a******h
发帖数: 1183 | 1 在下面的等式中，T, A_i, B 是独立随机变量，其中A_i的CDF已知：A_i, i=1,.. 是
iid，且服从指数分布，参数为1。 B的CDF也已知，那T的CDF能求出来吗？ | d****y
发帖数: 53 | 2 schematically (T is a discrete random variable; denote its prob's as p_T, T=
1,2,... \infinity)
compute the moment-generating-function on both sides, one gets
\sum_T=1^{\infinity} p_T [1/(1+u)]^T = MGF_of_B(u) = MGF_of_B( (1+u) -1 )
expanding both sides of the above equation as a power seris in y=(1+u) gives
you p_T.
Additional comments:
Interpretation of B: it is the arrival-time of the T-th event for a poisson
process.
【在 a******h 的大作中提到】
: 在下面的等式中，T, A_i, B 是独立随机变量，其中A_i的CDF已知：A_i, i=1,.. 是
: iid，且服从指数分布，参数为1。 B的CDF也已知，那T的CDF能求出来吗？
| a******h
发帖数: 1183 | 3 Thank you.
Could you explain a bit more about getting the moment-generating-function
at the left hand side?
T=
gives
poisson
【在 d****y 的大作中提到】
: schematically (T is a discrete random variable; denote its prob's as p_T, T=
: 1,2,... \infinity)
: compute the moment-generating-function on both sides, one gets
: \sum_T=1^{\infinity} p_T [1/(1+u)]^T = MGF_of_B(u) = MGF_of_B( (1+u) -1 )
: expanding both sides of the above equation as a power seris in y=(1+u) gives
: you p_T.
: Additional comments:
: Interpretation of B: it is the arrival-time of the T-th event for a poisson
: process.
| d****y
发帖数: 53 | 4 E[exp^{-ut}]=\int exp^{-t}exp^{-ut} \dt for an exponential dist. with rate=1
(your specification). This gives you the mgf on the lhs
【在 a******h 的大作中提到】
: Thank you.
: Could you explain a bit more about getting the moment-generating-function
: at the left hand side?
:
: T=
: gives
: poisson
| a******h
发帖数: 1183 | 5 Thanks. I got this part now.
Regarding the expansion on both sides, do you mean using taylor expansion?
If yes, then the problem is that on the lhs, the result is expressed with a
power series in 1/y, (y=(1+u)), while on the rhs, it gives you a power seris
in y. This way, we cannot obtain the expression of P_T, right?
=1
【在 d****y 的大作中提到】
: E[exp^{-ut}]=\int exp^{-t}exp^{-ut} \dt for an exponential dist. with rate=1
: (your specification). This gives you the mgf on the lhs
| d****y
发帖数: 53 | 6 think of it as a laurent expansion, not a taylor expansion: both negative
and positive powers on both sides must agree.
a
seris
【在 a******h 的大作中提到】
: Thanks. I got this part now.
: Regarding the expansion on both sides, do you mean using taylor expansion?
: If yes, then the problem is that on the lhs, the result is expressed with a
: power series in 1/y, (y=(1+u)), while on the rhs, it gives you a power seris
: in y. This way, we cannot obtain the expression of P_T, right?
:
: =1
| a******h
发帖数: 1183 | 7 Is it possible to express the CDF of T with A and B explicitly?
【在 d****y 的大作中提到】
: think of it as a laurent expansion, not a taylor expansion: both negative
: and positive powers on both sides must agree.
:
: a
: seris
|
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