Given a poisson process Z(t) with a given rate lamda, k and m nonnegative integers and t and c real and positive numbers, calculate the probability:

P(Z(t-c)=m | Z(t)=k)

thank you

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- Jun 21st 2011, 06:35 AMalehandr12Poisson process question
Given a poisson process Z(t) with a given rate lamda, k and m nonnegative integers and t and c real and positive numbers, calculate the probability:

P(Z(t-c)=m | Z(t)=k)

thank you - Jun 22nd 2011, 03:48 AMSpringFan25Re: Poisson process question
This is a poisson process so we have the following properties:

$\displaystyle [A]~Z_0 = 0$

$\displaystyle [B]~Z_t \sim Po(\lambda t)$

$\displaystyle [C]~Z_c - Z_b \sim Po(\lambda (c-b)) \text{ and is independent of } Z_b $

Start from the usual formula for conditional probabilities

$\displaystyle P(Z_{t-c} = m | Z_t =k) = \frac{P\left(Z_{t-c} = m \cap Z_{t} = k \right)}{P(Z_t)=k} $

The denominator is easy using property [B].

The numerator can be re-written as follows:

$\displaystyle P\left(Z_{t-c} = m \cap Z_{t} = k \right) = P\left(Z_{t-c} = m \cap Z_{t}-Z_{t-c} = k -m \right)$

using property [C], the events $\displaystyle Z_{t-c} =m$ and $\displaystyle Z_t -Z_{t-c} =k-m$ are independent, and hence

$\displaystyle P\left(Z_{t-c} = m \capZ_{t}-Z_{t-c} = k -m \right) = P\left(Z_{t-c} = m \right) \times P \left(Z_{t}-Z_{t-c} = k -m \right) $

Both of those terms follow a poisson distribution. (unless k-m is negative, then the probability is zero).

So the final answer is:

$\displaystyle P(Z_{t-c} = m | Z_t =k) = \frac{P\left(Z_{t-c} = m \right) \times P \left(Z_{t}-Z_{t-c} = k -m \right)}{P(Z_t = k)}$

Where:

$\displaystyle Z_t \sim Po(\lambda t)$

$\displaystyle Z_t - Z_{t-c} \sim Po(\lambda c)$

$\displaystyle Z_{t-c} \sim Po(\lambda (t-c))$

you can subsitute the formulae for the PMFs to finish. Dont forget that the probability is zero if m > k