# Relationship of continuous and discrete time models

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?

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 $$\to 0$$.

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 $$p=\lambda_d/d$$.