We have introduced the exponential distribution in Section 7.6.3, where we also pointed out its link with the Poisson distribution and the Poisson process. The standard use of exponential variables to model random time between events relies on its memoryless property, which we are now able to appreciate. Consider an exponential random variable X with parameter λ, and say that X models the random life of some equipment (or a lightbulb) whose average life is 1/λ. What is the probability that the random life of this device will exceed a threshold t? Given the CDF of the exponential distribution,7 we see that
This makes sense, as this probability goes to zero when t increases, with a speed that is high when expected life is short. Now suppose that, after lighting the bulb, we notice that it is alive and kicking at time t; we could wonder what its expected residual life is, given this information. In general, after a long timespan of work, the death of a piece of equipment getscloser and closer.8 To formalize the problem, we should consider the conditional probability that the overall life of the light bulb is larger than t + s:
We see a rather surprising result: The elapsed time t does not influence the residual life s and, after a timespan of length t, the lightbulb is statistically identical to a brand-new one; hence, it is “memoryless.”
The memoryless character of the exponential distribution can be a good reason to use or not to use it to model random phenomena. For instance, it is suitable for modeling certain “purely random” phenomena, but not situations such as failures due to wear. When its use is warranted, simple and manageable results are often obtained. One well-known property is the PASTA property, which stands for Poisson arrivals see time averages. In plain words, and cutting a few corners, if observers join a system according to a Poisson process, what they observe is the system as if it were in steady state. A full appreciation of this important property requires technical machinery that is beyond the scope of the book, but a simple example will shed some light on its relevance.
Example 8.9 (The confectioner’s shop puzzle) A mature lady loves confectionery. Every now and then, she feels the urgent need for a few pastries. Lucky her, just in front of her home there is a stop, where bus lines A and B stop. Both will get her to a confectioner’s shop, which will call shop A and shop B. The lady is quite unpredictable: Her pastry frenzy can burst out at any time in the day. So, let us say that the time at which she arrives at the bus stop is uniformly distributed over the working hours of a day. What is the probability that she will satisfy her need at shop A, rather than shop B? The easy answer is that it depends on the arrival frequencies of the two bus lines. If one frequency is much higher than the other one, this should have an effect. Let us assume that the arrivals of the two buses are quite regular, and that the frequencies are the same. To fix ideas, let us say that the time between bus arrivals is exactly 10 minutes for each bus line.
Fig. 8.6 Bus arrival schedule for Example 8.9.
Please: Think a bit before reading further and give your answer!
Since the two frequencies are the same, it is very tempting to say that the distribution should be 50–50, i.e., she could end up to either shop with the same probability. Now, I say that the actual probabilities are quite different. She reaches shop A with probability 90%, and shop B with probability 10%. How could you explain the mystery?
Please: Think a bit before reading further and give your answer!
While it is very tempting to assume a 50–50 distribution, we are neglecting the very peculiar nature of deterministic arrivals. On each line, a bus arrives exactly every 10 minutes, but we should consider how the arrivals are phased in time. Consider the time schedule illustrated in Fig. 8.6. The shift between the arrival times of the two bus lines is such that it is much more likely to get to shop A. To see this, imagine that the lady arrives just after the first bus of line A stopped; them she will catch the line B bus and reach shop B. If she arrives just after the first bus of line B stopped, she will reach shop A. The picture suggests that this second event is much more likely due to the odd shift between bus arrivals. However, if the bus arrival processes were Poisson processes with same rate, i.e., if the time between arrivals were two independent exponential variables with the same expected value, the probabilities would be 50–50. In fact, when the lady arrives, the time elapsed from the last arrival on each line would provide her with no information about which bus will arrive first.
The example is a bit pathological but quite instructive: The interaction between random and deterministic phenomena may be less trivial than one might expect.
Leave a Reply