Exponential random variable : Exercises

Introduction

The problems below all involve the exponential random variable in some way. Remember that this random variable is used to model the amount of time that we have to wait before some event occurs. Furthermore, this random variable has the all important property known as memorylessness: $$ P(T \gt x+y | T \gt y ) = P(T>x) \qquad \textrm{for all} \quad x,y > 0 $$ The cumulative probability distribution function for the exponential random variable is: $$ P(T \le x ) = 1 - e^{-\lambda x} $$

Example problems

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Problem 1

The alpha decay of Actinium-220 atoms to Francium-216 is a random process with a half life, $\gamma_{1/2}$, of 26~ms. The time taken for any individual atom to decay can be modelled using an exponentially distributed random variable, $T$. Give an expression for the cumulative probability distribution function for this variable and find the parameter, $\lambda$, that appears in this function. Remember that the half life is the time taken for the number of Actinium-220 atoms in the sample to half.

We model the decay of an Actinium-220 atom as an exponentially distributed random variable. We thus have \[ P(T \le t ) = 1 - e^{-\lambda t} \qquad P(T \gt t) = e^{-\lambda t} \] The half life is the value of $t$ such that $P(T \gt t)=0.5$ We can thus calculate $\lambda$ as follows: \[ 0.5 = e^{-26 \lambda} \quad \rightarrow \quad \lambda = -\frac{1}{26} \ln 0.5 \approx 0.0116~\textrm{ms}^{-1} \]

Problem 2

Bob is waiting to catch a bus. Alice meanwhile is visiting the library before going to catch the bus from the same bus stop as Bob. What is the probability that Alice will meet Bob at the bus stop if we are allowed to assume that the time she spends in the libarary, $T_A$, is an exponential random variable with mean $\frac{1}{\mu}$ and that Bob will on average wait $\frac{1}{\lambda}$ hours for a bus to arrive. Bob's waiting time, $T_B$, for a bus is also an exponetial random variable?

For Bob to meet Alice at the bus stop $T_B>T_A$ so the probability we are being asked to determine is $P(T_B>T_A)$. We are told two important things in the question:

Bob's waiting time for the bus, $T_B$, is exponentially distributed. Hence, $P(T_B>t) = e^{-\lambda t}$
The amount of time Bob would have to wait to meet Alice, $T_A$, is also exponentially distributed. Hence, $P(T_A>t) = e^{-\mu t}$

These in turn imply that:

The probability that Alice will have arrived by time $t$ is $P(T_A \le t) = 1 - e^{-\mu t}$
The probability that Bob's bus will have arrived by time $t$ is $P(T_B \le t) = 1 - e^{-\lambda t}$

These are the probability distribution functions for the random variables $T_A$ and $T_B$. From them we can determine the probability density functions: $$ P(T_A=t)=\frac{\textrm{d} }{\textrm{d}t} P(T_A \le t ) = \frac{\textrm{d} }{\textrm{d}t} \left( 1 - e^{-\mu t} \right) = \mu e^{-\mu t} $$ We can use this density function to determine the probability that the question asks for $P(T_B>T_A)$. We simply have to "sum" over each of the possible times at which Alice could have arrived multiplied by the probability that Bob is still at the bus stop. In other words we have to calculate the following integral: $$ \begin{aligned} P(T_B>T_A) & = \int_0^{\infty} P(T_B>s) P(T_A=s) \textrm{d}s \\ & = \int e^{-\lambda s} \mu e^{-\mu s} \textrm{d} s \\ & = \mu \int_0^{\infty} e^{-(\lambda + \mu ) s } \textrm{d}s \\ & = \mu \left[ -\frac{1}{\lambda + \mu} e^{-(\lambda + \mu)s} \right]_0^{\infty} = \frac{\mu}{\lambda + \mu } \end{aligned} $$

Problem 3

The departure of an airplane, $T$, is taken to be a continuous random variable. The probability density has an exponential distribution: $$ f_{T}(t) = \left\{ \begin{array}{ccc} \lambda e^{-\lambda t} & , & t \geq 0 \\ 0 & , & t \lt 0 \\ \end{array} \right. \qquad . $$ where, $t = 0$, is the scheduled departure time. Show that the probability it leaves at a time later than $t$ is: $P (T > t) = e^{-\lambda t} $, and show that, $\mathbb{E}(T)=1/\lambda $. Prove the {\em no memory} property: $$ P (T > t + s |T > s) = P (T > t) \qquad . $$ The queues for checkin are horrendous and as a result I am not at the gate ready for departure. If the time it takes me to reach the gate, $Y$, is an exponentially distributed random variable with parameter $\mu$ what is the probability I will catch my flight.

In the question above we are given the probability density function and told to calculate $P(T>t)$. The first step in doing so is to calculate the cumulative probability distribution function, $F_T(t)$, from the probability density. We can do this by integrating the probability density as shown below. $$ F_T(t) = P(T \le t) = \int_0^t f_T(t') \textrm{d}t' = \int_0^t \lambda e^{-\lambda t} \textrm{d}t = \left[ \frac{-\lambda e^{-\lambda t}}{\lambda} \right]_0^t = \left[ e^{-\lambda t} \right]_0^t = - e^{-\lambda t} + e^{0} = 1 - e^{-\lambda t} $$ Recall that the cumulative probability distribution function $F_T(t)=P(T\le t)$. We can thus write: $$ P(T>t) = 1 - P(T\le t) = 1 - F_T(t) = 1 - \left(1 - e^{-\lambda t} \right) = e^{-\lambda t} $$ To calculate the expectation for this random variable we need to do the following integral: $$ \mathbb{E}(T) = \int_0^\infty t f_T(t) \textrm{d}t = \int_0^\infty t \lambda e^{-\lambda t} \textrm{d} t \nonumber $$ We can solve the final integral above using integration by parts, which remember tells us that: $$ \int u \frac{\textrm{d}v}{\textrm{d}t} = uv - \int v \frac{\textrm{d}u}{\textrm{d}t} $$ In this case we will use: $$ \begin{aligned} u & = t \qquad \qquad \frac{\textrm{d}v}{\textrm{d}t} = \lambda e^{-\lambda t} \\ \frac{\textrm{d}u}{\textrm{d}t} & = 1 \qquad \qquad v = \frac{ -\lambda e^{-\lambda t }}{\lambda} = - e^{-\lambda t} \end{aligned} $$ when this is inserted into the formula above we find that: $$ \begin{aligned} \int_0^\infty t \lambda e^{-\lambda t} \textrm{d} t & = \left[ t e^{-\lambda t} \right]_0^\infty - \int_0^\infty 1 \left(-e^{-\lambda t}\right) \textrm{d}t \\ & = \left[ 0 - 0 \right] + \int_0^\infty e^{-\lambda t} \textrm{d}t \\ & = \left[ - \frac{ e^{-\lambda t} }{ \lambda } \right]_0^{\infty} \\ & = 0 + \frac{1}{\lambda} = \frac{1}{\lambda} \end{aligned} $$ To prove the no memory property we note that: $$ \begin{aligned} P(T > t +s \vert T > s ) & = \frac{ e^{-\lambda(t+s)} }{ e^{-\lambda s} } \\ & = \frac{ e^{-\lambda t} e^{- \lambda s} }{ e^{-\lambda s} } = e^{- \lambda t} \end{aligned} $$ The final result here is the same as the probability that $P(T>t)$, which we derived previously. Lastly the key to solving the final part of this problem is recognizing three things:

The amount of time taken for the plane to depart is a random variable, $T$, with cumulative probability distribution function: $P(T \le t) = 1 - e^{-\lambda t}$
The amount of time taken to get to the gate is a random variable, $Y$, with cumulative probability distribution function: $P(Y \le y) = 1 - e^{-\mu y}$
To catch the plane we have to arrive at the gate before the plane takes off. We are thus being asked to calculate the probability: $P(Y \lt T)$

Once you realise this the solution is the same as that for the above problem with Alice and Bob and the bus stop.

Problem 4

Secondary electrons generated during radiotherapy play an important role in damaging the DNA in tumours. These electrons are generated in water by the impact of the primary, ionizing radiation. They then diffuse through the solution before interacting with the DNA strand. However, whilst in solution electrons can also become solvated and hence inert. If the time taken for an electron to reach a DNA strand is an exponentially distributed random variable with parameter $\mu$ and the time taken for an electron to relax to a solvated (inert) state is an exponentially distributed random variable with parameter $\lambda$ show that the probability that any given electron will cause damage to DNA is given by: $$ \frac{\mu}{\lambda + \mu} $$

The key to solving this problem is recognizing three things:

The amount of time taken for the electron to become solvated is a random variable, $X$, with cumulative probability distribution function: $P(X \le x) = 1 - e^{-\lambda x}$
The amount of time taken for the electron to reach a DNA strand is a random variable, $Y$, with cumulative probability distribution function: $P(Y \le y) = 1 - e^{-\mu y}$
We have to calculate: $P(Y \lt X)$.