I am trying to understand two comments from a colleague. I have no clue about continuous time models. He said something like "just replace $\delta$ with $e^{-rdt}$ the Poisson process becomes a Bernoulli process."

Suppose I have a discrete time setting with discount factor $\delta$. A buyer receives a constant value $v$ for each period in which he uses a good. Then, for example, my total value today of using the good for two periods is $v(1+\delta)$. Would the value from having the good over a time interval $[0,d]$ be $\int_0^d v e^{-rt} dt$, or how does it work?

Next, if in one period of length $d$ in expectation $\lambda_d$ buyers arrive and the number of arrivals is Poisson distributed, how do I get to the Bernoulli process? What exactly is the probability? Whatever I wrote down seems to converge to zero.

Do you have a recommendation where I can read up on these basics?


1 Answer 1


Poisson process can be interpreted as a continuous case of Bernoulli process.

Taking your example, consider that the buyer is consuming the good in batches in fixed intervals of time with probability $p$. So the consumption is allowed only after fixed intervals, and the r.v. $X_t \sim Bern(p)$, where $X_t=1$ indicates that the buyer consumes the good at time $t$. Further, how much she consumes in $n$ periods is distributed as $Bin(n,p)$. The expected level of consumptions in $n$ periods as we know is $np$.

Now consider that the buyer can consume at any instant and just not in intervals. Unlike in previous case where consumption could happen only at $t=1,2,...n$, now it can take place at any $t\in \mathbb{R^+}$. Naturally the probability that consumption happens at some given $t$ is $0$. So comes the rate $\lambda$, i.e., we say that the buyer on average consumes $\lambda$ amount in a given length $n$ of time. Now the total consumption between $t=0$ to $t=n$ becomes $Pois(\lambda)$.

So, $Pois(\lambda)$ is a limiting case of $Bin(n,p)$ as time interval $\to 0$.

Coming to your second question:

The buyer gets value $v$ when he uses the good at a given time, and the value gets depreciated after one time interval by $\delta$. The total value, as you mention, is: $v(1+\delta+\delta^2+..)$

However, if we assume the value is depreciating continuously at rate $r$ we can say:

\begin{align} rv(t)&=\lim_{\Delta t \to 0} \frac{v(t+\Delta t) - v(t)}{\Delta t} \\ &=\frac{dv(t)}{dt} \\ \end{align}

From this you will get that the value from having the good over a time interval $[0,d]$ be $\int_0^d v e^{-rt} dt$

Lastly, to answer the last part: if you want to go to Bernoulli process from the described poisson process consider that arrivals can happen only in discrete time intervals such that on average in $d$ periods, $\lambda_d$ arrivals take place. So, this translates to the bernoulli process such that arrivals happens at a given time $t$ with probability $p=\lambda_d/d$.

  • $\begingroup$ Dayne's answer was excellent but just to make the OP understands, in your question, I think you meant ONE period rather than two periods when describing the discrete case. $\endgroup$
    – mark leeds
    Nov 19, 2020 at 10:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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