Probability Density Functions


by TranscendArcu
Tags: density, functions, probability
TranscendArcu
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#1
Apr6-12, 06:33 PM
P: 288
1. The problem statement, all variables and given/known data



3. 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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TranscendArcu
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#2
Apr6-12, 06:52 PM
P: 288
This is a very similar question that I'd like my work checked on.



[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?
Ray Vickson
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#3
Apr7-12, 02:48 AM
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Quote Quote by TranscendArcu View Post
1. The problem statement, all variables and given/known data



3. 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

TranscendArcu
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#4
Apr7-12, 09:46 AM
P: 288

Probability Density Functions


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?
TranscendArcu
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#5
Apr7-12, 10:05 AM
P: 288
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?
Ray Vickson
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#6
Apr7-12, 11:16 AM
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Quote Quote by TranscendArcu View Post
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 has a fairly gentle introduction to this area.

RGV
TranscendArcu
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#7
Apr7-12, 02:20 PM
P: 288
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.
Ray Vickson
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#8
Apr7-12, 05:23 PM
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Quote Quote by TranscendArcu View Post
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


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