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Shouldn't probability density functions be in the form of

$$P(X\in dx) = \cdots$$

Why does the one for gamma distributions divide by $dt$?

$T_r =$ time of $r^\text{th}$ arrival after time $0$ in a poisson arrival process with rate $\lambda$.

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And if I multiply both sides by $dt$, how am I supposed to calculate $dt$ on the right side?

Michael Hardy
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A_for_ Abacus
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  • How does the answer you saw fit to accept, actually solve the main problems you describe in the question, namely, "Why does the one for gamma distributions divide by dt?" and "And if I multiply both sides by dt, how am I supposed to calculate dt on the right side?"? These are not even mentioned in the post below, or am I confused? – Did Mar 11 '18 at 23:29

1 Answers1

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No, $$\mathbb P(X\in dx) = f_X(x)\cdot dx$$

Therefore the probability that $r$-th arrival occures at interval $(t,t+dt)$ $$\tag{1}\label{1} \mathbb P(T_r\in dt)=f_{T_r}(t)\cdot dt. $$ By the other side, the event that $r$-th arrival occures at interval $(t,t+dt)$ means that exactly $r-1$ arrivals occure before $t$, and exactly one - on the interval $(t,t+dt)$. This events are independent, and the probability of first one is $$ \mathbb P(N_t=r-1)=\frac{(\lambda t)^{r-1}}{(r-1)!}e^{-\lambda t}, $$ while the second one has the probability $$ \mathbb P(N_{dt}=1)=\lambda \cdot dt. $$ Therefore $$\tag{2}\label{2} \mathbb P(T_r\in dt)= \frac{(\lambda t)^{r-1}}{(r-1)!}e^{-\lambda t} \cdot \lambda\cdot dt $$ Compare (\ref{1}) and (\ref{2}).

NCh
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  • Thanks, I understand that the first probability is Poisson$(\lambda t)$ with $K=r-1$, but how did you get the second probability? – A_for_ Abacus Mar 11 '18 at 07:34
  • @A_for_Abacus The answer to this question depends on how was the Poisson process defined in your course. This can be either the part of definition: $P(N_{t+\delta}-N_t=1)=\lambda \delta+o(\delta)$ as $\delta\to 0$, $P(N_{t+\delta}-N_t>1)=o(\delta)$. Or we can comsider Poisson distribution $P(N_{t+\delta}-N_t=1)=\lambda \delta e^{-\lambda \delta}=\lambda \delta + o(\delta)$. – NCh Mar 11 '18 at 08:05
  • @NCh : If you haven't already, you may consider up-voting this question. (I have done so, but so far the total is $0.$) $\qquad$ – Michael Hardy Mar 11 '18 at 18:15
  • @MichaelHardy I doubt that it is interestng or showing research efforts to upvote it. Sorry. – NCh Mar 11 '18 at 18:19
  • @NCh : This seems like clearly a question in the mind of the poster, rather than a copied question that the poster might not even understand. Isn't the reason for "research efforts" primarily to assure that the poster actually has a question, rather than just copying a question that the poster doesn't understand? In this case, coming up with the question in the first place is creditable "research effort". $\qquad$ – Michael Hardy Mar 11 '18 at 18:39
  • @MichaelHardy Ok, agreed – NCh Mar 11 '18 at 18:40