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I have this problem that I think I've kind of solved but just wanted to make sure if it's correct.

The problem goes:

The homework submissions to the university computer center start at midnight (00:00). The number of homework submissions between midnight and any time t > 0 afterwords is distributed Poisson with mean λt, where λ > 0 is some constant. Let T be the time between midnight and the first submission. What is the distribution of T

I think the answer is: exponential distribution

because we have time that is continuous variable, and poisson is just for submissions that are between midnight that means that is for discrete variable

so the distribution of T needs to be exponential distribution.

Does that work?

Thank you in advance!

Cheul Hee Jason Yoo
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2 Answers2

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Yes, your answer is correct, but your explanation is not.

Problem with your explanation is, that you basically say, that $T$ has exponential distribution only because it is continuous. Since there are definitely more than one continuous distributions, how do we know, that $T$ does not have for example $\chi^2$ distribution (it is also continuous and positive)?

The right explanation is that $\{T_n\}$, where $T_n$ is time of n-th submission, is Poisson process and times between occurrences of Poisson process, $T_n-T_{n-1}$, have exponential distribution. In your case $T=T_1-T_0=T_1$.

iiivooo
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You're given that $$ P(n)=\frac{(\lambda t)^n}{n!} e^{-\lambda t} $$

The time to the first homework submission is the same as the "end" of having zero homework submissions, so $$ P(0)=e^{-\lambda T} $$

xidgel
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