0
$\begingroup$

for the case when $s < t$

Let $N(t)$ be the amount of arrivals occurring at time period t in a Poisson process.

When $N(t)$ is known, the number of events by time $s$ can then be shown to follow a binomial distribution with parameters $(N(t), \frac{s}{t})$ as follows:

$P(N(s) = k|N(t) = n) = \frac{p(N(s) = k, N(t) = n)}{P(N(t) = n)}$ =

$\frac{P(N(s) = k, N(t) - N(s) = n-k)}{P(N(t) = n)}$ =

$\frac{\frac{(\lambda s)^{k}}{k!}e^{-\lambda s} \cdot \frac{(\lambda(t-s))^{n-k}}{(n-k)!} e^{-\lambda(t-s)}}{\frac{(\lambda t)^{n}}{n!}e^{-\lambda t}}$ =

$\frac{s^{k}(t - s)^{n-k}}{t^{n}} \cdot \frac{n!}{k!(n-k)!} =$

$\binom{n}{k} (\frac{s}{t})^{k} (1 - \frac{s}{t})^{n-k}$

I understand all of the above-mentioned steps accept part of the last one.

Particularly, why does $\frac{s^{k}(t - s)^{n-k}}{t^{n}} = (\frac{s}{t})^{k} (1 - \frac{s}{t})^{n-k}$?

$\endgroup$
1
$\begingroup$

$$\bigg(\frac{s}{t}\bigg)^k\bigg(1-\frac{s}{t}\bigg)^{n-k}$$ $$=\bigg(\frac{1}{t}s\bigg)^k\bigg[\frac{1}{t}(t-s)\bigg]^{n-k}$$ $$=\bigg(\frac{1}{t}\bigg)^ks^k\bigg[\frac{1}{t}\bigg]^{n-k}(t-s)^{n-k}$$ $$=\bigg(\frac{1}{t}\bigg)^ns^k(t-s)^{n-k}$$ $$=\bigg(\frac{1}{t^n}\bigg)s^k(t-s)^{n-k}$$ $$=\frac{s^k(t-s)^{n-k}}{t^n}$$

Let me know if I have made any mistakes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.