# poisson processes as a function of a parameter

Problem: The number of customers arriving at a bank in a Poisson process $\lambda = 3$ per minute. What is the probability that no customer arrives in a $T$ minute interval?

Solution: This is a Poisson distribution problem with parameters $\lambda = 3$ and $x = T$. So I plug into the formula and get $\dfrac{ \lambda^T * e^{-\lambda} }{ T! }$.

So $P(\text{no customers arrive in a }T\text{ minute interval}) = P( Xt=0 ) = ( 3\times5 )^0 \times e^{-3}\times\frac{T}{0!} = e^{-3}T$.

So solving this, I can only solve the problem as a function of $T$. I have no way of verifying if this is indeed the correct solution. Is anyone familiar with Poisson processes? Any help would be great, thanks in advance!

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Hi Jake, do you know how to use latex to write the formulas in a readable way? Anyway the answer you get ($e^{-3T}$) is correct but it seems you have some confusions. When you write "( Lambda^T * e^-Lambda ) / ( T! )" for the formula $(\lambda t)^x \frac{e^{-\lambda t}}{x!}$ x is integer and represents the numbers of costumers. In your problem T is a positive real meaning a time. In this case you want $P(X_T=0)$ and since $X_T \sim Poi(\lambda T)$ you get the result. Note also that this shows that the time arrival of the events has an exponential distribution –  Kolmo Apr 30 '12 at 15:05
Awesome, thanks! I'm new here so I'm not familiar with how to input formulas and edit text to improve readability. Sorry for the confusion –  EulerChild Apr 30 '12 at 15:13
Hard to know, the intermediate step $\lambda^Te^{-\lambda}/T!$ is wrong, getting ultimately the right answer looks purely accidental. The crucial fact is that if the number of events in a certain time interval $\tau$ has Poisson distribution with parameter $\lambda$, then the number of events in the time interval $T\tau$ has Poisson distribution with parameter $\lambda T$. (Here $\tau$ is $1$ minute.) –  André Nicolas Apr 30 '12 at 15:34