Proof φ x u e e iux e x k 0 iu k k x k x k 0 iu k k

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Proof φ X ( u ) = E [ e iux ] = E X k =0 ( iu ) k k ! x k = X k =0 ( iu ) k k ! E [ x k ] . Then d j du j φ X ( u ) = j ! i j j ! E [ x j ] + X k>j u k - j ( other stuff ) so at u = 0 , d j du j φ X ( j ) u =0 = i j E [ X j ] . 2 Example 4 X ∼ N ( μ, σ 2 ) . Then it can be shown (homework!) that φ X ( u ) = e - u 2 σ 2 / 2+ iuμ Then it is straightforward to verity (homework!) that E [ X ] = μ E [ X 2 ] = σ 2 + μ 2 so that var( X ) = σ 2 . 2 Definition 11 For a joint r.v. ( X, Y ) we define a joint characteristic function φ XY ( u, v ) = E [ e iuX + ivY ] . 2 Then φ XY and F XY are uniquely related (two-dimensional Fourier transforms). Definition 12 The n th order moments of two random variables are the quantities of the form μ k,l = E [ X k Y l ] k 0 , l 0 , k + l = n. The nth order central moments are m kl = E [( X - E [ X ]) k ( Y - E [ Y ]) l k 0 , l 0 , k + l = n. 2 Example 5 For n = 2 , the second order moments are E [ X 2 ] , E [ Y 2 ] and E [ XY ] . The central moments are cov( X, Y ) , var( X ) and var( Y ) . 2 Properties: 1. Moments: μ k,l = n ∂u k ∂v l φ XY ( u, v ) u,v =0 i - ( k + l ) 2. X and Y are independent if and only if φ X,Y ( u, v ) = φ X ( u ) φ Y ( v ) for all ( u, v ) R 2 .
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ECE 6010: Lecture 2 – More on Random Variables 10 Sums of independent random variables Let X and Y be independent r.v.s, and let Z = X + Y. Then φ Z ( u ) = E [exp( iuz )] = E [exp( iuX + iuY )] = φ X,Y ( u, u ) . But also E [exp( iuX + iuY )] = E [exp( iuX ) exp( iuY )] = φ X ( u ) φ Y ( u ) . So φ Z ( u ) = φ X ( u ) φ Y ( u ) If X and Y are continuous r.v.s, then so is Z . f Z ( z ) = F - 1 [ φ Z ( u )] = F - 1 [ φ X ( z ) φ Y ( z )] = f X ( z ) * f Y ( z ) by the convolution theorem . Thus, when continuous independent random variables are added, the p.d.f of the sum is the convolution of the p.d.f.s (and respectively p.m.f. for discrete independent r.v.s). An example: Jointly Gaussian If ( X, Y ) ∼ N ( μ x , μ y , σ 2 x , σ 2 y , ρ ) , then φ X,Y ( u, v ) = exp[ i ( x + y ) - 1 2 ( u 2 σ 2 x + v 2 σ 2 y + 2 uvρσ x σ y )] We make an observation here: the “form” of the Gaussian p.d.f. is the exponential of quadratics. The form of the Fourier transform of the eponential of quadratics is of the form exponential of quadratics. This little fact gives rise to much of the analytical and practical usefulness of Gaussian r.v.s. Characteristic functions marginals We observe that φ X,Y ( u, 0) = φ X ( u ) In our Gaussian example, we have φ X ( u ) = φ X,Y ( u, 0) = exp( iuμ x - σ 2 x u 2 / 2) which is the ch.f. for a Gaussian, X ∼ N ( μ x , σ 2 x ) . We could, of course, have obtained a similar result via integration, but this is much easier. Some important inequalities In general, when we observe an outcome of a random variable, we “expect” it to be near the mean (that is, near the expected value). Further, the farther the outcome is from the mean, the less likely we expect the outcome to be. There are some very useful probabilities which quantize these intuitive “expectations.” These are the Markov inequality, and its
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ECE 6010: Lecture 2 – More on Random Variables 11 consequences, the Chebyshev inequality and the Chernoff bound. We will introduce these here.
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  • Fall '08
  • Stites,M
  • Probability theory, lim

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