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I am trying to think about the summed duration of waiting times of i.i.d random variables with exponentially distributed wiating times, and particularly my question is how many such variables will yield a total duration equal to some number of interest. For example, if there are events that occur at a rate according to an exponential distribution with rate parameter $\lambda$, such as say phone calls to a calling center on average every 20 mins and according to an exponential distribution, then how many total calls could be expected during any fixed period of time, say 24 hours? And what would be the distribution of this number?

My understanding is that the probability of the sum of the wait times of $k$ such events (ie. independent and identically distributed with exponential distribution) is given by the probability density function of the k-Erlang distribution. So I think this let's me compute for a given number of events, the probability of the value of the total duration of the sum of their wait times. However, I am interested in computing the other way: given a duration of interest (e.g. 24 hours) and a known rate characteristic of an exponential distribution, how many iid events to expect during that time?

Please advise on how to approach this.

Thanks! -Dan

Daniel Naftalovich
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2 Answers2

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Let $X$ be the number of events in $24$ hours. Then $X$ has Poisson distribution with parameter $\lambda=\frac{24}{1/3}=72$. Thus $$\Pr(X=k)=e^{-72}\frac{72^k}{k!}.$$ For such a largish $\lambda$, the normal approximation is reasonably satisfactory.

André Nicolas
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The exponential distribution tells you the distribution waiting times between events. If you want the actual number of events, then you can use the fact that the number of observed events in a given time frame with known rate is a Poisson random variable. If the rate of observed events is $\lambda$ events/hour then the expected number of events in one day is $\lambda\times 24$. Using your example, if a call occurs every $20$ minutes, then the rate of calls is obviously $0.05 \,\text{calls/minute} = 3\,\text{calls/hour}$. Therefore, in one day we expect $(3\,\text{calls/hour})(24\,\text{hours/day}) = 72\,\text{calls/day}$. We could write the distribution of calls per day as $C\sim \text{Poisson}(x; \lambda = 72)$. See here for some facts about this distribution.

Now suppose that there are $k$ call centers each with call rate $\lambda_{i}$ for $1 \leq i \leq k$. Then, the total number of calls is

$$ \text{C}_{\text{tot}} = \sum_{i = 1}^{k}C_{i} $$

We can use the addition theorem for Poisson random variables, which states that if $X \sim \text{Poisson}(x; \lambda_{1})$ and $Y \sim \text{Poisson}(x; \lambda_{2})$ then $Z = X + Y \sim \text{Poisson}(x; \lambda_{1} + \lambda_{2})$. In other words, we can just add the means. Therefore, the total number of calls is distributed as

$$ C_{\text{tot}} \sim \text{Poisson}\left(x; \sum_{i = 1}^{k} \lambda_{i} \right) $$

Let us assume for the sake of computation that there are $10$ call centers and that the means of all the call centers are identical and equal to $72\,\text{calls/day}$ as before. Then

$$ C_{\text{tot}} \sim \text{Poisson}\left(x; k \lambda \right) = \text{Poisson}\left(x; 720 \right) $$

In other words, we expect $720$ total calls per day, which makes sense. The distribution of the total number of calls is again Poisson.

Mr. G
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