<p>Poisson process can be interpreted as a continuous case of Bernoulli process.</p>
<p>Taking your example, consider that the buyer is consuming the good in batches in fixed intervals of time with probability <span class="math-container">$p$</span>. So the consumption is allowed only after fixed intervals, and the r.v. <span class="math-container">$X_t \sim Bern(p)$</span>, where <span class="math-container">$X_t=1$</span> indicates that the buyer consumes the good at time <span class="math-container">$t$</span>. Further, how much she consumes in <span class="math-container">$n$</span> periods is distributed as <span class="math-container">$Bin(n,p)$</span>. The expected level of consumptions in <span class="math-container">$n$</span> periods as we know is <span class="math-container">$np$</span>.</p>
<p>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 <span class="math-container">$t=1,2,...n$</span>, now it can take place at any <span class="math-container">$t\in \mathbb{R^+}$</span>. Naturally the probability that consumption happens at some given <span class="math-container">$t$</span> is <span class="math-container">$0$</span>. So comes the rate <span class="math-container">$\lambda$</span>, i.e., we say that the buyer on average consumes <span class="math-container">$\lambda$</span> amount in a given <em>length</em> <span class="math-container">$n$</span> of time. Now the total consumption between <span class="math-container">$t=0$</span> to <span class="math-container">$t=n$</span> becomes <span class="math-container">$Pois(\lambda)$</span>.</p>
<p>So, <span class="math-container">$Pois(\lambda)$</span> is a limiting case of <span class="math-container">$Bin(n,p)$</span> as time interval <span class="math-container">$\to 0$</span>.</p>
<p>Coming to your second question:</p>
<p>The buyer gets value <span class="math-container">$v$</span> when he uses the good at a given time, and the value gets depreciated after <em>one</em> time interval by <span class="math-container">$\delta$</span>. The total value, as you mention, is: <span class="math-container">$v(1+\delta+\delta^2+..)$</span></p>
<p>However, if we assume the value is depreciating continuously at rate <span class="math-container">$r$</span> we can say:</p>
<p><span class="math-container">\begin{align}
rv(t)&amp;=\lim_{\Delta t \to 0} \frac{v(t+\Delta t) - v(t)}{\Delta t} \\
&amp;=\frac{dv(t)}{dt} \\
\end{align}</span></p>
<p>From this you will get that the value from having the good over a time interval <span class="math-container">$[0,d]$</span> be <span class="math-container">$\int_0^d v e^{-rt} dt$</span></p>
<p>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 <span class="math-container">$d$</span> periods, <span class="math-container">$\lambda_d$</span> arrivals take place. So, this translates to the bernoulli process such that arrivals happens at a given time <span class="math-container">$t$</span> with probability <span class="math-container">$p=\lambda_d/d$</span>.</p>

CC BY-SA 4.0
2020-11-19T10:41:30.693Z
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.
mark leeds
CC BY-SA 4.0
2020-11-18T16:48:52.300Z
2020-11-18T16:48:52.300Z
Dayne
<p>I am trying to understand two comments from a colleague. I have no clue about continuous time models. He said something like &quot;just replace <span class="math-container">$\delta$</span> with <span class="math-container">$e^{-rdt}$</span> the Poisson process becomes a Bernoulli process.&quot;</p>
<p>Suppose I have a discrete time setting with discount factor <span class="math-container">$\delta$</span>. A buyer receives a constant value <span class="math-container">$v$</span> for each period in which he uses a good.
Then, for example, my total value today of using the good for two periods is <span class="math-container">$v(1+\delta)$</span>. Would the value from having the good over a time interval <span class="math-container">$[0,d]$</span> be <span class="math-container">$\int_0^d v e^{-rt} dt$</span>, or how does it work?</p>
<p>Next, if in one period of length <span class="math-container">$d$</span> in expectation <span class="math-container">$\lambda_d$</span> 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.</p>
<p>Do you have a recommendation where I can read up on these basics?</p>

<continuous-time>
Relationship of continuous and discrete time models
Max
# Relationship of continuous and discrete time models

(Asked by: username_0 on 2020-11-16)

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?

**Tags:** continuous-time

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## 1 answers

### Answer by username_1 on 2020-11-18. Score: 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 $p=\lambda\_d/d$.

> **Comments:**
> **username_2**: 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.  

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