+ If events are occurring sequentially with the same mean rate of occurrence after each event, + an Erlang random variable measures the amount of time until the $k^{th}$ event occurs. + The random variable is the summation of $k$ independent and identically distributed (IID) Exponential random variables. +
+ +<% + include('templates/rvCards/erlang.html') +%> + + ++If you set $k$ equal to 50 up above, that is the equivalent of summing together 50 Exponential random variables with a mean rate of $\lambda$. Notice +how the resulting PDF resembles that of a Gaussian. We will explore why that is when we cover the Central Limit Theorem. +
+ +Example: + Agner Krarup Erlang wants to reward the 10th customer who walks into his gag foam-phone store, "Phoney Foam Phones". + Agner needs 15 minutes to prepare the prize. Since a YouTube video went viral showcasing how foam-phones can be used to clean dishes, + the store has become surprisingly popular. People are walking in at a mean rate + of one person per minute. What is the probability that Erlang will have the time to prepare his prize? +
+ Let $X$ be the number of minutes until the 10th customer walks in. Since a customer walking in is an event that is + occurring at a mean rate of once per minute, $X \sim {\rm Erlang}(k = 10, \lambda = 1)$. + The question is asking us to calculate P(X > 15): + + \begin{align*} + P(X > 15) &= 1 - P(X < 15) \\ + &= 1 - F_{X}(15) \\ + &= 1 - (1 - \sum_{n=0}^{10-1}{\frac{1}{n!} e^{-1 \cdot 15} (1 \cdot 15)^{n}}) \\ + &= 1 - (1 - \sum_{n=0}^{9}{\frac{1}{n!} e^{-15} (15)^{n}}) \\ + &= 0.070 + \end{align*} + + Sorry Agner! +Erlang Random Variable
+ + +| Notation: + | $X \sim {\rm Erlang}(k, \lambda)$ | +
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| Description: + | Time until $k^{th}$ event occurs if (a) the events occur with a constant mean rate and (b) they occur independently of time since last event. | +
| Parameters: + | $k \in \{1, 2, \dots\}$, occurrence of event + $\lambda \in \{0, 1, \dots\}$, the constant average rate |
+
| Support: + | $x \in \mathbb{R}^+$ | +
| PDF equation: | +$$f(x) = \frac{\lambda^{k}x^{k-1}e^{- \lambda x}}{(k-1)!}$$ | +
| CDF equation: | +$$F(x) = 1 - \sum_{n=0}^{k-1}{\frac{1}{n!} e^{-\lambda x} (\lambda x )^{n}}$$ | +
| Expectation: | +$\E[X] = k/\lambda$ | +
| Variance: | +$\var(X) = k/\lambda^2$ | +
| PDF graph: | +