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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?

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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 $p=\lambda_d/d$.

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  • $\begingroup$ 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 '20 at 10:41

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