# Exponential Random Variables and Conditional Probability Problem

• crazy_craig
In summary, the submarine can remain at sea for a maximum of 1.35 years if at least two of its three navigational devices are working.
crazy_craig

## Homework Statement

A submarine has three navigational devices but can remain at sea if at least two are working. Suppose that the failure times are exponential with means 1 year, 1.5 years, and 3 years. What is the average length of time the boat can remain at sea?

## Homework Equations

Density for an exponential random variable: f(x)= λe-λx for x>=0
E(T)=1/λ if T is an exponential random variable

Maybe relevant: P(S<T)=λS / (λS + λT) for exponential random variables S and T, and this is similar for many exponential random variables.

## The Attempt at a Solution

The boat can remain at sea until 2 parts break.
Let T be the time that the boat can remain at sea.

Since the sample space can be divided into the order in which the parts fail, we have:

E(T)= Sum where 1<=i,j,k<=3 of: E(T|Ti<Tj<Tk)P(Ti<Tj<Tk)

where E(T|Ti<Tj<Tk)=E(Tj|Ti<Tj<Tk) since the boat can remain at sea until two parts fail.

Now, this would be a similar method that we've used with discrete random variables in class. Unfortunately, I'm a little rusty with my continuous probability. Is there somewhere to go from here, or maybe an easier way to approach the problem?

Thanks!

We ask ourselves: What is the probability the boat lasts less than or equal to t years? If we have an expression for this probability, we can find the expectation as the weighted sum of all years by the probability for that year (in a continuum through integration).

$$P(T \le t) = P((T_1 \le t \cap T_2 \le t) \cup (T_1 \le t \cap T_3 \le t) \cup (T_2 \le t \cap T_3 \le t))$$

Using set theory. Let A B and C be events:
$$P(A \cup B \cup C)=P(A) + P(B)+P(C)-P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$$

This results in, I believe, if T_1, T_2, and T_3 are independent variables:
$$E\{T\} = e_{t1}+e_{t2}+e_{t3}-e_{t1}e_{t2}-e_{t1}e_{t3}-e_{t2}e_{t3}+e_{t1}e_{t2}e_{t3}$$
where
$$E\{ T_j \} = e_{tj}$$

edit: Upon further thought, I don't think I am right. GL.

Last edited:
Yea, in that case, the E[T]=1 when the answer in the back of the book is E[T]=1.35 years

Thanks for trying!

crazy_craig said:
Yea, in that case, the E[T]=1 when the answer in the back of the book is E[T]=1.35 years

Thanks for trying!

The problem is quite easy if you know the memoryless and decomposition properties of exponential processes, and harder if you don't. Here is what I mean: let the three independent failure times be exponential with rates a, b and c. The rates have u nits of "per year" and are a = 1. b = 2/3 and c = 1/3 in your case; they are the reciprocals of the means. For simplicity of writing, denote the exponential density u*exp(-u*t) as fe(t,u).

Anyway, the time to first failure is the minimum of the three, so is exponential with rate = sum of all the rates; that is, T1~ has density f1(t) = fe(t,a+b+c). The time to the next failure depends on which of the three units failed first. By the "decomposition property", the probability that unit A failed first is a/s, where s = a+b+c. Given that A failed first, the times to failure of B and C are still exponential with their original parameters; that is, B and C are "like new"---that is the memoryless property. So, the time to the next failure is the min of (Tb,Tc), so has density fe(t,b+c). Similarly if B or C fail first, etc. Thus, the time to the next failure is a mixture, with density f2(t) = (a/s)*fe(t,b+c) + (b/s)*fe(t,a+c) + (c/s)*fe(t,a+b).
The time to failure of the sub has density fs(t) which is the convolution of f1 and f2. However, all you really want is the expected value, which is a bit simpler.

Anyway, you now have all the machinery you need to get the answer; doing convolutions of exponentials is easy. When you work it all out you will, indeed, obtain the answer that the mean time to failure of the sub is 1.35 years.

RGV

Thanks Ray! You're very good at explaining things, and I appreciate your help (again).

## What is an exponential random variable?

An exponential random variable is a type of continuous probability distribution that is often used to model the time between events that occur randomly and independently at a constant rate. It is characterized by a single parameter, lambda (λ), which represents the average number of events per unit time.

## How do you calculate the probability of an event occurring in a given time interval?

The probability of an event occurring in a given time interval can be calculated using the exponential probability density function (PDF) formula, which is P(X ≤ x) = 1 - e^(-λx), where X is the random variable and λ is the parameter. This formula gives the probability of an event occurring in a time interval of length x or less.

## What is the relationship between exponential random variables and Poisson processes?

Exponential random variables are often used to model the time between events in a Poisson process, which is a type of random process where events occur at a constant average rate and are independent of each other. The time between events in a Poisson process follows an exponential distribution, with λ representing the average number of events per unit time.

## How do you use conditional probability to solve problems involving exponential random variables?

Conditional probability is used to find the probability of an event occurring given that another event has already occurred. In the context of exponential random variables, conditional probability can be used to find the probability of an event occurring within a certain time interval, given that another event has already occurred within a different time interval. This can be calculated using the conditional probability formula, P(A|B) = P(A ∩ B) / P(B).

## What are some real-world applications of exponential random variables?

Exponential random variables are commonly used in a variety of fields, including engineering, finance, and telecommunications. They can be used to model the time between equipment failures, the time between customer arrivals, and the time between network packet arrivals, among others. They are also frequently used in survival analysis to model the time until an event (such as death or failure) occurs.

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The text passage is a discussion about a probability problem involving the failure times of navigational devices on a submarine. The problem is framed as a homework question, and the solution involves the use of exponential random variables and their properties.
single_question_with_answer

the passage contains a single mathematical question about the average length of time a submarine can remain at sea given the failure rates of its navigational devices. the answer to this question is provided in the text, stating that the average length of time the boat can remain at sea is 1.35 years.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Exponential Random Variables and Conditional Probability Problem

• crazy_craig
In summary, the submarine can remain at sea for a maximum of 1.35 years if at least two of its three navigational devices are working.
crazy_craig

## Homework Statement

A submarine has three navigational devices but can remain at sea if at least two are working. Suppose that the failure times are exponential with means 1 year, 1.5 years, and 3 years. What is the average length of time the boat can remain at sea?

## Homework Equations

Density for an exponential random variable: f(x)= λe-λx for x>=0
E(T)=1/λ if T is an exponential random variable

Maybe relevant: P(S<T)=λS / (λS + λT) for exponential random variables S and T, and this is similar for many exponential random variables.

## The Attempt at a Solution

The boat can remain at sea until 2 parts break.
Let T be the time that the boat can remain at sea.

Since the sample space can be divided into the order in which the parts fail, we have:

E(T)= Sum where 1<=i,j,k<=3 of: E(T|Ti<Tj<Tk)P(Ti<Tj<Tk)

where E(T|Ti<Tj<Tk)=E(Tj|Ti<Tj<Tk) since the boat can remain at sea until two parts fail.

Now, this would be a similar method that we've used with discrete random variables in class. Unfortunately, I'm a little rusty with my continuous probability. Is there somewhere to go from here, or maybe an easier way to approach the problem?

Thanks!

We ask ourselves: What is the probability the boat lasts less than or equal to t years? If we have an expression for this probability, we can find the expectation as the weighted sum of all years by the probability for that year (in a continuum through integration).

$$P(T \le t) = P((T_1 \le t \cap T_2 \le t) \cup (T_1 \le t \cap T_3 \le t) \cup (T_2 \le t \cap T_3 \le t))$$

Using set theory. Let A B and C be events:
$$P(A \cup B \cup C)=P(A) + P(B)+P(C)-P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$$

This results in, I believe, if T_1, T_2, and T_3 are independent variables:
$$E\{T\} = e_{t1}+e_{t2}+e_{t3}-e_{t1}e_{t2}-e_{t1}e_{t3}-e_{t2}e_{t3}+e_{t1}e_{t2}e_{t3}$$
where
$$E\{ T_j \} = e_{tj}$$

edit: Upon further thought, I don't think I am right. GL.

Last edited:
Yea, in that case, the E[T]=1 when the answer in the back of the book is E[T]=1.35 years

Thanks for trying!

crazy_craig said:
Yea, in that case, the E[T]=1 when the answer in the back of the book is E[T]=1.35 years

Thanks for trying!

The problem is quite easy if you know the memoryless and decomposition properties of exponential processes, and harder if you don't. Here is what I mean: let the three independent failure times be exponential with rates a, b and c. The rates have u nits of "per year" and are a = 1. b = 2/3 and c = 1/3 in your case; they are the reciprocals of the means. For simplicity of writing, denote the exponential density u*exp(-u*t) as fe(t,u).

Anyway, the time to first failure is the minimum of the three, so is exponential with rate = sum of all the rates; that is, T1~ has density f1(t) = fe(t,a+b+c). The time to the next failure depends on which of the three units failed first. By the "decomposition property", the probability that unit A failed first is a/s, where s = a+b+c. Given that A failed first, the times to failure of B and C are still exponential with their original parameters; that is, B and C are "like new"---that is the memoryless property. So, the time to the next failure is the min of (Tb,Tc), so has density fe(t,b+c). Similarly if B or C fail first, etc. Thus, the time to the next failure is a mixture, with density f2(t) = (a/s)*fe(t,b+c) + (b/s)*fe(t,a+c) + (c/s)*fe(t,a+b).
The time to failure of the sub has density fs(t) which is the convolution of f1 and f2. However, all you really want is the expected value, which is a bit simpler.

Anyway, you now have all the machinery you need to get the answer; doing convolutions of exponentials is easy. When you work it all out you will, indeed, obtain the answer that the mean time to failure of the sub is 1.35 years.

RGV

Thanks Ray! You're very good at explaining things, and I appreciate your help (again).

## What is an exponential random variable?

An exponential random variable is a type of continuous probability distribution that is often used to model the time between events that occur randomly and independently at a constant rate. It is characterized by a single parameter, lambda (λ), which represents the average number of events per unit time.

## How do you calculate the probability of an event occurring in a given time interval?

The probability of an event occurring in a given time interval can be calculated using the exponential probability density function (PDF) formula, which is P(X ≤ x) = 1 - e^(-λx), where X is the random variable and λ is the parameter. This formula gives the probability of an event occurring in a time interval of length x or less.

## What is the relationship between exponential random variables and Poisson processes?

Exponential random variables are often used to model the time between events in a Poisson process, which is a type of random process where events occur at a constant average rate and are independent of each other. The time between events in a Poisson process follows an exponential distribution, with λ representing the average number of events per unit time.

## How do you use conditional probability to solve problems involving exponential random variables?

Conditional probability is used to find the probability of an event occurring given that another event has already occurred. In the context of exponential random variables, conditional probability can be used to find the probability of an event occurring within a certain time interval, given that another event has already occurred within a different time interval. This can be calculated using the conditional probability formula, P(A|B) = P(A ∩ B) / P(B).

## What are some real-world applications of exponential random variables?

Exponential random variables are commonly used in a variety of fields, including engineering, finance, and telecommunications. They can be used to model the time between equipment failures, the time between customer arrivals, and the time between network packet arrivals, among others. They are also frequently used in survival analysis to model the time until an event (such as death or failure) occurs.

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user
The text passage is a discussion about a probability problem involving the failure times of navigational devices on a submarine. The problem is framed as a homework question, and the solution involves the use of exponential random variables and their properties.

TYPE: SINGLE_QUESTION_WITH_ANSWER

The passage contains a single mathematical question about the average length of time a submarine can remain at sea given the failure rates of its navigational devices. The answer to this question is provided in the text, stating that the average length of time the boat can remain at sea is 1.35 years.