Exponential Distribution(HARD)

Cyril

2009-04-04 2:21 am

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....

回答 (1)

myisland8132

2009-04-04 6:24 am

✔ 最佳答案

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)]

參考: 結果之中必然有原因

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