Lesson 44 – Keep waiting: The memoryless property of exponential distribution

Bob Waits for the Bus

As the building entrance door closes behind, Bob glances at his post-it note. It has the directions and address of the car dealer. Bob is finally ready to buy his first (used) car. He walks to the nearby bus stop jubilantly thinking he will seldom use the bus again. Bob is tired of the waiting. Throughout these years the one thing he could establish is that the average wait time for his inbound 105 at the Cross St @ Main St is 15 minutes.

Bob may not care, but we know that his wait time follows an exponential distribution that has a probability density function $f(t) = \lambda e^{-\lambda t}$ .

The random variable T, the wait time between buses is an exponential distribution with parameter $\lambda$ . He waits 15 minutes on average. Some days he boards the bus earlier than 15 minutes, and some days he waits much longer.

Looking at the function $f(t) = \lambda e^{-\lambda t}$ , and the typical information we have for exponential distribution, i.e., the average wait time, it will be useful to relate the parameter $\lambda$ to the average wait time.

The average wait time is the average of the distribution — the expected value E[.].

E[X] for a continuous distribution, as you know from lesson 24 is $E[X] = \int x f(x) dx$ .

Applying this using the limits of the exponential distribution, we can derive the following.

$E[T] = \int_{0}^{\infty} t f(t) dt$

$E[T] = \int_{0}^{\infty} t \lambda e^{-\lambda t} dt$

$E[T] = \lambda \int_{0}^{\infty} t e^{-\lambda t} dt$

The definite integral is $\frac{1}{\lambda^{2}}$ .

So we have

$E[T] = \frac{1}{\lambda}$

The parameter $\lambda$ is a non-negative real number ( $\lambda > 0$ ), and represents the reciprocal of the expected value of T.

In Bob’s case, since the average wait time (E[T]) is 15 minutes, the parameter $\lambda$ is 0.066.

Bob gets to the bus shelter, greets the person next to him and thinks to himself “Hope the wait will not exceed 10 minutes today.”

Please tell him the probability he waits more than 10 minutes is 0.5134.

$P(T > 10) = e^{-\lambda t} = e^{-10/15} = 0.5134$

Bob is visibly anxious. He turns his hand and looks at his wristwatch. “10 minutes. The wait won’t be much longer.”

Please tell him about the memoryless property of the exponential distribution. The probability that he waits for another ten minutes, given he already waited 10 minutes is also 0.5134.

Let’s see how. We will assume t represents the first ten minutes and s represents the second ten minutes.

$P(T > t + s \mid T > t) = \frac{P(T > t \cap T > t+s)}{P(T > t)}$

$\hspace{10cm} = \frac{P(T > t+s)}{P(T > t)}$

$\hspace{10cm} = \frac{e^{-\lambda (t+s)}}{e^{-\lambda t}}$

$\hspace{10cm} = \frac{e^{-\lambda t} e^{-\lambda s}}{e^{-\lambda t}}$

$\hspace{10cm} = e^{-\lambda s}$

$P(T > 10 + 10 \mid T > 10) = e^{-10\lambda} = 0.5134$

The probability distribution of the remaining time until the event occurs is always the same regardless of the time that passed.

There is no memory in the process. The history is not relevant. The time to next arrival is not influenced by when the last event or arrival occurred.

This property is unique to the strictly decreasing functions: exponential and the geometric distributions.

The probability that Bob has to wait another s minutes (t + s) given that he already waited t minutes is the same as the probability that Bob waited the first s minutes. It is independent of the current wait time.

Bob Gets His First Chevy

Bob arrives at the dealers. He loves the look of the red 1997 Chevy. He looks over the window pane; “Ah, manual shift!” That made up his mind. He knows what he is getting. The price was reasonable. A good running engine is all he needed to drive it away.

The manager was young, a Harvard alum, as Bob identified from things in the room. “There is no guarantee these days with academic inflation, … the young lad is running a family business, … or his passion is to sell cars,” he thought to himself.

The manager tells him that the engine is in perfect running condition and the average breakdown time is four years. Bob does some estimates ($$$$) in his mind while checking out the car. He is happy with what he is getting and closes the deal.

Please tell Bob that there is a 22% likelihood that his Chevy manual shift will break down in the first year.

The number of years this car will run ~ exponential
distribution with a rate ( $\lambda$ ) of 1/4.

Since the average breakdown time (expected value E[T]) is four years, the parameter $\lambda$ = 1/4.

$P(T \le 1) = 1 - e^{-\lambda t} = 1 - e^{-(1/4)} = 0.22$

Bob should also know that there is a 37% chance that his car will still be running fine after four years.

$P(T > 4) = e^{-4\lambda} = e^{-4/4} = 0.37$

Bob in Four Years

Bob used the car for four years now with regular servicing, standard oil changes, and tire rotations. The engine is great.

Since the average lifetime has passed, should he think about a new car? How long should we expect his car to continue without a breakdown? Another four years?

Since he used it for four years, what is the probability that there will be no breakdown until the next four years?

You guessed it, 37%.

$P(T > 8 \mid T > 4) = \frac{P(T > 8)}{P(T > 4)} = e^{-4\lambda} = 0.37$

Now let’s have a visual interpretation of this memoryless property.

The probability distribution of the wait time (engine breakdown) for $\lambda$ = 1/4 looks like this.

Let us assume another random variable $T_{2} = T - 4$ , as the breakdown time after four years of usage. The lower bound for $T_{2}$ is 0 (since we measure from four years), and the upper bound is $\infty$ .

For any values $T > 4$ , the distribution is another exponential function — it is shifted by four years.

$f(t_{2}) = \lambda e^{-\lambda t_{2}} = \lambda e^{-\lambda (t-4)}$

Watch this animation, you will understand it better.

The original distribution is represented using the black line. The conditional distribution $P(T > 4+s \mid T > 4)$ is shown as a red line using links.

The same red line with links (truncated at 4) is shown as the shifted exponential distribution ( $f(t_{2})=\lambda e^{-\lambda (t-4)}$ ). So, the red line with links from t = 4 is the same as the original function from t = 0. It is just shifted.

The average value of $T$ is four years. The average value of $T_{2} = T - 4$ is also four. They have the same distribution.

If Bob reads our lessons, he’d understand that his Chevy will serve him, on the average, another four years.

Just like the car dealer’s four-year liberal arts degree from Harvard is forgotten, Bob’s four-year car usage history is forgotten — Memoryless.

As the saying goes, some memories are best forgotten, but the lessons from our classroom are never forgotten.

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Bob Waits for the Bus

Bob Gets His First Chevy

Bob in Four Years

Now let’s have a visual interpretation of this memoryless property.

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