³€ ª 1€ñ ‚Ø·€ˆ ¹ تنÙÙŠ M ص€U â€ð ‚دقX H †‚ƒ ¨‚‚ ²ÙŠÙ„ƒˆ„ pƒédiv„ €0 åpubtype="footnote" p†œsiz€Ð1"æac Dubai"‡V19€ ‰†Ê § 2‚ filepos=0€ ‚ï‚ï‚ïŠ=079cd6"> 90‰è€@ŠVŠ Š·Š³„'„' ‹Ú‰r ˆØ«ˆ° ‚‹È• Similarly for the asymptotic distribution of ρˆ(h), eg, is ρ(1) = 0?

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Then, U = g(X) and V = h(Y) are also independent for any function g and h We will come back to various properties of functions of random variables at the end of this chapter 2 2 Moments and Conditional Expectation Using expectation, we can define the moments and other special functions of a random variable Definition 2 Let X and Y be random variables with their expectations µ X = E(XNotation ∼ AN µn,σ2 n) means 'asymptotically normal' −µn σn →d Z, where Z∼ N(0,1) 15 Estimating µ for a linear process Asymptotically normal Theorem (A5) For a linear process Xt = µ P j ψjWt−j, if P ψj 6= 0 , then X¯ n ∼ AN µx, V n , where V = X∞ h=−∞ γ(hI,j =1, 2,,n be the variance–covariance

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X(t) G y(t) For the sinusoidal input x(t) = cos(!t) the output takes the form y(t) = Acos(!t ') = acos(!t)bsin(!t) and we want to determine the amplitude A and phase ' 30 As the phase shift introduces a combination of cosines and sines, the problem can be simplifled by embedding the input signal into a larger class of signals involving both a cosine and a sine component It turnsTitle Microsoft Word modulo passeggeri GRITdocx Author andreacetta Created Date PMY ¦ Ó ³ ã ï A ¨ ¯ y , G O $ y § å Ì æ ¤ ³ ã ï INDEX Â å µ t ® ` T l h ¯ î q ` ` h {*9* x B ` h { Â å µ t ® ` T l h ¯ K s \ q \ s \ q { { h w x z µ Ð Ã p ¶ s ª » Q { f ` o A t ` o F ü s ;

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E (y)= a i E (x i)= µ a i and V (y)= a 2 i V (x i)= σ 2 a 2 i Any linear function of a set of normally distributed variables is normally distributed If x i ∼ N (µ,σ 2);Title 13_â ¡5PCRï¼ è¡¨ï¼ æ §â ¢3ç­ ï¼ _ Author yukarichaya Created Date 11/9/ AMÓ ) ò ' 5 µ é à F (à Ô * ¼ = = ù ª Q M J j k & J ® ´ Z T 7 T a m k A ö 9 & ¢ Ñ ß} y h V ¥ ð N 2VDND O ¦ 2VDND O y h V ¥ N ª ù ¯ 2 q O Q x 4 * % = Ö ú J?

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ó Q E M t ` X Ð è b à ² ï { h X ^ w § Ë O U µ » ï ¼ Å { s y z f w 7¼ I C M < 0b O Z o K 8 \ î 8 r M Q ^#Õ b$Ù I #Õ c ²  µZ = f(x;y) g(x;y) zg 2 Find the CDF F Z(z) = P(Z z) = P(g(X;Y) z) = P(f(x;y) g(x;y) zg) = Z Z Az p X;Y(x;y)dxdy 3 The pdf is p Z(z) = F0 Z (z) Example 5 Practice problem Let (X;Y) be uniform on the unit square Let Z= X=Y Find the density of Z 5 Important Distributions Normal (Gaussian) X˘N( ;˙ 2) if p(x) = 1 ˙ p 2ˇ e (x )2=(2˙2 If X2Rd then X˘N( ;) if p(x) = 1 (2ˇ)d=2j j

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H / h ¶ W ñ / Ï Ú ð ý / ¢ Ô x m W ñ v b 9 y / Ñ 6 * P c , æ ý v ¦ r / Ú ;Called so because its natural logarithm Y = ln(X) yields a normal rv X has density f(x) = (1 xσ √ 2π e −(ln(x)−µ)2 2σ2, if x ≥ 0;EX2jY = y = 1 25 (y 1)2 4 25 (y 1) Thus EX2jY = 1 25 (Y 1)2 4 25 (Y 1) = 1 25 (Y2 2Y 3) Once again, EX2jY is a function of Y Intuition EXjY is the function of Y that bests approximates X This is a vague statement since we have not said what \best" means We consider two extreme cases First suppose that X is itself a function of

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