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Exponential Distribution(HARD)

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Let X~ Exp(lemda)a. Define a quantization function q : R ->Z (all real number maps to interger) such that Q = q(X) = floor(X). Find the PMF of Q.b. Describe the distribtion of Q in terms of a standard distribution.c. Find the expectation of Q.d. Now define the error from this quantization as R =... 顯示更多 Let X~ Exp(lemda) a. Define a quantization function q : R ->Z (all real number maps to interger) such that Q = q(X) = floor(X). Find the PMF of Q. b. Describe the distribtion of Q in terms of a standard distribution. c. Find the expectation of Q. d. Now define the error from this quantization as R = X-Q. Find the CDF of R. I have no idea how to do this question....

最佳解答:

X~ Exp(λ) (a) Q= floor(X). So the probability of Q= 0 is F(1)-F(0)=1- e^(-λ) Similarly, the probability of Q= k is F(k+1)-F(k)=e^(-λk)-e^-λ(k+1)=e^(-λk)(1-e^(-λ)) (b) Q~Geometric (1-e^(-λ)) (c) E(Q)=1/[1-e^(-λ)] (d) For particular k, the CDF of R is F(R|k) F(X)-F(Q) =F(Q+R)-F(Q) =[1- e^(-λ)(k+R)]-[1- e^(-λk)] =e^(-λk)-e^(-λ)(k+R) =e^(-λk)[1-e^(-λR)] The whole CDF of R is F(R) =Σ(k fro 0 to infinity) e^(-λk)[1-e^(-λR)] =[1-e^(-λR)]Σ(k fro 0 to infinity) e^(-λk) =[1-e^(-λR)] / [1-e^(-λ)] Check: When R=0., F(R)=0 When R=1, F(R)=1 So our CDF actually makes sense That's all 2009-04-04 00:33:06 補充： (c) E(Q)=e^(-λ)/[1-e^(-λ)]

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