Let number of mosquitos landing on a victim be a Poisson process with intensity $\lambda$. Each mosquito bites victim with probability $p$ independently of others. Prove that total number of bites is a Poisson process with intensity $p \lambda$.
Let $X_t$ be a total number of bites until the moment $t$. To be a Poisson process it should satisfy
- $X_0 = 0$
- if $t<s\leq u<r$ then $X_r-X_u$ and $X_s-X_t$ are independent
- for $t\geq s$ $X_t-X_s$ is Poisson distributed
1 is obvious but 2 and 3 aren't clear for me. Since the Poisson distribution with intensity $\lambda$ is $\frac{(\lambda t)^k}{k!}e^{-\lambda t}$ then probability that total number of mosquitoes will be $k$ at the moment $t$ is given by $P(X_t=k) = \int\limits_0^t \frac{(\lambda x)^k}{k!}e^{-\lambda x}dx$.
Am I right? If yes, how to evaluate such integral without computer algebra systems? Then, define the result of integration by $I(t)$ I woild like to say that total number of bites until the moment $t$ is $\int\limits_0^t p^{I(x)} dx$, since at every moment $t_0$ there are $I(t_0)$ mosquitoes, each biting independently with probability $p$.
I would be very glad if anyone explain me how it should be proved.