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$P_{n} (t)$ is the probability of $n$ events occurring during a time interval $t$. How will you express $P_{0} (t + h)$ in terms of $P_{0} (h)$, if $P_{0} (t)$ has stationary independent increments? (Note: $P_{t} (t)$is the probability density function).

@HitechGa Can you explain your solution as why did you take $P_0(t)= e^{-rt}$ and not $P_0(t)= (rt)^{r} e^{-rt}/r!$ ?

@Abhrajyoti00 Hello, you need to revisit the definition of Poisson distribution from a standard book. Well, the probability mass function of Poisson distribution can be given as:

Let $r$ be the average rate of an event happening in unit time.

Hence, the average rate of happening of that event in time $t$ units =$rt$.

So this becomes our mean ($\lambda$)= $rt$

Now, let $X$ be the random variable indicating the number of occurrences of that event and it is this random variable $X$ which follows the Poisson distribution. So the probability mass function is given as:

$$P(X=x)= \frac{\lambda^x\times e^{-\lambda}}{x!}$$

In the given GATE question, $x=0$, note the subscript of $P$ which denotes a $0$, so we have:
$$P(X=0)=e^{-\lambda}$$

$$\implies P(X=0)=e^{-rt}$$

Thanks @HitechGa. Got to know that the equation can be adapted if, instead of the average number of events $\lambda$, we are given the average rate $r$ at which events occur. Then $\lambda = rt$

Till this time I just knew about $\lambda = np$

$P_0(t)$ denote the probability that no events happened in an interval of length $t.$$P_0(t + h) = P_0(t) P_0(h)$$ This is because if there are no events in interval$[0,t+h]$then there are no events in intervals 1.$[0,t]$2.$[t, t+h]$These two intervals are non overlapping and it is given in question that$P_0(t)\$ has stationary independent increments and so their probabilities are independent.

PS: One of the axioms of Poisson distribution is that the numbers of events in two nonoverlapping regions are independent.

by

What does stationary independent increment mean here?