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This list of all twoletter combinations includes 1352 (2 × 26 2) of the possible 2704 (52 2) combinations of upper and lower case from the modern core Latin alphabetA twoletter combination in bold means that the link links straight to a Wikipedia article (not a disambiguation page) As specified at WikipediaDisambiguation#Combining_terms_on_disambiguation_pages,Remark 12 For nonnegative rvs X, it is sometimes more common to use the Laplace transform, L(s) = E(e−sX), s ≥ 0, which is always finite, and then (−1)nL(n)(0) = E(), n ≥ 1 For discrete rvs X, it is sometimes more common to use M(z) = E(zX) = X∞ k=−∞ zkp(k), z ≤ 1Where the interchange of integrals is justi ed by Fubini's theorem for improper Riemann integrals

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2 LORENTZ FORCE LAW 2 2 Lorentz Force Law The Lorentz force in Gaussian Units is given by F~ = Q ˆ E~ ~v c £B~!;Problem 3 Show that f z e z is an analytic Function Solution f z u iv e z ex iy e ex iy e y i yx cos sin u e y v e y x xcos , sin x x, u e cosy v e siny x x x x, u e siny v e cosy y y ie, u v u v x y y x , Hence CR equations are satisfied f z e z is analyticDepartment of Computer Science and Engineering University of Nevada, Reno Reno, NV 557 Email Qipingataolcom Website wwwcseunredu/~yanq I came to the US

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2 KEITH CONRAD Instead of using polar coordinates, set x= ytin the inner integral (yis xed) Then dx= ydtand (21) J2 = Z 1 0 Z 1 0 e 2y2(t21)ydt dy= Z 1 0 Z 1 0 ye y2(t 1) dy dt;1 Answer1 One wants to show that E ( T ∣ X, Z) = T with T = E ( Y ∣ X) This holds true in full generality since (i) the random variable T is σ ( X) measurable by definition hence T is σ ( X, Z) measurable, and (ii) E ( U ∣ X, Z) = U for every σ ( X, Z) measurable random variable U Recall that E ( U ∣ V) is defined as theDefinition of Z = E(X) X X σ µ Rule 7b E(a ± X) * b = (a ± E(X)) * b Remember, a = µX, b = 1/σX 0 Remember E(X) = µX, so the numerator = 0 QED Intuitively, the above makes sense;

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Factorial Product of all integers up¦ x = Sum of x scores V Sigma Population Standard Deviation N ¦ x ( P)2 V = E(x P) ¦ (x P)2 P(x) Measures of dispersion V2 Sigma square Population variance N ¦ x 2 ( P)2 V Measures of dispersion Mathematical Statistical Symbols Symbol Text Equivalent Meaning Formula Link to Glossary (if appropriate) !Solution Let f (z¯)=f (x −i y)=u(x,y)iv(x,y) where u and v are real Then fx =f ′(z¯)=ux ivx, fy =−i f ′(z¯)=uy ivy Thus, f ′(z¯)=ux ivx =i(uy ivy), which implies ux =−vy, vx =uy Differentiatingthese relations, one gets uxx uyy =−vxy vyx =0, vxx vyy =uxy −uyx =0 Foranalyticity, f must satisfyCR relations ux =vy, vx =−uyBut thenux =vy =−vy =0etc, ie ux

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Z xfX(x)dx = EX Consider (v) Suppose that the random variables are discrete We need to compute the expected value of the random variable EXjY It is a function of Y and it takes on the value EXjY = y when Y = y So by the law of the unconscious whatever, EEXjY = X ySubtract the mean from every case and the new mean becomes zero Now, for the variance, Equation Explanation = X V(Z) = V X XY respectively The covariance, denoted with cov(X;Y), is a measure of the association between Xand Y

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(a) (4 points) Show that E(x) = µ E( x ) = E (∑ = n i xi n 1 1) = n 1 E(∑) = = n i xi 1 n 1 ∑ = n i E xi 1 ( ) = n 1 n µ = µ It is possible to show that Var (x) = n σ2, but you need not prove it (b) (4 points) Let Z = σ n(x −µ) Show that E(Z) = 0 E(Z) = E(σ n(x −µ)) = σ n E( x −µ) = 0 because E( x ) = µ (c) (4The divergence of F is e x z 2 x z e x z 2 x z If F were the curl of vector field G, then div F = div curl G = 0 div F = div curl G = 0 But, the divergence of F is not zero, and therefore F is not the curl of any other vector fieldV u } o v µ u v u o } À v X Z o Á Ç Z µ u v } *29(510(17 2) ,1',$ 0,1,675< 2) 5$,/$

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