# Now let i r t s ? q dq and i ? t therefore d dt p k t

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Now, let I = R t s λ ( q ) dq and I 0 = λ ( t ) Therefore, d dt p k ( t, s ) = - I 0 e I I k + e I kI k - 1 I 0 k ! = - I 0 e I I k k ! + e I I k - 1 I 0 ( k - 1)! = e I I k k ! λ ( t ) + e I I k - 1 ( k - 1)! λ ( t ) = - p k ( t, s ) λ ( t ) + p k - 1 ( t, s ) λ ( t ) So that p k ( t, s ) does satisfy. And X t - X s is Poisson. (b) Find the mean and autocorrelation functions of { X t } . We have, E [ X t - X s ] = k =0 kp k ( t, s ). So, E [ X t ] = E [ X t - X 0 ] = X k =0 kp k ( t, 0) = X k =0 k e - R t 0 λ ( q ) dq R t 0 λ ( q ) dq k k ! = Z t 0 λ ( q ) dq 2

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Now, E [ X t X s ] = E [ X t ( X t + X s - X t )] = E [ X 2 t ] + E [ X t ( X s - X t )] = E [ X 2 t ] + E [ X t Y t ] = E [ X 2 t ] + E [ X t ] E [ Y t ] = λ 2 X + λ X + λ X λ Y = λ X ( λ X + 1 + λ Y ) = Z t 0 λ ( q ) dq Z t 0 λ ( q ) dq + 1 + Z s t λ ( q ) dq So, R X ( t, s ) = Z t 0 λ ( q ) dq Z s 0 λ ( q ) dq + 1 (for s > t ) and R X ( t, s ) = Z s 0 λ ( q ) dq Z t 0 λ ( q ) dq + 1 (for t > s ) 4. Suppose that { X t , t R } is a w.s.s., zero-mean, Gaussian random process with auto-correlation function R X ( τ ) , τ R and power spectral density S X ( ω ) , ω R . Define the random process { Y t , t R } by Y t = ( X t ) 2 , t R . find the mean, autocorrelation, and powerspectral density of { Y t , t R } . Mean : μ y ( t ) = E [ X 2 t ] = R X (0) = σ 2 Autocorrelation : R Y ( t, s ) = E [ Y t Y s ] = E [ X 2 t X 2 s ] = 4 ∂u 2 ∂v 2 Φ X t X s ( u, v ) u = v =0 1 i 4 .
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• Fall '08
• Stites,M
• Probability theory, Cos, Stochastic process, Autocorrelation, Stationary process, Xt

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