Math 311

Class 17

“Memoryless” Variables

Given a continuous random variable \(X \ge 0\), we say that \(X\) is memoryless if for each \(s \ge 0\) and for each \(t \ge 0\)

\[\operatorname{P}\left(X > t + s\mid X > t\right) = \operatorname{P}\left(X > s\right)\]

If we interpret \(X\) as the time spent waiting for something to happen, this says that if we already waited \(t\) time units, the probability we will have to wait more than additional \(s\) units is the same as the probability of having to wait more than \(s\) units from the start.

Every time \(t\) is a new start.

Distribution of \(X\)

\[\begin{align} \operatorname{P}\left(X > t + s\mid X > t\right) &= \operatorname{P}\left(X > s\right)\\[6pt] \class{fragment}{\data{fragment-index=1}{\frac{\operatorname{P}\left(X > t + s\class{fragment highlight-red}{\data{fragment-index=2}{\text{ and } X > t}}\right)}{\operatorname{P}\left(X > t\right)}}} & \class{fragment}{\data{fragment-index=1}{{} = \operatorname{P}\left(X > s\right)}}\\[6pt] \class{fragment}{\data{fragment-index=3}{\frac{\operatorname{P}\left(X > t + s\right)}{\operatorname{P}\left(X > t\right)}}} & \class{fragment}{\data{fragment-index=3}{{} = \operatorname{P}\left(X > s\right)}}\\[6pt] \class{fragment}{\data{fragment-index=4}{\operatorname{P}\left(X > t + s\right)}} & \class{fragment}{\data{fragment-index=4}{{} = \operatorname{P}\left(X > s\right)\operatorname{P}\left(X > t\right)}}\\[6pt] \class{fragment}{\data{fragment-index=5}{\color{green}\text{Define } G(x) = 1 - F(x) = \operatorname{P}(X > x).}}&\\[6pt] \class{fragment}{\data{fragment-index=6}{G\left(t + s\right)}} & \class{fragment}{\data{fragment-index=6}{{} = G\left(s\right)G\left(t\right)}}\\[6pt] \class{fragment}{\data{fragment-index=7}{G(x)}}&\class{fragment}{\data{fragment-index=7}{{} = e^{-\lambda x} \text{ for some } \lambda}}\\[6pt] \class{fragment}{\data{fragment-index=8}{F(x)}}&\class{fragment}{\data{fragment-index=8}{{} = 1 - e^{-\lambda x}}}\\[6pt] \class{fragment}{\data{fragment-index=9}{X}}&\class{fragment}{\data{fragment-index=9}{{} \sim \operatorname{Exp}(\lambda)}}\\[6pt] \end{align}\]

Exponential Distributions

\(\displaystyle f(x) = \begin{cases} 0 & \text{ if } x < 0\\\lambda e^{-\lambda x} & \text{ if } x \ge 0\end{cases}\)

\(\displaystyle F(x) = \begin{cases} 0 & \text{ if } x < 0\\1 - e^{-\lambda x} & \text{ if } x \ge 0\end{cases}\)

\(\displaystyle \operatorname{P}(X > x) = G(x) = 1 - F(x) = \begin{cases} 1 & \text{ if } x < 0\\ e^{-\lambda x} & \text{ if } x \ge 0\end{cases}\)

\(\operatorname{E} X = \frac{1}{\lambda}\)

Example 1

• Suppose time one spends in a bank is exponentially distributed with mean \(\frac{1}{\lambda} = 10\).
• What is the probability that a customer will spend more than 15 minutes at the bank?
• What is the probability that a customer will spend total of more than 15 minutes at the bank, given that they are still there after 10 minutes?

Example 2

• Suppose time one spends with a clerk in a bank is exponentially distributed with mean \(\frac{1}{\lambda}\).
• The bank has two clerks.
• Suppose that when Mr. Smith enters the bank, both clerks are busy, one with Mr. Jones, one with Mr. Brown.
• When one of Mr. Jones and Mr. Brown is done, the clerk will immediately start serving Mr. Smith.
• What is the probability that of the three customers, Mr. Smith will be the last to leave the bank?

The Failure Rate

Let \(X \ge 0\) is a continuous random variable with CDF \(F\) and PDF \(f\). Define the failure rate function for \(X\) as \[r(t) = \frac{f(t)}{1 - F(t)} = \frac{f(t)}{G(t)}.\]

\(\displaystyle F(t) = 1 - \exp\left(-\int_0^t r(x)\;dx\right)\)

 

 

\(X \sim \operatorname{Exp}(\lambda) \Leftrightarrow r(t) = \lambda\)

Example 3

• A bin contains \(n\) types of batteries.
• The lifetime of a battery of type \(j\) is exponentially distributed with rate \(\lambda_j\).
• The proportion of batteries of type \(j\) in the bin is \(P_j\).
• Let \(X\) be the lifetime of a battery that is randomly chosen from the box.
• \(\displaystyle F_X(t) = 1 - \sum_{j = 1}^n P_je^{-\lambda_j t}\)
• \(\displaystyle f_X(t) = \sum_{j = 1}^n \lambda_j P_je^{-\lambda_j t}\)
• \(X\) is a hyperexponential variable
• What is \(\displaystyle \lim_{t \to \infty} r(t)\)?