People normally talk about the expected number of occurrences (k) for a fixed time λ, which is given as E(k) = λ. I would like to know the inverse: for a fixed number of occurrences, what is thr expected time E(λ) in which it will be reached?
I tried using the integral formula for the expected value and got an improper integral represented by the gamma function, but I don't know if that's correct.
On
You don’t need to know the gamma distribution of the sum of exponential variables to find this expectation. By the linearity of expectation, this is just $k$ times the expected waiting time for a single event.
For a homogeneous Poisson process with intensity $\lambda$, the random variable that describes the number of events observed in an interval of length $t$ is $$X(t) \sim \operatorname{Poisson}(\lambda t), \\ \Pr[X(t) = x] = e^{-\lambda t} \frac{(\lambda t)^x}{x!}, \quad x \in \{0, 1, 2, \ldots \}. \tag{1}$$ The random variable that describes the event time of the $x^{\rm th}$ event is $$T(x) \sim \operatorname{Gamma}(x, \lambda), \\ f_{T(x)}(t) = \frac{\lambda^x t^{x-1} e^{-\lambda t}}{\Gamma(x)}, \quad t > 0. \tag{2}$$ The reason the event time is gamma distributed is because the interarrival time between consecutive events is exponentially distributed with rate $\lambda$; thus, the $x^{\rm th}$ event time is the sum of $x$ IID exponential random variables, which is gamma with shape $x$ and rate $\lambda$.
Therefore, the expected time of the $x^{\rm th}$ event is simply $\operatorname{E}[T(x)] = x/\lambda$.