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Memory-less Processes

January 10th, 2017


Today we will investigate a very interesting property of the exponential distribution. Often used to model waiting times, the exponential distribution has a probability density function of

$$P(x) = \lambda e^{-\lambda x}$$

It has an expected value of $\frac{1}{\lambda}$. If we use this model for a waiting time for a bus, then we expect to wait $\frac{1}{\lambda}$ minutes (for our model, we will use minutes, but it really doesn’t matter).

 

I. Expected Value

We start by computing the expected value. Let $T$ have a distribution of $$P(T) = \lambda e^{-\lambda T}$$

We compute the expected value of $T$, $\langle T\rangle$ via

$$\langle T\rangle = \int_{T_{\text{min}}}^{T_{\text{max}}}T\cdot P(T) \text{d}T$$

We could wait a minimum of 0 minutes (the bus is there when we arrive), and could wait at most, well, infinity, because the bus may break and never arrive. The exponential distribution is defined for $T\in[0,\infty)$, so our integral becomes

$$\langle T\rangle = \int_{0}^{\infty}T\lambda e^{-\lambda T} \text{d}T $$

$$= -Te^{-\lambda T}+\int e^{-\lambda T}\text{d}T\hspace{4mm}|_{T=0}^{T=\infty}$$

$$ = -Te^{-\lambda T}-\frac{1}{\lambda}e^{-\lambda T}\hspace{4mm}|_{T=0}^{T=\infty}$$

$$ = 0 – (-\frac{1}{\lambda}) = \frac{1}{\lambda}$$

II. Conditional Distributions Intro

Now we will examine a conditional distribution. This sounds confusing, but it’s a pretty easy concept to understand. Let’s say you have a red ball and a yellow ball, and you pick a ball from a hat, then pick the other ball out of the hat. What’s the probability that you pick the red ball second? Hopefully it’s obvious that it’s 50%. Well, what if I tell you that you pick the yellow ball first, now, what’s the probability that you pick the red ball second given that you pick the yellow ball first? 100%. This is what a conditional distribution is; the probability of some event occurring, given some prior knowledge.

In our case, our event is the probability that we wait $T$ minutes, and the knowledge we are given is that we’ve already waited $W$ minutes. Let’s say you are waiting for a bus, how long do you expect until the next bus comes given that you have already waited $W$ minutes and no bus has arrived?

We will use Bayes’ Theorem, which states that the probability of $A$ given $B$, or $P(A|B)$, can be expressed as

$$P(A|B) = \frac{P(A\hspace{1mm}\text{and}\hspace{1mm}B)}{P(B)}$$

III. Conditional Distribution Calculations

In our case, we want the probability that we wait $T$ minutes given we’ve waited $W$ minutes already. It is very important to understand how to interpret the information given. By waiting $W$ minutes, we know that $T\geq W$. We know nothing else, other than the fact that $T$ must be greater than or equal to $W$, since we have already waited that long. Therefore,

$$P(B) = P(T\geq W) = \int_{W}^{\infty} \lambda e^{-\lambda T}\text{d}T$$

$$ = -e^{-\lambda T}\hspace{4mm}|_{T=W}^{T=\infty}$$

$$ = e^{-\lambda W}$$

Now, we need to calculate $P(A\hspace{1mm}\text{and}\hspace{1mm}B)$. But in fact, we don’t, we already know it! The probability that we wait $T$ minutes AND $T\geq W$ is just the probability that we wait $T$ minutes for $T\geq W$ and 0 for $T<W$.

Therefore, we have

$$P(T|T\geq W) = \frac{e^{-\lambda T}}{e^{-\lambda W}} \hspace{5mm} T\geq W$$

$$= e^{-\lambda (T-W)}\hspace{5mm} T\geq W$$

IV. Conditional Expected Value

So, how long do we expect to wait, given that we’ve already waited $W$ minutes? The expected value can be calculated the same way as before.

$$\langle T|T\geq W\rangle = \int_{W}^{\infty}T\cdot P(T|T\geq W) \text{d}T$$

$$ = \int_{W}^{\infty}T\lambda  \frac{e^{-\lambda T}}{e^{-\lambda W}}\text{d}T $$

$$= e^{\lambda W}\int_{W}^{\infty}T\lambda e^{-\lambda T}\text{d}T $$

$$ = e^{\lambda W}\cdot (-Te^{-\lambda T}-\frac{1}{\lambda}e^{-\lambda T}\hspace{4mm}|_{T=W}^{T=\infty})$$

$$ = e^{\lambda W}\cdot (We^{-\lambda W}+ \frac{e^{-\lambda W}}{\lambda})$$

$$ = W + \frac{1}{\lambda}$$

This shows that we expect to wait $W + \frac{1}{\lambda}$ minutes. What’s significant about this? Well, given no information, we expected to wait $\frac{1}{\lambda}$ minutes. Given we’ve already waited $W$ minutes, we now expect to wait $W + \frac{1}{\lambda}$ minutes. This means that we simply expect to wait another $\frac{1}{\lambda}$ minutes. No matter how long we have already waited, we are no more likely to encounter a bus any quicker than if we hadn’t been waiting. Pretty interesting, I think.

That’s all folks. Thanks for reading!


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