Poisson Distribution* I have a confusion in this,if there exists a Poisson distribution with λ as its parameter, let p be a probability of a related event happening, then is the distribution of the second event of probability p also a Poisson distribution with parameter p*λ.
 A: Suppose that $X$ is a Poisson random variable with parameter $\lambda$; i.e., $$\Pr[X = x] = e^{-\lambda} \frac{\lambda^x}{x!}, \quad x = 0, 1, 2, \ldots.$$  This random variable models the number of events occurring in a given fixed time period when the average rate of events in such a time period is $\lambda$.
Now, suppose we have an associated event, in the sense that whenever the original event occurs, the associated event has a probability $p$ of occurring at the same time; and the random variable $Y$ counts the random number of associated events in the same given time period as $X$.  Or, perhaps the model is that there are events of two types, and that $X$ counts the number of events of either type, whereas $Y$ counts the number of events of the first type, where, given that an event is observed, the probability it is of the first type is $p$.
Then, the goal is to show that the unconditional distribution of $Y$ is also Poisson, but with rate $p \lambda$.
To this end, we note that the conditional distribution of $Y$ given $X$ is binomial with parameters $n = X$ and $p$, because given that we have observed $X$ events, the number of $Y$-type events among these is equivalent to a sum of IID Bernoulli trials on each of the observed events, with probability of success $p$; that is to say, $$\Pr[Y = y \mid X] = \binom{X}{y} p^y (1-p)^{X-y}, \quad y = 0, 1, \ldots, X.$$  This is the key observation.
Now we condition $Y$ on $X$:
$$\begin{align*} \Pr[Y = y] 
&= \sum_{x=0}^\infty \Pr[Y = y \mid X = x]\Pr[X = x] \\
&= \sum_{x=y}^\infty \Pr[Y = y \mid X = x]e^{-\lambda} \frac{\lambda^x}{x!} \\
&= e^{-\lambda} p^y \sum_{x=y}^\infty \binom{x}{y} (1-p)^{x-y} \frac{\lambda^x}{x!} \\
&= e^{-\lambda} (p\lambda)^y \sum_{m=0}^\infty \binom{y+m}{y} \frac{((1-p)\lambda)^m}{(y+m)!} \\
&= e^{-\lambda} \frac{(p\lambda)^y}{y!} \sum_{m=0}^\infty \frac{((1-p)\lambda)^m}{m!} \\
&= e^{-p\lambda} \frac{(p\lambda)^y}{y!} \sum_{m=0}^\infty e^{-(1-p)\lambda} \frac{((1-p)\lambda)^m}{m!} \\
&= e^{-p\lambda} \frac{(p\lambda)^y}{y!}.
\end{align*}$$
Note here that we used the fact that the last sum in our derivation was the sum of a Poisson random variable with parameter $(1-p)\lambda$ over its support, and therefore equals $1$.  And the final result we clearly recognize as a Poisson PMF with rate $p\lambda$, as claimed.
This proof demonstrates a phenomenon known as Poisson thinning.
