Probability Density Functions

In summary, the conversation discusses the concept of evaluating integrals in various problems, such as radioactive decay and machine failure. The first problem involves using a Poisson random variable with a mean of λ(b-a) to calculate the probability of decay occurring after a certain time t. However, this concept is not mentioned until two chapters later in the textbook, causing confusion for the person solving the problem. The second problem involves using a different method to calculate the probability of machine failure, but is unrelated to the first problem. It is recommended to review and make note of the definitions
  • #1
TranscendArcu
285
0

Homework Statement



Screen_shot_2012_04_06_at_7_15_49_PM.png


The Attempt at a Solution


I think I'm just evaluating integrals in this problem, not so? For part a)

[itex]\int_0 ^1 λe^{-λt}dt = \int_0 ^1 e^{-t}dt = -e^{-t} |^1 _0 = \frac{-1}{e} + 1[/itex]

For part b)
[itex]\int_0 ^3 e^{-t}dt = -e^{-t} |^3 _0 = \frac{-1}{e^3} + 1[/itex]

For part c)
[itex]\int_3 ^4 e^{-t}dt = -e^{-t} |^4 _3 = \frac{-1}{e^4} + \frac{1}{e^3}[/itex]

For part d)
Consider the limit as R goes to ∞ of [itex]\int_4 ^R e^{-t}dt = -e^{-t} | ^R _4 = 0 + \frac{1}{e^4}[/itex]

Does that seem right?
 
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  • #2
This is a very similar question that I'd like my work checked on.

Screen_shot_2012_04_06_at_7_36_28_PM.png


[itex]\int_0 ^T λe^{-λt}dt = \int_0 ^T (.01)e^{-(.01)t}dt =-e^{-(.01)t} |^T _0 =\frac{-1}{e^{\frac{T}{100}}} + 1 = Q[/itex]. I think this is the probability that the lightbulb will fail within the T hours. Thus, the probability that it will not fail should be given by [itex]1-Q[/itex]; that is, the probability of all outcomes minus the probability that the lightbulb fails.

For part b, I can solve the equation [itex]\frac{-1}{e^{\frac{t}{100}}} + 1 = \frac{1}{2}[/itex] for t. Write
[itex]e^{\frac{-t}{100}} = \frac{1}{2}[/itex]
[itex]\frac{t}{100} = ln(2)[/itex]
[itex]t = 100ln(2)[/itex]

So 100*ln(2) hours is the time when the reliability of the bulb is 1/2.

Is that right?
 
  • #3
TranscendArcu said:

Homework Statement



Screen_shot_2012_04_06_at_7_15_49_PM.png


The Attempt at a Solution


I think I'm just evaluating integrals in this problem, not so? For part a)

[itex]\int_0 ^1 λe^{-λt}dt = \int_0 ^1 e^{-t}dt = -e^{-t} |^1 _0 = \frac{-1}{e} + 1[/itex]

For part b)
[itex]\int_0 ^3 e^{-t}dt = -e^{-t} |^3 _0 = \frac{-1}{e^3} + 1[/itex]

For part c)
[itex]\int_3 ^4 e^{-t}dt = -e^{-t} |^4 _3 = \frac{-1}{e^4} + \frac{1}{e^3}[/itex]

For part d)
Consider the limit as R goes to ∞ of [itex]\int_4 ^R e^{-t}dt = -e^{-t} | ^R _4 = 0 + \frac{1}{e^4}[/itex]

Does that seem right?

No, it does not seem right.

The question is very badly worded. I think what they are saying is that the times between decays are iid exponential random variables with rate λ. That means that the count of the number of decays in a time interval [a,b] is a Poisson random variable with mean λ(b-a).

In particular, the probability that a decay occurs after time t = 4 is 1, since the probability of 0 decays in the interval [4,T] is exp(-λ(T-4)), which → 0 at T → ∞.

RGV
 
  • #4
How about these integrals then:

a) [itex]\int_0 ^1 (1-0)e^{(-t)} dt = 1 - \frac{1}{e}[/itex]
b) [itex]\int_0 ^1 (3)e^{-3t} dt = 1 - \frac{1}{e^3}[/itex]
c) [itex]\int_0 ^1 (4-3)e^{-(4-3)t} dt = 1 - \frac{1}{e}[/itex]
d) [itex]\int_0 ^1 (R-4)e^{-(R-4)t} dt = 1 -e^{4-R}[/itex], which, as you indicated, goes to one as R goes to infinity.

How is that?
 
  • #5
I guess this changes my answer to question six slightly. Would I write instead:

Calculate the probability that it will burn out: [itex]\int_0 ^1 (.01)(T-0)e^{(.01)(T-0)t} dt = e^{(.01)T} - 1[/itex]

Thus, the probability that it will not burn out is given by [itex]1 - e^{(.01)T} + 1 = 2 - e^{(.01)T}[/itex]

We can set this expression equal to 2 and solve for T, which we find to be ln(3/2)/.01

Is that better?
 
  • #6
TranscendArcu said:
I guess this changes my answer to question six slightly. Would I write instead:

Calculate the probability that it will burn out: [itex]\int_0 ^1 (.01)(T-0)e^{(.01)(T-0)t} dt = e^{(.01)T} - 1[/itex]

Thus, the probability that it will not burn out is given by [itex]1 - e^{(.01)T} + 1 = 2 - e^{(.01)T}[/itex]

We can set this expression equal to 2 and solve for T, which we find to be ln(3/2)/.01

Is that better?

When you say "this changes my answer...", I don't know what the 'this' refers to, because you are replying without quoting. Anyway, I would say that the first question (about radioactive decay) and the second one (about machine failure) are completely unrelated. The Poisson process has nothing at all to do with the machine failure problem.

So, I would go with your original answers to the machine question.

Let me ask you a serious question here: do you actually know what the different models and methods are about, or are you just guessing? You seem very uncertain. I would recommend that you do additional reading about these topics, to eliminate your sources of apparent confusion. If your textbook does not have a decent discussion, the web certainly does.
The source http://www.ssc.upenn.edu/~rwright/courses/poisson.pdf [Broken] has a fairly gentle introduction to this area.

RGV
 
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  • #7
I am just guessing. My textbook is just a pdf, so just out of curiosity I did a search for "poisson." The results that came up were in chapter five called "Important Distributions and Densities." The chapter that I pulled these questions from were from chapter two called "Continuous Probability Densities." It's certainly clear how the two are related, but poisson isn't mentioned in my text for another two chapters! So when I sat down to these problems, I thought they were just integral evaluations, and when you told me that I had done the first problem wrong, I assumed that I had misunderstood how to properly integrate these functions. That is, that I have to readjust the bounds and include (b-a) terms. That's why I reworked the second problem: it seemed very similar, and at this point in the text, I only knew one way to properly solve, and it made no mention of poisson.
 
  • #8
TranscendArcu said:
I am just guessing. My textbook is just a pdf, so just out of curiosity I did a search for "poisson." The results that came up were in chapter five called "Important Distributions and Densities." The chapter that I pulled these questions from were from chapter two called "Continuous Probability Densities." It's certainly clear how the two are related, but poisson isn't mentioned in my text for another two chapters! So when I sat down to these problems, I thought they were just integral evaluations, and when you told me that I had done the first problem wrong, I assumed that I had misunderstood how to properly integrate these functions. That is, that I have to readjust the bounds and include (b-a) terms. That's why I reworked the second problem: it seemed very similar, and at this point in the text, I only knew one way to properly solve, and it made no mention of poisson.

As I said, the wording of the first problem is very poor, so I had to guess what they wanted, using past knowledge of radioactivity and probability, plus some verbal clues left by the author of the problem (viz., the phrase "not necessarily the first"). That seemed to me to indicate that you could have decays happening over and over again; and while the problem did not specify it explicitly, the only sensible interpretation I could come with is that the exponential density given in the problem is the probability density of times between successive decays. That puts the whole thing into the province of the Poisson process (note the importance of that extra word 'process' when you do a search).

My interpretation could be all wrong, but in that case I have no clue at all what the author could possible want, and the problem would be describing something very different from actual, real-world radioactivity.

The link I gave you in my previous reply explains the material in a not-too-complicated way. Have you tried reading it?

RGV
 

1. What is a probability density function (PDF)?

A probability density function (PDF) is a mathematical function that describes the probability of a continuous random variable taking on a certain value. It represents the relative likelihood of different outcomes occurring within a given range of values.

2. How is a PDF different from a probability distribution function (PDF)?

A PDF and a probability distribution function (PDF) are two different ways of representing the same information. A PDF shows the probability of a continuous random variable taking on a certain value, while a probability distribution function shows the probability of a discrete random variable taking on a certain value.

3. How do you interpret a PDF?

The area under a PDF curve represents the probability of a random variable falling within a specific range of values. The height of the curve at a given point represents the likelihood of the random variable taking on that specific value.

4. How do you calculate the mean and variance of a PDF?

The mean of a PDF is the average value of all possible outcomes and can be calculated by integrating the product of the random variable and the PDF. The variance of a PDF is a measure of the spread of the data and can be calculated by integrating the squared difference between the random variable and the mean.

5. What are some common applications of PDFs?

PDFs are widely used in statistics, economics, and many other fields to model and analyze continuous data. They are also used in machine learning and data science for data visualization, statistical modeling, and predictive analysis.

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